标签归档:dataframe

什么格式是保存 Pandas 数据的最好格式?

在数据分析相关项目工作时,我通常使用Jupyter笔记本和pandas库来处理和移动我的数据。对于中等大小的数据集来说,这是一个非常直接的过程,你甚至可以将其存储为纯文本文件而没有太多的开销。

然而,当你的数据集中的观测数据数量较多时,保存和加载数据回内存的过程就会变慢,现在程序的重新启动都会迫使你等待数据重新加载。所以最终,CSV文件或任何其他纯文本格式都会失去吸引力。

我们可以做得更好。有很多二进制格式可以用来将数据存储到磁盘上,其中有很多格式pandas都支持。我们怎么能知道哪一种更适合我们的目的呢?

来吧,我们尝试其中的几个,然后进行对比!这就是我决定在这篇文章中要做的:通过几种方法将 pandas.DataFrame 保存到磁盘上,看看哪一种在I/O速度、内存消耗和磁盘空间方面做的更好。

在这篇文章中,我将展示我的测试结果。

1.要比较的格式

我们将考虑采用以下格式来存储我们的数据:

1. CSV — 数据科学家的一个好朋友
2. Pickle — 一种Python的方式来序列化事物
3. MessagePack — 它就像JSON,但又快又小
4. HDF5 — 一种设计用于存储和组织大量数据的文件格式
5. Feather — 一种快速、轻量级、易于使用的二进制文件格式,用于存储数据框架
6. Parquet — Apache Hadoop的柱状存储格式

所有这些格式都是被广泛使用的,而且(也许除了MessagePack)在你做一些数据分析的事情时非常经常遇到。

为了追求找到最好的缓冲格式来存储程序会话之间的数据,我选择了以下指标进行比较。

1. size_mb – 文件大小(Mb)。
2. save_time – 将数据帧保存到磁盘上所需的时间量。
3. load_time – 将之前转储的数据帧加载到内存中所需要的时间量。
4. save_ram_delta_mb – 数据帧保存过程中最大的内存消耗增长量。
5. load_ram_delta_mb – 数据帧加载过程中的最大内存消耗增长量。

请注意,当我们使用高效压缩的二进制数据格式,如 Parquet 时,最后两个指标变得非常重要。它们可以帮助我们估计加载序列化数据所需的内存量,此外还有数据大小本身。我们将在接下来的章节中更详细地讨论这个问题。

2.测试及结果

我决定使用一个合成数据集进行测试,以便更好地控制序列化的数据结构和属性。

另外,我在我的基准中使用了两种不同的方法:

(a) 将生成的分类变量保留为字符串。

(b) 在执行任何I/O之前将它们转换为 pandas.Categorical 数据类型。

函数generate_dataset显示了我在基准中是如何生成数据集的:

def generate_dataset(n_rows, num_count, cat_count, max_nan=0.1, max_cat_size=100):
    """
    随机生成具有数字和分类特征的数据集。
    
    数字特征取自正态分布X ~ N(0, 1)。
    分类特征则被生成为随机的uuid4字符串。
    
    此外,数字和分类特征的max_nan比例被替换为NaN值。
    """
    dataset, types = {}, {}
    
    def generate_categories():
        from uuid import uuid4
        category_size = np.random.randint(2, max_cat_size)
        return [str(uuid4()) for _ in range(category_size)]
    
    for col in range(num_count):
        name = f'n{col}'
        values = np.random.normal(0, 1, n_rows)
        nan_cnt = np.random.randint(1, int(max_nan*n_rows))
        index = np.random.choice(n_rows, nan_cnt, replace=False)
        values[index] = np.nan
        dataset[name] = values
        types[name] = 'float32'
        
    for col in range(cat_count):
        name = f'c{col}'
        cats = generate_categories()
        values = np.array(np.random.choice(cats, n_rows, replace=True), dtype=object)
        nan_cnt = np.random.randint(1, int(max_nan*n_rows))
        index = np.random.choice(n_rows, nan_cnt, replace=False)
        values[index] = np.nan
        dataset[name] = values
        types[name] = 'object'
    
    return pd.DataFrame(dataset), types

我们将CSV文件的保存和加载性能作为一个基准。

五个随机生成的具有一百万个观测值的数据集被转储到CSV中,并读回内存以获得平均指标。

每种二进制格式都针对20个随机生成的具有相同行数的数据集进行测试。

这些数据集包括15个数字特征和15个分类特征。你可以在这个资源库中找到带有基准测试功能和所需的完整源代码:

https://github.com/devforfu/pandas-formats-benchmark

或在Python实用宝典后台回复 Pandas IO对比 ,下载完整代码。

(a) 数据为字符串特征时的性能

下图显示了每种数据格式的平均I/O时间。一个有趣的观察是,hdf显示出比csv更慢的加载速度,而其他二进制格式的表现明显更好。其中最令人印象深刻的是feather和parquet。

在保存数据和从磁盘上读取数据时,内存开销如何?

下一张图片告诉我们,hdf 的表现就不是那么好了。可以肯定的是,csv在保存/加载纯文本字符串时不需要太多的额外内存,而Feather和parquet则相当接近:

最后,让我们看看文件的大小。这次parquet显示了一个令人印象深刻的结果,考虑到这种格式是为有效存储大量数据而开发的,这并不令人惊讶。

(b) 字符串特征转换为数字时的性能

在上一节中,我们没有尝试有效地存储我们的分类特征而是使用普通的字符串。让我们来弥补这个遗漏吧! 这一次我们使用一个专门的 pandas.Categorical 类型,转字符串特征为数字特征。

看看现在与纯文本的csv相比,它看起来如何!

现在所有的二进制格式都显示出它们的真正力量。Csv的基准结果已经远远落后了,所以让我们把它去掉,以便更清楚地看到各种二进制格式之间的差异:

Feather 和 Pickle 显示了最好的 I/O 速度,而 hdf 仍然显示了明显的性能开销。

现在是时候比较数据进程加载时的内存消耗了。下面的柱状图显示了我们之前提到的关于parquet格式的一个重要事实。

可以看到 parquet 读写时的内存空间差距有多大,你有可能你无法将比较大的 parquet 文件加载到内存中。

最后的图显示了各格式的文件大小。所有的格式都显示出良好的效果,除了hdf仍然需要比其他格式多得多的空间:

3.结论

正如我们的测试所显示的,似乎 feather 格式是存储Python会话数据的理想候选者。它显示了很快的I/O速度,在磁盘上不占用太多内存,并且在加载回RAM时不需要消耗太大的内存。

当然,这种比较并不意味着你应该在每个可能的情况下使用这种格式。例如,feather格式一般不会被用作长期文件存储的格式。

另外,某些特定情况下也无法使用 feather,这由你的整个程序架构决定。然而,就如本帖开头所述的目的,它在不被任何特殊事项限制的情况下是一个很好的选择。

本文译自 towardsdatascience
作者: Ilia Zaitsev
有部分修改。

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Pandarallel 一个能让你的Pandas计算火力拉满的工具

没有使用Pandarallel
使用了Pandarallel

众所周知,由于GIL的存在,Python单进程中的所有操作都是在一个CPU核上进行的,所以为了提高运行速度,我们一般会采用多进程的方式。而多进程无非就是以下几种方案:

  • 1、multiprocessing
  • 2、concurrent.futures.ProcessPoolExecutor()
  • 3、joblib
  • 4、ppserver
  • 5、celery

这些方案对于普通Pandas玩家来说都不是特别友好,怎样才能算作一个友好的并行处理方案?就是原来的逻辑我基本不用变,仅修改需要计算的那行就能完成我们目标的方案,而 pandarallel 就是一个这样友好的工具。

没有并行计算(原始pandas) pandarallel
df.apply(func)df.parallel_apply(func)
df.applymap(func)df.parallel_applymap(func)
df.groupby(args).apply(func)df.groupby(args).parallel_apply(func)
df.groupby(args1).col_name.rolling(args2).apply(func)df.groupby(args1).col_name.rolling(args2).parallel_apply(func)
df.groupby(args1).col_name.expanding(args2).apply(func)df.groupby(args1).col_name.expanding(args2).parallel_apply(func)
series.map(func)series.parallel_map(func)
series.apply(func)series.parallel_apply(func)
series.rolling(args).apply(func)series.rolling(args).parallel_apply(func)

可以看到,在 pandarallel 的世界里,你只需要替换原有的 pandas 处理语句就能实现多CPU并行计算。非常方便、非常nice.

在4核CPU的性能测试上,它比原始语句快了接近4倍。测试条件(OS: Linux Ubuntu 16.04,Hardware: Intel Core i7 @ 3.40 GHz – 4 cores),这就是我所说的,它把CPU充分利用了起来。

下面就给大家介绍这个模块怎么用,其实非常简单,任何代码只需要加几行代码就能实现质的飞跃。

1.准备

开始之前,你要确保Python和pip已经成功安装在电脑上,如果没有,请访问这篇文章:超详细Python安装指南 进行安装。

(可选1) 如果你用Python的目的是数据分析,可以直接安装Anaconda:Python数据分析与挖掘好帮手—Anaconda,它内置了Python和pip.

(可选2) 此外,推荐大家用VSCode编辑器来编写小型Python项目:Python 编程的最好搭档—VSCode 详细指南

Windows环境下打开Cmd(开始—运行—CMD),苹果系统环境下请打开Terminal(command+空格输入Terminal),输入命令安装依赖:

pip install pandarallel

对于windows用户,有一个不好的消息是,它只能在Windows的linux子系统上运行(WSL),你可以在微软官网上找到安装教程:

https://docs.microsoft.com/zh-cn/windows/wsl/about

2.使用Pandarallel

使用前,需要对Pandarallel进行初始化:

from pandarallel import pandarallel
pandarallel.initialize()

这样才能调用并行计算的API,不过 initialize 中有一个重要参数需要说明,那就是 nb_workers ,它将指定并行计算的Worker数,如果没有设置,所有CPU的核都会用上。

Pandarallel一共支持8种Pandas操作,下面是一个apply方法的例子。

import pandas as pd
import time
import math
import numpy as np
from pandarallel import pandarallel

# 初始化
pandarallel.initialize()
df_size = int(5e6)
df = pd.DataFrame(dict(a=np.random.randint(1, 8, df_size),
                       b=np.random.rand(df_size)))
def func(x):
    return math.sin(x.a**2) + math.sin(x.b**2)

# 正常处理
res = df.apply(func, axis=1)

# 并行处理
res_parallel = df.parallel_apply(func, axis=1)

# 查看结果是否相同
res.equals(res_parallel)

其他方法使用上也是类似的,在原始的函数名称前加上 parallel_。比如DataFrame.groupby.apply:

import pandas as pd
import time
import math
import numpy as np
from pandarallel import pandarallel

# 初始化
pandarallel.initialize()
df_size = int(3e7)
df = pd.DataFrame(dict(a=np.random.randint(1, 1000, df_size),
                       b=np.random.rand(df_size)))
def func(df):
    dum = 0
    for item in df.b:
        dum += math.log10(math.sqrt(math.exp(item**2)))
        
    return dum / len(df.b)

# 正常处理
res = df.groupby("a").apply(func)
# 并行处理
res_parallel = df.groupby("a").parallel_apply(func)
res.equals(res_parallel)

又比如 DataFrame.groupby.rolling.apply:

import pandas as pd
import time
import math
import numpy as np
from pandarallel import pandarallel

# 初始化
pandarallel.initialize()
df_size = int(1e6)
df = pd.DataFrame(dict(a=np.random.randint(1, 300, df_size),
                       b=np.random.rand(df_size)))
def func(x):
    return x.iloc[0] + x.iloc[1] ** 2 + x.iloc[2] ** 3 + x.iloc[3] ** 4

# 正常处理
res = df.groupby('a').b.rolling(4).apply(func, raw=False)
# 并行处理
res_parallel = df.groupby('a').b.rolling(4).parallel_apply(func, raw=False)
res.equals(res_parallel)

案例都是类似的,这里就直接列出表格,不浪费大家宝贵的时间去阅读一些重复的例子了:

没有并行计算(原始pandas) pandarallel
df.apply(func)df.parallel_apply(func)
df.applymap(func)df.parallel_applymap(func)
df.groupby(args).apply(func)df.groupby(args).parallel_apply(func)
df.groupby(args1).col_name.rolling(args2).apply(func)df.groupby(args1).col_name.rolling(args2).parallel_apply(func)
df.groupby(args1).col_name.expanding(args2).apply(func)df.groupby(args1).col_name.expanding(args2).parallel_apply(func)
series.map(func)series.parallel_map(func)
series.apply(func)series.parallel_apply(func)
series.rolling(args).apply(func)series.rolling(args).parallel_apply(func)

3.注意事项

1. 我有 8 个 CPU,但 parallel_apply 只能加快大约4倍的计算速度。为什么?

答:正如我前面所言,Python中每个进程占用一个核,Pandarallel 最多只能加快到你所拥有的核心的总数,一个 4 核的超线程 CPU 将向操作系统显示 8 个 CPU,但实际上只有 4 个核心,因此最多加快4倍。

2. 并行化是有成本的(实例化新进程,通过共享内存发送数据,…),所以只有当并行化的计算量足够大时,并行化才是有意义的。对于很少量的数据,使用 Pandarallel 并不总是值得的。

我们的文章到此就结束啦,如果你喜欢今天的 Python 教程,请持续关注Python实用宝典。

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如何获取熊猫DataFrame的最后N行?

问题:如何获取熊猫DataFrame的最后N行?

我有熊猫数据帧df1df2(df1是vanila数据帧,df2由’STK_ID’和’RPT_Date’索引):

>>> df1
    STK_ID  RPT_Date  TClose   sales  discount
0   000568  20060331    3.69   5.975       NaN
1   000568  20060630    9.14  10.143       NaN
2   000568  20060930    9.49  13.854       NaN
3   000568  20061231   15.84  19.262       NaN
4   000568  20070331   17.00   6.803       NaN
5   000568  20070630   26.31  12.940       NaN
6   000568  20070930   39.12  19.977       NaN
7   000568  20071231   45.94  29.269       NaN
8   000568  20080331   38.75  12.668       NaN
9   000568  20080630   30.09  21.102       NaN
10  000568  20080930   26.00  30.769       NaN

>>> df2
                 TClose   sales  discount  net_sales    cogs
STK_ID RPT_Date                                             
000568 20060331    3.69   5.975       NaN      5.975   2.591
       20060630    9.14  10.143       NaN     10.143   4.363
       20060930    9.49  13.854       NaN     13.854   5.901
       20061231   15.84  19.262       NaN     19.262   8.407
       20070331   17.00   6.803       NaN      6.803   2.815
       20070630   26.31  12.940       NaN     12.940   5.418
       20070930   39.12  19.977       NaN     19.977   8.452
       20071231   45.94  29.269       NaN     29.269  12.606
       20080331   38.75  12.668       NaN     12.668   3.958
       20080630   30.09  21.102       NaN     21.102   7.431

我可以通过以下方式获得df2的最后3行:

>>> df2.ix[-3:]
                 TClose   sales  discount  net_sales    cogs
STK_ID RPT_Date                                             
000568 20071231   45.94  29.269       NaN     29.269  12.606
       20080331   38.75  12.668       NaN     12.668   3.958
       20080630   30.09  21.102       NaN     21.102   7.431

同时df1.ix[-3:]给出所有行:

>>> df1.ix[-3:]
    STK_ID  RPT_Date  TClose   sales  discount
0   000568  20060331    3.69   5.975       NaN
1   000568  20060630    9.14  10.143       NaN
2   000568  20060930    9.49  13.854       NaN
3   000568  20061231   15.84  19.262       NaN
4   000568  20070331   17.00   6.803       NaN
5   000568  20070630   26.31  12.940       NaN
6   000568  20070930   39.12  19.977       NaN
7   000568  20071231   45.94  29.269       NaN
8   000568  20080331   38.75  12.668       NaN
9   000568  20080630   30.09  21.102       NaN
10  000568  20080930   26.00  30.769       NaN

为什么呢 如何获得df1(索引的数据帧)的最后3行?熊猫0.10.1

I have pandas dataframe df1 and df2 (df1 is vanila dataframe, df2 is indexed by ‘STK_ID’ & ‘RPT_Date’) :

>>> df1
    STK_ID  RPT_Date  TClose   sales  discount
0   000568  20060331    3.69   5.975       NaN
1   000568  20060630    9.14  10.143       NaN
2   000568  20060930    9.49  13.854       NaN
3   000568  20061231   15.84  19.262       NaN
4   000568  20070331   17.00   6.803       NaN
5   000568  20070630   26.31  12.940       NaN
6   000568  20070930   39.12  19.977       NaN
7   000568  20071231   45.94  29.269       NaN
8   000568  20080331   38.75  12.668       NaN
9   000568  20080630   30.09  21.102       NaN
10  000568  20080930   26.00  30.769       NaN

>>> df2
                 TClose   sales  discount  net_sales    cogs
STK_ID RPT_Date                                             
000568 20060331    3.69   5.975       NaN      5.975   2.591
       20060630    9.14  10.143       NaN     10.143   4.363
       20060930    9.49  13.854       NaN     13.854   5.901
       20061231   15.84  19.262       NaN     19.262   8.407
       20070331   17.00   6.803       NaN      6.803   2.815
       20070630   26.31  12.940       NaN     12.940   5.418
       20070930   39.12  19.977       NaN     19.977   8.452
       20071231   45.94  29.269       NaN     29.269  12.606
       20080331   38.75  12.668       NaN     12.668   3.958
       20080630   30.09  21.102       NaN     21.102   7.431

I can get the last 3 rows of df2 by:

>>> df2.ix[-3:]
                 TClose   sales  discount  net_sales    cogs
STK_ID RPT_Date                                             
000568 20071231   45.94  29.269       NaN     29.269  12.606
       20080331   38.75  12.668       NaN     12.668   3.958
       20080630   30.09  21.102       NaN     21.102   7.431

while df1.ix[-3:] give all the rows:

>>> df1.ix[-3:]
    STK_ID  RPT_Date  TClose   sales  discount
0   000568  20060331    3.69   5.975       NaN
1   000568  20060630    9.14  10.143       NaN
2   000568  20060930    9.49  13.854       NaN
3   000568  20061231   15.84  19.262       NaN
4   000568  20070331   17.00   6.803       NaN
5   000568  20070630   26.31  12.940       NaN
6   000568  20070930   39.12  19.977       NaN
7   000568  20071231   45.94  29.269       NaN
8   000568  20080331   38.75  12.668       NaN
9   000568  20080630   30.09  21.102       NaN
10  000568  20080930   26.00  30.769       NaN

Why ? How to get the last 3 rows of df1 (dataframe without index) ? Pandas 0.10.1


回答 0

别忘了DataFrame.tail!例如df1.tail(10)

Don’t forget DataFrame.tail! e.g. df1.tail(10)


回答 1

这是因为使用整数索引(通过-3而不是positionix通过标签选择索引,这是设计使然:请参见pandas“ gotchas” *中的整数索引)。

*在较新版本的熊猫中,建议使用loc或iloc删除ix作为位置或标签的歧义:

df.iloc[-3:]

请参阅文档

正如Wes所指出的,在这种特定情况下,您应该只使用tail!

This is because of using integer indices (ix selects those by label over -3 rather than position, and this is by design: see integer indexing in pandas “gotchas”*).

*In newer versions of pandas prefer loc or iloc to remove the ambiguity of ix as position or label:

df.iloc[-3:]

see the docs.

As Wes points out, in this specific case you should just use tail!


回答 2

如何获取熊猫DataFrame的最后N行?

如果您按位置进行切片,__getitem__(即使用进行切片[])效果很好,并且是我针对该问题找到的最简洁的解决方案。

pd.__version__
# '0.24.2'

df = pd.DataFrame({'A': list('aaabbbbc'), 'B': np.arange(1, 9)})
df

   A  B
0  a  1
1  a  2
2  a  3
3  b  4
4  b  5
5  b  6
6  b  7
7  c  8

df[-3:]

   A  B
5  b  6
6  b  7
7  c  8

例如,这与调用相同df.iloc[-3:]iloc内部委托__getitem__)。


顺便说一句,如果要查找每个组的最后N行,请使用groupbyGroupBy.tail

df.groupby('A').tail(2)

   A  B
1  a  2
2  a  3
5  b  6
6  b  7
7  c  8

How to get the last N rows of a pandas DataFrame?

If you are slicing by position, __getitem__ (i.e., slicing with[]) works well, and is the most succinct solution I’ve found for this problem.

pd.__version__
# '0.24.2'

df = pd.DataFrame({'A': list('aaabbbbc'), 'B': np.arange(1, 9)})
df

   A  B
0  a  1
1  a  2
2  a  3
3  b  4
4  b  5
5  b  6
6  b  7
7  c  8

df[-3:]

   A  B
5  b  6
6  b  7
7  c  8

This is the same as calling df.iloc[-3:], for instance (iloc internally delegates to __getitem__).


As an aside, if you want to find the last N rows for each group, use groupby and GroupBy.tail:

df.groupby('A').tail(2)

   A  B
1  a  2
2  a  3
5  b  6
6  b  7
7  c  8

如何在Spark DataFrame中添加常量列?

问题:如何在Spark DataFrame中添加常量列?

我想在中添加DataFrame具有任意值的列(每行相同)。使用withColumn以下内容时出现错误:

dt.withColumn('new_column', 10).head(5)
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-50-a6d0257ca2be> in <module>()
      1 dt = (messages
      2     .select(messages.fromuserid, messages.messagetype, floor(messages.datetime/(1000*60*5)).alias("dt")))
----> 3 dt.withColumn('new_column', 10).head(5)

/Users/evanzamir/spark-1.4.1/python/pyspark/sql/dataframe.pyc in withColumn(self, colName, col)
   1166         [Row(age=2, name=u'Alice', age2=4), Row(age=5, name=u'Bob', age2=7)]
   1167         """
-> 1168         return self.select('*', col.alias(colName))
   1169 
   1170     @ignore_unicode_prefix

AttributeError: 'int' object has no attribute 'alias'

似乎我可以通过添加和减去其他一列(这样它们加到零)然后添加我想要的数字(在这种情况下为10)来欺骗该函数按我的意愿工作:

dt.withColumn('new_column', dt.messagetype - dt.messagetype + 10).head(5)
[Row(fromuserid=425, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=47019141, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=49746356, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=93506471, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=80488242, messagetype=1, dt=4809600.0, new_column=10)]

这绝对是骇客,对吧?我认为还有一种更合法的方法吗?

I want to add a column in a DataFrame with some arbitrary value (that is the same for each row). I get an error when I use withColumn as follows:

dt.withColumn('new_column', 10).head(5)
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-50-a6d0257ca2be> in <module>()
      1 dt = (messages
      2     .select(messages.fromuserid, messages.messagetype, floor(messages.datetime/(1000*60*5)).alias("dt")))
----> 3 dt.withColumn('new_column', 10).head(5)

/Users/evanzamir/spark-1.4.1/python/pyspark/sql/dataframe.pyc in withColumn(self, colName, col)
   1166         [Row(age=2, name=u'Alice', age2=4), Row(age=5, name=u'Bob', age2=7)]
   1167         """
-> 1168         return self.select('*', col.alias(colName))
   1169 
   1170     @ignore_unicode_prefix

AttributeError: 'int' object has no attribute 'alias'

It seems that I can trick the function into working as I want by adding and subtracting one of the other columns (so they add to zero) and then adding the number I want (10 in this case):

dt.withColumn('new_column', dt.messagetype - dt.messagetype + 10).head(5)
[Row(fromuserid=425, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=47019141, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=49746356, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=93506471, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=80488242, messagetype=1, dt=4809600.0, new_column=10)]

This is supremely hacky, right? I assume there is a more legit way to do this?


回答 0

Spark 2.2+

Spark 2.2引入typedLit了support SeqMapTuplesSPARK-19254),并且应该支持以下调用(Scala):

import org.apache.spark.sql.functions.typedLit

df.withColumn("some_array", typedLit(Seq(1, 2, 3)))
df.withColumn("some_struct", typedLit(("foo", 1, 0.3)))
df.withColumn("some_map", typedLit(Map("key1" -> 1, "key2" -> 2)))

Spark 1.3以上lit),1.4以上arraystruct),2.0以上map):

的第二个参数DataFrame.withColumn应该是a,Column因此您必须使用文字:

from pyspark.sql.functions import lit

df.withColumn('new_column', lit(10))

如果您需要复杂的列,则可以使用以下代码块构建这些列array

from pyspark.sql.functions import array, create_map, struct

df.withColumn("some_array", array(lit(1), lit(2), lit(3)))
df.withColumn("some_struct", struct(lit("foo"), lit(1), lit(.3)))
df.withColumn("some_map", create_map(lit("key1"), lit(1), lit("key2"), lit(2)))

可以在Scala中使用完全相同的方法。

import org.apache.spark.sql.functions.{array, lit, map, struct}

df.withColumn("new_column", lit(10))
df.withColumn("map", map(lit("key1"), lit(1), lit("key2"), lit(2)))

为了提供名称structs或者使用alias上的每个字段:

df.withColumn(
    "some_struct",
    struct(lit("foo").alias("x"), lit(1).alias("y"), lit(0.3).alias("z"))
 )

cast整个对象

df.withColumn(
    "some_struct", 
    struct(lit("foo"), lit(1), lit(0.3)).cast("struct<x: string, y: integer, z: double>")
 )

尽管较慢,也可以使用UDF。

注意事项

可以使用相同的构造将常量参数传递给UDF或SQL函数。

Spark 2.2+

Spark 2.2 introduces typedLit to support Seq, Map, and Tuples (SPARK-19254) and following calls should be supported (Scala):

import org.apache.spark.sql.functions.typedLit

df.withColumn("some_array", typedLit(Seq(1, 2, 3)))
df.withColumn("some_struct", typedLit(("foo", 1, 0.3)))
df.withColumn("some_map", typedLit(Map("key1" -> 1, "key2" -> 2)))

Spark 1.3+ (lit), 1.4+ (array, struct), 2.0+ (map):

The second argument for DataFrame.withColumn should be a Column so you have to use a literal:

from pyspark.sql.functions import lit

df.withColumn('new_column', lit(10))

If you need complex columns you can build these using blocks like array:

from pyspark.sql.functions import array, create_map, struct

df.withColumn("some_array", array(lit(1), lit(2), lit(3)))
df.withColumn("some_struct", struct(lit("foo"), lit(1), lit(.3)))
df.withColumn("some_map", create_map(lit("key1"), lit(1), lit("key2"), lit(2)))

Exactly the same methods can be used in Scala.

import org.apache.spark.sql.functions.{array, lit, map, struct}

df.withColumn("new_column", lit(10))
df.withColumn("map", map(lit("key1"), lit(1), lit("key2"), lit(2)))

To provide names for structs use either alias on each field:

df.withColumn(
    "some_struct",
    struct(lit("foo").alias("x"), lit(1).alias("y"), lit(0.3).alias("z"))
 )

or cast on the whole object

df.withColumn(
    "some_struct", 
    struct(lit("foo"), lit(1), lit(0.3)).cast("struct<x: string, y: integer, z: double>")
 )

It is also possible, although slower, to use an UDF.

Note:

The same constructs can be used to pass constant arguments to UDFs or SQL functions.


回答 1

在spark 2.2中,有两种方法可以在DataFrame的列中添加常量值:

1)使用 lit

2)使用typedLit

两者之间的区别在于typedLit还可以处理参数化的Scala类型,例如List,Seq和Map

样本数据框:

val df = spark.createDataFrame(Seq((0,"a"),(1,"b"),(2,"c"))).toDF("id", "col1")

+---+----+
| id|col1|
+---+----+
|  0|   a|
|  1|   b|
+---+----+

1)使用lit在名为newcol的新列中添加常量字符串值:

import org.apache.spark.sql.functions.lit
val newdf = df.withColumn("newcol",lit("myval"))

结果:

+---+----+------+
| id|col1|newcol|
+---+----+------+
|  0|   a| myval|
|  1|   b| myval|
+---+----+------+

2)使用typedLit

import org.apache.spark.sql.functions.typedLit
df.withColumn("newcol", typedLit(("sample", 10, .044)))

结果:

+---+----+-----------------+
| id|col1|           newcol|
+---+----+-----------------+
|  0|   a|[sample,10,0.044]|
|  1|   b|[sample,10,0.044]|
|  2|   c|[sample,10,0.044]|
+---+----+-----------------+

In spark 2.2 there are two ways to add constant value in a column in DataFrame:

1) Using lit

2) Using typedLit.

The difference between the two is that typedLit can also handle parameterized scala types e.g. List, Seq, and Map

Sample DataFrame:

val df = spark.createDataFrame(Seq((0,"a"),(1,"b"),(2,"c"))).toDF("id", "col1")

+---+----+
| id|col1|
+---+----+
|  0|   a|
|  1|   b|
+---+----+

1) Using lit: Adding constant string value in new column named newcol:

import org.apache.spark.sql.functions.lit
val newdf = df.withColumn("newcol",lit("myval"))

Result:

+---+----+------+
| id|col1|newcol|
+---+----+------+
|  0|   a| myval|
|  1|   b| myval|
+---+----+------+

2) Using typedLit:

import org.apache.spark.sql.functions.typedLit
df.withColumn("newcol", typedLit(("sample", 10, .044)))

Result:

+---+----+-----------------+
| id|col1|           newcol|
+---+----+-----------------+
|  0|   a|[sample,10,0.044]|
|  1|   b|[sample,10,0.044]|
|  2|   c|[sample,10,0.044]|
+---+----+-----------------+

查找名称包含特定字符串的列

问题:查找名称包含特定字符串的列

我有一个带有列名称的数据框,我想找到一个包含特定字符串但与之不完全匹配的数据框。我在寻找'spike'列名喜欢'spike-2''hey spike''spiked-in'(该'spike'部分总是连续)。

我希望列名以字符串或变量的形式返回,因此我以后可以使用df['name']df[name]照常访问列。我试图找到方法,但没有成功。有小费吗?

I have a dataframe with column names, and I want to find the one that contains a certain string, but does not exactly match it. I’m searching for 'spike' in column names like 'spike-2', 'hey spike', 'spiked-in' (the 'spike' part is always continuous).

I want the column name to be returned as a string or a variable, so I access the column later with df['name'] or df[name] as normal. I’ve tried to find ways to do this, to no avail. Any tips?


回答 0

只需遍历DataFrame.columns,这是一个示例,在此示例中,您将获得匹配的列名称列表:

import pandas as pd

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)

spike_cols = [col for col in df.columns if 'spike' in col]
print(list(df.columns))
print(spike_cols)

输出:

['hey spke', 'no', 'spike-2', 'spiked-in']
['spike-2', 'spiked-in']

说明:

  1. df.columns 返回列名列表
  2. [col for col in df.columns if 'spike' in col]df.columns使用变量遍历列表col并将其添加到结果列表(如果col包含)'spike'。此语法是列表理解

如果只希望结果数据集的列匹配,则可以执行以下操作:

df2 = df.filter(regex='spike')
print(df2)

输出:

   spike-2  spiked-in
0        1          7
1        2          8
2        3          9

Just iterate over DataFrame.columns, now this is an example in which you will end up with a list of column names that match:

import pandas as pd

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)

spike_cols = [col for col in df.columns if 'spike' in col]
print(list(df.columns))
print(spike_cols)

Output:

['hey spke', 'no', 'spike-2', 'spiked-in']
['spike-2', 'spiked-in']

Explanation:

  1. df.columns returns a list of column names
  2. [col for col in df.columns if 'spike' in col] iterates over the list df.columns with the variable col and adds it to the resulting list if col contains 'spike'. This syntax is list comprehension.

If you only want the resulting data set with the columns that match you can do this:

df2 = df.filter(regex='spike')
print(df2)

Output:

   spike-2  spiked-in
0        1          7
1        2          8
2        3          9

回答 1

此答案使用DataFrame.filter方法执行此操作而无需列表理解:

import pandas as pd

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6]}
df = pd.DataFrame(data)

print(df.filter(like='spike').columns)

将仅输出“ spike-2”。您还可以使用正则表达式,如某些人在上面的评论中建议的那样:

print(df.filter(regex='spike|spke').columns)

将输出两列:[‘spike-2’,’hey spke’]

This answer uses the DataFrame.filter method to do this without list comprehension:

import pandas as pd

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6]}
df = pd.DataFrame(data)

print(df.filter(like='spike').columns)

Will output just ‘spike-2’. You can also use regex, as some people suggested in comments above:

print(df.filter(regex='spike|spke').columns)

Will output both columns: [‘spike-2’, ‘hey spke’]


回答 2

您也可以使用 df.columns[df.columns.str.contains(pat = 'spike')]

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)

colNames = df.columns[df.columns.str.contains(pat = 'spike')] 

print(colNames)

这将输出列名称: 'spike-2', 'spiked-in'

有关pandas.Series.str.contains的更多信息。

You can also use df.columns[df.columns.str.contains(pat = 'spike')]

data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)

colNames = df.columns[df.columns.str.contains(pat = 'spike')] 

print(colNames)

This will output the column names: 'spike-2', 'spiked-in'

More about pandas.Series.str.contains.


回答 3

# select columns containing 'spike'
df.filter(like='spike', axis=1)

您还可以按名称选择正则表达式。请参阅:pandas.DataFrame.filter

# select columns containing 'spike'
df.filter(like='spike', axis=1)

You can also select by name, regular expression. Refer to: pandas.DataFrame.filter


回答 4

df.loc[:,df.columns.str.contains("spike")]
df.loc[:,df.columns.str.contains("spike")]

回答 5

您还可以使用以下代码:

spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]

You also can use this code:

spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]

回答 6

根据“开始”,“包含”和“结束”获取名称和子集:

# from: /programming/21285380/find-column-whose-name-contains-a-specific-string
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.contains.html
# from: https://cmdlinetips.com/2019/04/how-to-select-columns-using-prefix-suffix-of-column-names-in-pandas/
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.filter.html




import pandas as pd



data = {'spike_starts': [1,2,3], 'ends_spike_starts': [4,5,6], 'ends_spike': [7,8,9], 'not': [10,11,12]}
df = pd.DataFrame(data)



print("\n")
print("----------------------------------------")
colNames_contains = df.columns[df.columns.str.contains(pat = 'spike')].tolist() 
print("Contains")
print(colNames_contains)



print("\n")
print("----------------------------------------")
colNames_starts = df.columns[df.columns.str.contains(pat = '^spike')].tolist() 
print("Starts")
print(colNames_starts)



print("\n")
print("----------------------------------------")
colNames_ends = df.columns[df.columns.str.contains(pat = 'spike$')].tolist() 
print("Ends")
print(colNames_ends)



print("\n")
print("----------------------------------------")
df_subset_start = df.filter(regex='^spike',axis=1)
print("Starts")
print(df_subset_start)



print("\n")
print("----------------------------------------")
df_subset_contains = df.filter(regex='spike',axis=1)
print("Contains")
print(df_subset_contains)



print("\n")
print("----------------------------------------")
df_subset_ends = df.filter(regex='spike$',axis=1)
print("Ends")
print(df_subset_ends)

Getting name and subsetting based on Start, Contains, and Ends:

# from: https://stackoverflow.com/questions/21285380/find-column-whose-name-contains-a-specific-string
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.contains.html
# from: https://cmdlinetips.com/2019/04/how-to-select-columns-using-prefix-suffix-of-column-names-in-pandas/
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.filter.html




import pandas as pd



data = {'spike_starts': [1,2,3], 'ends_spike_starts': [4,5,6], 'ends_spike': [7,8,9], 'not': [10,11,12]}
df = pd.DataFrame(data)



print("\n")
print("----------------------------------------")
colNames_contains = df.columns[df.columns.str.contains(pat = 'spike')].tolist() 
print("Contains")
print(colNames_contains)



print("\n")
print("----------------------------------------")
colNames_starts = df.columns[df.columns.str.contains(pat = '^spike')].tolist() 
print("Starts")
print(colNames_starts)



print("\n")
print("----------------------------------------")
colNames_ends = df.columns[df.columns.str.contains(pat = 'spike$')].tolist() 
print("Ends")
print(colNames_ends)



print("\n")
print("----------------------------------------")
df_subset_start = df.filter(regex='^spike',axis=1)
print("Starts")
print(df_subset_start)



print("\n")
print("----------------------------------------")
df_subset_contains = df.filter(regex='spike',axis=1)
print("Contains")
print(df_subset_contains)



print("\n")
print("----------------------------------------")
df_subset_ends = df.filter(regex='spike$',axis=1)
print("Ends")
print(df_subset_ends)

熊猫数据框获取每个组的第一行

问题:熊猫数据框获取每个组的第一行

我有DataFrame下面的熊猫。

df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4,5,6,6,6,7,7],
                'value'  : ["first","second","second","first",
                            "second","first","third","fourth",
                            "fifth","second","fifth","first",
                            "first","second","third","fourth","fifth"]})

我想通过[“ id”,“ value”]对此分组,并获得每个分组的第一行。

        id   value
0        1   first
1        1  second
2        1  second
3        2   first
4        2  second
5        3   first
6        3   third
7        3  fourth
8        3   fifth
9        4  second
10       4   fifth
11       5   first
12       6   first
13       6  second
14       6   third
15       7  fourth
16       7   fifth

预期结果

    id   value
     1   first
     2   first
     3   first
     4  second
     5  first
     6  first
     7  fourth

我尝试了以下操作,仅给出的第一行DataFrame。任何有关此的帮助表示赞赏。

In [25]: for index, row in df.iterrows():
   ....:     df2 = pd.DataFrame(df.groupby(['id','value']).reset_index().ix[0])

I have a pandas DataFrame like following.

df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4,5,6,6,6,7,7],
                'value'  : ["first","second","second","first",
                            "second","first","third","fourth",
                            "fifth","second","fifth","first",
                            "first","second","third","fourth","fifth"]})

I want to group this by [“id”,”value”] and get the first row of each group.

        id   value
0        1   first
1        1  second
2        1  second
3        2   first
4        2  second
5        3   first
6        3   third
7        3  fourth
8        3   fifth
9        4  second
10       4   fifth
11       5   first
12       6   first
13       6  second
14       6   third
15       7  fourth
16       7   fifth

Expected outcome

    id   value
     1   first
     2   first
     3   first
     4  second
     5  first
     6  first
     7  fourth

I tried following which only gives the first row of the DataFrame. Any help regarding this is appreciated.

In [25]: for index, row in df.iterrows():
   ....:     df2 = pd.DataFrame(df.groupby(['id','value']).reset_index().ix[0])

回答 0

>>> df.groupby('id').first()
     value
id        
1    first
2    first
3    first
4   second
5    first
6    first
7   fourth

如果需要id作为列:

>>> df.groupby('id').first().reset_index()
   id   value
0   1   first
1   2   first
2   3   first
3   4  second
4   5   first
5   6   first
6   7  fourth

要获取n条第一条记录,可以使用head():

>>> df.groupby('id').head(2).reset_index(drop=True)
    id   value
0    1   first
1    1  second
2    2   first
3    2  second
4    3   first
5    3   third
6    4  second
7    4   fifth
8    5   first
9    6   first
10   6  second
11   7  fourth
12   7   fifth
>>> df.groupby('id').first()
     value
id        
1    first
2    first
3    first
4   second
5    first
6    first
7   fourth

If you need id as column:

>>> df.groupby('id').first().reset_index()
   id   value
0   1   first
1   2   first
2   3   first
3   4  second
4   5   first
5   6   first
6   7  fourth

To get n first records, you can use head():

>>> df.groupby('id').head(2).reset_index(drop=True)
    id   value
0    1   first
1    1  second
2    2   first
3    2  second
4    3   first
5    3   third
6    4  second
7    4   fifth
8    5   first
9    6   first
10   6  second
11   7  fourth
12   7   fifth

回答 1

这将为您提供每组的第二行(零索引,nth(0)与first()相同):

df.groupby('id').nth(1) 

文档:http : //pandas.pydata.org/pandas-docs/stable/groupby.html#taking-the-nth-row-of-each-group

This will give you the second row of each group (zero indexed, nth(0) is the same as first()):

df.groupby('id').nth(1) 

Documentation: http://pandas.pydata.org/pandas-docs/stable/groupby.html#taking-the-nth-row-of-each-group


回答 2

我建议使用.nth(0)而不是.first()如果您需要获得第一行。

它们之间的区别在于它们处理NaN的方式,因此.nth(0)无论该行中的值是什么,都将返回组的第一行,而.first()最终将返回每列中的第一个not NaN值。

例如,如果您的数据集是:

df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4],
            'value'  : ["first","second","third", np.NaN,
                        "second","first","second","third",
                        "fourth","first","second"]})

>>> df.groupby('id').nth(0)
    value
id        
1    first
2    NaN
3    first
4    first

>>> df.groupby('id').first()
    value
id        
1    first
2    second
3    first
4    first

I’d suggest to use .nth(0) rather than .first() if you need to get the first row.

The difference between them is how they handle NaNs, so .nth(0) will return the first row of group no matter what are the values in this row, while .first() will eventually return the first not NaN value in each column.

E.g. if your dataset is :

df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4],
            'value'  : ["first","second","third", np.NaN,
                        "second","first","second","third",
                        "fourth","first","second"]})

>>> df.groupby('id').nth(0)
    value
id        
1    first
2    NaN
3    first
4    first

And

>>> df.groupby('id').first()
    value
id        
1    first
2    second
3    first
4    first

回答 3

也许这就是你想要的

import pandas as pd
idx = pd.MultiIndex.from_product([['state1','state2'],   ['county1','county2','county3','county4']])
df = pd.DataFrame({'pop': [12,15,65,42,78,67,55,31]}, index=idx)
                pop
state1 county1   12
       county2   15
       county3   65
       county4   42
state2 county1   78
       county2   67
       county3   55
       county4   31
df.groupby(level=0, group_keys=False).apply(lambda x: x.sort_values('pop', ascending=False)).groupby(level=0).head(3)

> Out[29]: 
                pop
state1 county3   65
       county4   42
       county2   15
state2 county1   78
       county2   67
       county3   55

maybe this is what you want

import pandas as pd
idx = pd.MultiIndex.from_product([['state1','state2'],   ['county1','county2','county3','county4']])
df = pd.DataFrame({'pop': [12,15,65,42,78,67,55,31]}, index=idx)
                pop
state1 county1   12
       county2   15
       county3   65
       county4   42
state2 county1   78
       county2   67
       county3   55
       county4   31
df.groupby(level=0, group_keys=False).apply(lambda x: x.sort_values('pop', ascending=False)).groupby(level=0).head(3)

> Out[29]: 
                pop
state1 county3   65
       county4   42
       county2   15
state2 county1   78
       county2   67
       county3   55

回答 4

如果只需要我们可以处理的每个组的第一行drop_duplicates,请注意函数default方法keep='first'

df.drop_duplicates('id')
Out[1027]: 
    id   value
0    1   first
3    2   first
5    3   first
9    4  second
11   5   first
12   6   first
15   7  fourth

If you only need the first row from each group we can do with drop_duplicates, Notice the function default method keep='first'.

df.drop_duplicates('id')
Out[1027]: 
    id   value
0    1   first
3    2   first
5    3   first
9    4  second
11   5   first
12   6   first
15   7  fourth

用sklearn缩放的pandas数据框列

问题:用sklearn缩放的pandas数据框列

我有一个带有混合类型列的pandas数据框,我想将sklearn的min_max_scaler应用于某些列。理想情况下,我想就地进行这些转换,但还没有找到一种方法来进行。我编写了以下有效的代码:

import pandas as pd
import numpy as np
from sklearn import preprocessing

scaler = preprocessing.MinMaxScaler()

dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
min_max_scaler = preprocessing.MinMaxScaler()

def scaleColumns(df, cols_to_scale):
    for col in cols_to_scale:
        df[col] = pd.DataFrame(min_max_scaler.fit_transform(pd.DataFrame(dfTest[col])),columns=[col])
    return df

dfTest

    A   B   C
0    14.00   103.02  big
1    90.20   107.26  small
2    90.95   110.35  big
3    96.27   114.23  small
4    91.21   114.68  small

scaled_df = scaleColumns(dfTest,['A','B'])
scaled_df

A   B   C
0    0.000000    0.000000    big
1    0.926219    0.363636    small
2    0.935335    0.628645    big
3    1.000000    0.961407    small
4    0.938495    1.000000    small

我很好奇这是否是进行此转换的首选/最有效的方法。有没有办法可以使用df.apply更好呢?

我也很惊讶我无法使用以下代码:

bad_output = min_max_scaler.fit_transform(dfTest['A'])

如果我将整个数据帧传递给缩放器,则它可以工作:

dfTest2 = dfTest.drop('C', axis = 1) good_output = min_max_scaler.fit_transform(dfTest2) good_output

我很困惑为什么将系列传递给定标器会失败。在上面的完整工作代码中,我希望只将一个系列传递给缩放器,然后将dataframe column =设置为缩放的序列。我已经看到这个问题在其他几个地方问过,但找不到一个好的答案。任何帮助了解这里发生的事情将不胜感激!

I have a pandas dataframe with mixed type columns, and I’d like to apply sklearn’s min_max_scaler to some of the columns. Ideally, I’d like to do these transformations in place, but haven’t figured out a way to do that yet. I’ve written the following code that works:

import pandas as pd
import numpy as np
from sklearn import preprocessing

scaler = preprocessing.MinMaxScaler()

dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
min_max_scaler = preprocessing.MinMaxScaler()

def scaleColumns(df, cols_to_scale):
    for col in cols_to_scale:
        df[col] = pd.DataFrame(min_max_scaler.fit_transform(pd.DataFrame(dfTest[col])),columns=[col])
    return df

dfTest

    A   B   C
0    14.00   103.02  big
1    90.20   107.26  small
2    90.95   110.35  big
3    96.27   114.23  small
4    91.21   114.68  small

scaled_df = scaleColumns(dfTest,['A','B'])
scaled_df

A   B   C
0    0.000000    0.000000    big
1    0.926219    0.363636    small
2    0.935335    0.628645    big
3    1.000000    0.961407    small
4    0.938495    1.000000    small

I’m curious if this is the preferred/most efficient way to do this transformation. Is there a way I could use df.apply that would be better?

I’m also surprised I can’t get the following code to work:

bad_output = min_max_scaler.fit_transform(dfTest['A'])

If I pass an entire dataframe to the scaler it works:

dfTest2 = dfTest.drop('C', axis = 1) good_output = min_max_scaler.fit_transform(dfTest2) good_output

I’m confused why passing a series to the scaler fails. In my full working code above I had hoped to just pass a series to the scaler then set the dataframe column = to the scaled series. I’ve seen this question asked a few other places, but haven’t found a good answer. Any help understanding what’s going on here would be greatly appreciated!


回答 0

我不确定以前的版本是否pandas阻止了此操作,但现在以下代码段对我来说效果很好,并且无需使用就可以产生所需的内容apply

>>> import pandas as pd
>>> from sklearn.preprocessing import MinMaxScaler


>>> scaler = MinMaxScaler()

>>> dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],
                           'B':[103.02,107.26,110.35,114.23,114.68],
                           'C':['big','small','big','small','small']})

>>> dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A', 'B']])

>>> dfTest
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

I am not sure if previous versions of pandas prevented this but now the following snippet works perfectly for me and produces exactly what you want without having to use apply

>>> import pandas as pd
>>> from sklearn.preprocessing import MinMaxScaler


>>> scaler = MinMaxScaler()

>>> dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],
                           'B':[103.02,107.26,110.35,114.23,114.68],
                           'C':['big','small','big','small','small']})

>>> dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A', 'B']])

>>> dfTest
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

回答 1

像这样?

dfTest = pd.DataFrame({
           'A':[14.00,90.20,90.95,96.27,91.21],
           'B':[103.02,107.26,110.35,114.23,114.68], 
           'C':['big','small','big','small','small']
         })
dfTest[['A','B']] = dfTest[['A','B']].apply(
                           lambda x: MinMaxScaler().fit_transform(x))
dfTest

    A           B           C
0   0.000000    0.000000    big
1   0.926219    0.363636    small
2   0.935335    0.628645    big
3   1.000000    0.961407    small
4   0.938495    1.000000    small

Like this?

dfTest = pd.DataFrame({
           'A':[14.00,90.20,90.95,96.27,91.21],
           'B':[103.02,107.26,110.35,114.23,114.68], 
           'C':['big','small','big','small','small']
         })
dfTest[['A','B']] = dfTest[['A','B']].apply(
                           lambda x: MinMaxScaler().fit_transform(x))
dfTest

    A           B           C
0   0.000000    0.000000    big
1   0.926219    0.363636    small
2   0.935335    0.628645    big
3   1.000000    0.961407    small
4   0.938495    1.000000    small

回答 2

正如pir的评论中提到的那样-该.apply(lambda el: scale.fit_transform(el))方法将产生以下警告:

DeprecationWarning:在0.17中弃用1d数组作为数据,它将在0.19中引发ValueError。如果数据具有单个功能,则使用X.reshape(-1,1)来重塑数据,如果包含单个样本,则使用X.reshape(1,-1)来重塑数据。

将您的列转换为numpy数组应该可以完成这项工作(我更喜欢StandardScaler):

from sklearn.preprocessing import StandardScaler
scale = StandardScaler()

dfTest[['A','B','C']] = scale.fit_transform(dfTest[['A','B','C']].as_matrix())

编辑 2018年11月(已针对熊猫0.23.4测试)-

作为罗布·默里提到的意见,大熊猫的电流(v0.23.4)版本.as_matrix()的回报FutureWarning。因此,应将其替换为.values

from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()

scaler.fit_transform(dfTest[['A','B']].values)

编辑 2019年5月(已针对熊猫0.24.2测试)-

正如joelostblom在评论中提到的那样:“因此0.24.0,建议使用.to_numpy()代替.values。”

更新的示例:

import pandas as pd
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
dfTest = pd.DataFrame({
               'A':[14.00,90.20,90.95,96.27,91.21],
               'B':[103.02,107.26,110.35,114.23,114.68],
               'C':['big','small','big','small','small']
             })
dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A','B']].to_numpy())
dfTest
      A         B      C
0 -1.995290 -1.571117    big
1  0.436356 -0.603995  small
2  0.460289  0.100818    big
3  0.630058  0.985826  small
4  0.468586  1.088469  small

As it is being mentioned in pir’s comment – the .apply(lambda el: scale.fit_transform(el)) method will produce the following warning:

DeprecationWarning: Passing 1d arrays as data is deprecated in 0.17 and will raise ValueError in 0.19. Reshape your data either using X.reshape(-1, 1) if your data has a single feature or X.reshape(1, -1) if it contains a single sample.

Converting your columns to numpy arrays should do the job (I prefer StandardScaler):

from sklearn.preprocessing import StandardScaler
scale = StandardScaler()

dfTest[['A','B','C']] = scale.fit_transform(dfTest[['A','B','C']].as_matrix())

Edit Nov 2018 (Tested for pandas 0.23.4)–

As Rob Murray mentions in the comments, in the current (v0.23.4) version of pandas .as_matrix() returns FutureWarning. Therefore, it should be replaced by .values:

from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()

scaler.fit_transform(dfTest[['A','B']].values)

Edit May 2019 (Tested for pandas 0.24.2)–

As joelostblom mentions in the comments, “Since 0.24.0, it is recommended to use .to_numpy() instead of .values.”

Updated example:

import pandas as pd
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
dfTest = pd.DataFrame({
               'A':[14.00,90.20,90.95,96.27,91.21],
               'B':[103.02,107.26,110.35,114.23,114.68],
               'C':['big','small','big','small','small']
             })
dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A','B']].to_numpy())
dfTest
      A         B      C
0 -1.995290 -1.571117    big
1  0.436356 -0.603995  small
2  0.460289  0.100818    big
3  0.630058  0.985826  small
4  0.468586  1.088469  small

回答 3

df = pd.DataFrame(scale.fit_transform(df.values), columns=df.columns, index=df.index)

这应该在没有折旧警告的情况下起作用。

df = pd.DataFrame(scale.fit_transform(df.values), columns=df.columns, index=df.index)

This should work without depreciation warnings.


回答 4

您只能使用以下方法进行操作 pandas

In [235]:
dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
df = dfTest[['A', 'B']]
df_norm = (df - df.min()) / (df.max() - df.min())
print df_norm
print pd.concat((df_norm, dfTest.C),1)

          A         B
0  0.000000  0.000000
1  0.926219  0.363636
2  0.935335  0.628645
3  1.000000  0.961407
4  0.938495  1.000000
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

You can do it using pandas only:

In [235]:
dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
df = dfTest[['A', 'B']]
df_norm = (df - df.min()) / (df.max() - df.min())
print df_norm
print pd.concat((df_norm, dfTest.C),1)

          A         B
0  0.000000  0.000000
1  0.926219  0.363636
2  0.935335  0.628645
3  1.000000  0.961407
4  0.938495  1.000000
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

回答 5

我知道这是一个很老的评论,但仍然:

不要使用单括号(dfTest['A']),而应使用双括号(dfTest[['A']])

即:min_max_scaler.fit_transform(dfTest[['A']])

我相信这会取得理想的结果。

I know it’s a very old comment, but still:

Instead of using single bracket (dfTest['A']), use double brackets (dfTest[['A']]).

i.e: min_max_scaler.fit_transform(dfTest[['A']]).

I believe this will give the desired result.


创建两个熊猫数据框列的字典的最有效方法是什么?

问题:创建两个熊猫数据框列的字典的最有效方法是什么?

组织以下熊猫数据框的最有效方法是什么:

数据=

Position    Letter
1           a
2           b
3           c
4           d
5           e

变成字典一样alphabet[1 : 'a', 2 : 'b', 3 : 'c', 4 : 'd', 5 : 'e']

What is the most efficient way to organise the following pandas Dataframe:

data =

Position    Letter
1           a
2           b
3           c
4           d
5           e

into a dictionary like alphabet[1 : 'a', 2 : 'b', 3 : 'c', 4 : 'd', 5 : 'e']?


回答 0

In [9]: pd.Series(df.Letter.values,index=df.Position).to_dict()
Out[9]: {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e'}

速度比较(使用Wouter方法)

In [6]: df = pd.DataFrame(randint(0,10,10000).reshape(5000,2),columns=list('AB'))

In [7]: %timeit dict(zip(df.A,df.B))
1000 loops, best of 3: 1.27 ms per loop

In [8]: %timeit pd.Series(df.A.values,index=df.B).to_dict()
1000 loops, best of 3: 987 us per loop
In [9]: pd.Series(df.Letter.values,index=df.Position).to_dict()
Out[9]: {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e'}

Speed comparion (using Wouter’s method)

In [6]: df = pd.DataFrame(randint(0,10,10000).reshape(5000,2),columns=list('AB'))

In [7]: %timeit dict(zip(df.A,df.B))
1000 loops, best of 3: 1.27 ms per loop

In [8]: %timeit pd.Series(df.A.values,index=df.B).to_dict()
1000 loops, best of 3: 987 us per loop

回答 1

我找到了解决问题的更快方法,至少在使用以下方法的大型数据集上: df.set_index(KEY).to_dict()[VALUE]

50,000行的证明:

df = pd.DataFrame(np.random.randint(32, 120, 100000).reshape(50000,2),columns=list('AB'))
df['A'] = df['A'].apply(chr)

%timeit dict(zip(df.A,df.B))
%timeit pd.Series(df.A.values,index=df.B).to_dict()
%timeit df.set_index('A').to_dict()['B']

输出:

100 loops, best of 3: 7.04 ms per loop  # WouterOvermeire
100 loops, best of 3: 9.83 ms per loop  # Jeff
100 loops, best of 3: 4.28 ms per loop  # Kikohs (me)

I found a faster way to solve the problem, at least on realistically large datasets using: df.set_index(KEY).to_dict()[VALUE]

Proof on 50,000 rows:

df = pd.DataFrame(np.random.randint(32, 120, 100000).reshape(50000,2),columns=list('AB'))
df['A'] = df['A'].apply(chr)

%timeit dict(zip(df.A,df.B))
%timeit pd.Series(df.A.values,index=df.B).to_dict()
%timeit df.set_index('A').to_dict()['B']

Output:

100 loops, best of 3: 7.04 ms per loop  # WouterOvermeire
100 loops, best of 3: 9.83 ms per loop  # Jeff
100 loops, best of 3: 4.28 ms per loop  # Kikohs (me)

回答 2

在Python 3.6中,最快的方法仍然是WouterOvermeire。Kikohs的提议比其他两个方案要慢。

import timeit

setup = '''
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randint(32, 120, 100000).reshape(50000,2),columns=list('AB'))
df['A'] = df['A'].apply(chr)
'''

timeit.Timer('dict(zip(df.A,df.B))', setup=setup).repeat(7,500)
timeit.Timer('pd.Series(df.A.values,index=df.B).to_dict()', setup=setup).repeat(7,500)
timeit.Timer('df.set_index("A").to_dict()["B"]', setup=setup).repeat(7,500)

结果:

1.1214002349999777 s  # WouterOvermeire
1.1922008498571748 s  # Jeff
1.7034366211428602 s  # Kikohs

In Python 3.6 the fastest way is still the WouterOvermeire one. Kikohs’ proposal is slower than the other two options.

import timeit

setup = '''
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randint(32, 120, 100000).reshape(50000,2),columns=list('AB'))
df['A'] = df['A'].apply(chr)
'''

timeit.Timer('dict(zip(df.A,df.B))', setup=setup).repeat(7,500)
timeit.Timer('pd.Series(df.A.values,index=df.B).to_dict()', setup=setup).repeat(7,500)
timeit.Timer('df.set_index("A").to_dict()["B"]', setup=setup).repeat(7,500)

Results:

1.1214002349999777 s  # WouterOvermeire
1.1922008498571748 s  # Jeff
1.7034366211428602 s  # Kikohs

回答 3

TL; DR

>>> import pandas as pd
>>> df = pd.DataFrame({'Position':[1,2,3,4,5], 'Letter':['a', 'b', 'c', 'd', 'e']})
>>> dict(sorted(df.values.tolist())) # Sort of sorted... 
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
>>> from collections import OrderedDict
>>> OrderedDict(df.values.tolist())
OrderedDict([('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', 5)])

在长

解决方案说明: dict(sorted(df.values.tolist()))

鉴于:

df = pd.DataFrame({'Position':[1,2,3,4,5], 'Letter':['a', 'b', 'c', 'd', 'e']})

[出]:

 Letter Position
0   a   1
1   b   2
2   c   3
3   d   4
4   e   5

尝试:

# Get the values out to a 2-D numpy array, 
df.values

[出]:

array([['a', 1],
       ['b', 2],
       ['c', 3],
       ['d', 4],
       ['e', 5]], dtype=object)

然后(可选):

# Dump it into a list so that you can sort it using `sorted()`
sorted(df.values.tolist()) # Sort by key

要么:

# Sort by value:
from operator import itemgetter
sorted(df.values.tolist(), key=itemgetter(1))

[出]:

[['a', 1], ['b', 2], ['c', 3], ['d', 4], ['e', 5]]

最后,将2个元素的列表转换成字典。

dict(sorted(df.values.tolist())) 

[出]:

{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}

有关

回答@sbradbio评论:

如果一个特定的键有多个值,而您想保留所有值,那么这不是最有效,但最直观的方法是:

from collections import defaultdict
import pandas as pd

multivalue_dict = defaultdict(list)

df = pd.DataFrame({'Position':[1,2,4,4,4], 'Letter':['a', 'b', 'd', 'e', 'f']})

for idx,row in df.iterrows():
    multivalue_dict[row['Position']].append(row['Letter'])

[出]:

>>> print(multivalue_dict)
defaultdict(list, {1: ['a'], 2: ['b'], 4: ['d', 'e', 'f']})

TL;DR

>>> import pandas as pd
>>> df = pd.DataFrame({'Position':[1,2,3,4,5], 'Letter':['a', 'b', 'c', 'd', 'e']})
>>> dict(sorted(df.values.tolist())) # Sort of sorted... 
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
>>> from collections import OrderedDict
>>> OrderedDict(df.values.tolist())
OrderedDict([('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', 5)])

In Long

Explaining solution: dict(sorted(df.values.tolist()))

Given:

df = pd.DataFrame({'Position':[1,2,3,4,5], 'Letter':['a', 'b', 'c', 'd', 'e']})

[out]:

 Letter Position
0   a   1
1   b   2
2   c   3
3   d   4
4   e   5

Try:

# Get the values out to a 2-D numpy array, 
df.values

[out]:

array([['a', 1],
       ['b', 2],
       ['c', 3],
       ['d', 4],
       ['e', 5]], dtype=object)

Then optionally:

# Dump it into a list so that you can sort it using `sorted()`
sorted(df.values.tolist()) # Sort by key

Or:

# Sort by value:
from operator import itemgetter
sorted(df.values.tolist(), key=itemgetter(1))

[out]:

[['a', 1], ['b', 2], ['c', 3], ['d', 4], ['e', 5]]

Lastly, cast the list of list of 2 elements into a dict.

dict(sorted(df.values.tolist())) 

[out]:

{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}

Related

Answering @sbradbio comment:

If there are multiple values for a specific key and you would like to keep all of them, it’s the not the most efficient but the most intuitive way is:

from collections import defaultdict
import pandas as pd

multivalue_dict = defaultdict(list)

df = pd.DataFrame({'Position':[1,2,4,4,4], 'Letter':['a', 'b', 'd', 'e', 'f']})

for idx,row in df.iterrows():
    multivalue_dict[row['Position']].append(row['Letter'])

[out]:

>>> print(multivalue_dict)
defaultdict(list, {1: ['a'], 2: ['b'], 4: ['d', 'e', 'f']})

熊猫:如何将一列中的文本分成多行?

问题:熊猫:如何将一列中的文本分成多行?

我正在处理一个较大的csv文件,并且最后一列的旁边是一串文本,我想用一个特定的分隔符来分割它。我想知道是否有使用pandas或python的简单方法?

CustNum  CustomerName     ItemQty  Item   Seatblocks                 ItemExt
32363    McCartney, Paul      3     F04    2:218:10:4,6                   60
31316    Lennon, John        25     F01    1:13:36:1,12 1:13:37:1,13     300

我想先按空格(' ')(':')Seatblocks列中按冒号分开,但每个单元格将导致列数不同。我具有重新排列列的功能,因此Seatblocks列位于工作表的末尾,但是我不确定从那里开始如何做。我可以使用内置text-to-columns函数和快速宏在excel中完成此操作,但是我的数据集记录太多,无法处理excel。

最终,我想记录约翰·列侬的记录并创建多行,并将每组座位的信息放在单独的行上。

I’m working with a large csv file and the next to last column has a string of text that I want to split by a specific delimiter. I was wondering if there is a simple way to do this using pandas or python?

CustNum  CustomerName     ItemQty  Item   Seatblocks                 ItemExt
32363    McCartney, Paul      3     F04    2:218:10:4,6                   60
31316    Lennon, John        25     F01    1:13:36:1,12 1:13:37:1,13     300

I want to split by the space(' ') and then the colon(':') in the Seatblocks column, but each cell would result in a different number of columns. I have a function to rearrange the columns so the Seatblocks column is at the end of the sheet, but I’m not sure what to do from there. I can do it in excel with the built in text-to-columns function and a quick macro, but my dataset has too many records for excel to handle.

Ultimately, I want to take records such John Lennon’s and create multiple lines, with the info from each set of seats on a separate line.


回答 0

这将座垫按空间划分,并给每个单独的行。

In [43]: df
Out[43]: 
   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0    32363  McCartney, Paul        3  F04               2:218:10:4,6       60
1    31316     Lennon, John       25  F01  1:13:36:1,12 1:13:37:1,13      300

In [44]: s = df['Seatblocks'].str.split(' ').apply(Series, 1).stack()

In [45]: s.index = s.index.droplevel(-1) # to line up with df's index

In [46]: s.name = 'Seatblocks' # needs a name to join

In [47]: s
Out[47]: 
0    2:218:10:4,6
1    1:13:36:1,12
1    1:13:37:1,13
Name: Seatblocks, dtype: object

In [48]: del df['Seatblocks']

In [49]: df.join(s)
Out[49]: 
   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    32363  McCartney, Paul        3  F04       60  2:218:10:4,6
1    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13

或者,将每个冒号分隔的字符串放在自己的列中:

In [50]: df.join(s.apply(lambda x: Series(x.split(':'))))
Out[50]: 
   CustNum     CustomerName  ItemQty Item  ItemExt  0    1   2     3
0    32363  McCartney, Paul        3  F04       60  2  218  10   4,6
1    31316     Lennon, John       25  F01      300  1   13  36  1,12
1    31316     Lennon, John       25  F01      300  1   13  37  1,13

这有点丑陋,但也许有人会用更漂亮的解决方案。

This splits the Seatblocks by space and gives each its own row.

In [43]: df
Out[43]: 
   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0    32363  McCartney, Paul        3  F04               2:218:10:4,6       60
1    31316     Lennon, John       25  F01  1:13:36:1,12 1:13:37:1,13      300

In [44]: s = df['Seatblocks'].str.split(' ').apply(Series, 1).stack()

In [45]: s.index = s.index.droplevel(-1) # to line up with df's index

In [46]: s.name = 'Seatblocks' # needs a name to join

In [47]: s
Out[47]: 
0    2:218:10:4,6
1    1:13:36:1,12
1    1:13:37:1,13
Name: Seatblocks, dtype: object

In [48]: del df['Seatblocks']

In [49]: df.join(s)
Out[49]: 
   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    32363  McCartney, Paul        3  F04       60  2:218:10:4,6
1    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13

Or, to give each colon-separated string in its own column:

In [50]: df.join(s.apply(lambda x: Series(x.split(':'))))
Out[50]: 
   CustNum     CustomerName  ItemQty Item  ItemExt  0    1   2     3
0    32363  McCartney, Paul        3  F04       60  2  218  10   4,6
1    31316     Lennon, John       25  F01      300  1   13  36  1,12
1    31316     Lennon, John       25  F01      300  1   13  37  1,13

This is a little ugly, but maybe someone will chime in with a prettier solution.


回答 1

与Dan不同的是,我认为他的回答相当优雅……但是不幸的是,它的效率也非常低下。因此,由于问题提到“大的csv文件”,因此我建议尝试使用Shell Dan的解决方案:

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print df['col'].apply(lambda x : pd.Series(x.split(' '))).head()"

…与这种替代方案相比:

time python -c "import pandas as pd;
from scipy import array, concatenate;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(concatenate(df['col'].apply( lambda x : [x.split(' ')]))).head()"

… 还有这个:

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(dict(zip(range(3), [df['col'].apply(lambda x : x.split(' ')[i]) for i in range(3)]))).head()"

第二个简单地避免了分配10万个序列,这足以使它快10倍左右。但是,第三种解决方案有点讽刺地浪费了对str.split()的调用(每行每列调用一次,因此比其他两种解决方案多三倍),它比第一种解决方案快40倍,因为它甚至避免实例化100000个列表。是的,这确实有点丑陋…

编辑: 此答案建议如何使用“ to_list()”并避免使用lambda。结果是像

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(df.col.str.split().tolist()).head()"

这比第三个解决方案更有效,而且肯定更优雅。

编辑:更简单

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(list(df.col.str.split())).head()"

也可以,并且几乎一样有效。

编辑: 更简单!并处理NaN(但效率较低):

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print df.col.str.split(expand=True).head()"

Differently from Dan, I consider his answer quite elegant… but unfortunately it is also very very inefficient. So, since the question mentioned “a large csv file”, let me suggest to try in a shell Dan’s solution:

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print df['col'].apply(lambda x : pd.Series(x.split(' '))).head()"

… compared to this alternative:

time python -c "import pandas as pd;
from scipy import array, concatenate;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(concatenate(df['col'].apply( lambda x : [x.split(' ')]))).head()"

… and this:

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(dict(zip(range(3), [df['col'].apply(lambda x : x.split(' ')[i]) for i in range(3)]))).head()"

The second simply refrains from allocating 100 000 Series, and this is enough to make it around 10 times faster. But the third solution, which somewhat ironically wastes a lot of calls to str.split() (it is called once per column per row, so three times more than for the others two solutions), is around 40 times faster than the first, because it even avoids to instance the 100 000 lists. And yes, it is certainly a little ugly…

EDIT: this answer suggests how to use “to_list()” and to avoid the need for a lambda. The result is something like

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(df.col.str.split().tolist()).head()"

which is even more efficient than the third solution, and certainly much more elegant.

EDIT: the even simpler

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(list(df.col.str.split())).head()"

works too, and is almost as efficient.

EDIT: even simpler! And handles NaNs (but less efficient):

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print df.col.str.split(expand=True).head()"

回答 2

import pandas as pd
import numpy as np

df = pd.DataFrame({'ItemQty': {0: 3, 1: 25}, 
                   'Seatblocks': {0: '2:218:10:4,6', 1: '1:13:36:1,12 1:13:37:1,13'}, 
                   'ItemExt': {0: 60, 1: 300}, 
                   'CustomerName': {0: 'McCartney, Paul', 1: 'Lennon, John'}, 
                   'CustNum': {0: 32363, 1: 31316}, 
                   'Item': {0: 'F04', 1: 'F01'}}, 
                    columns=['CustNum','CustomerName','ItemQty','Item','Seatblocks','ItemExt'])

print (df)
   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0    32363  McCartney, Paul        3  F04               2:218:10:4,6       60
1    31316     Lennon, John       25  F01  1:13:36:1,12 1:13:37:1,13      300

链接的另一个类似解决方案是use reset_indexrename

print (df.drop('Seatblocks', axis=1)
             .join
             (
             df.Seatblocks
             .str
             .split(expand=True)
             .stack()
             .reset_index(drop=True, level=1)
             .rename('Seatblocks')           
             ))

   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    32363  McCartney, Paul        3  F04       60  2:218:10:4,6
1    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13

如果in列中不是NOT NaN值,则最快的解决方案是listDataFrame构造函数使用理解:

df = pd.DataFrame(['a b c']*100000, columns=['col'])

In [141]: %timeit (pd.DataFrame(dict(zip(range(3), [df['col'].apply(lambda x : x.split(' ')[i]) for i in range(3)]))))
1 loop, best of 3: 211 ms per loop

In [142]: %timeit (pd.DataFrame(df.col.str.split().tolist()))
10 loops, best of 3: 87.8 ms per loop

In [143]: %timeit (pd.DataFrame(list(df.col.str.split())))
10 loops, best of 3: 86.1 ms per loop

In [144]: %timeit (df.col.str.split(expand=True))
10 loops, best of 3: 156 ms per loop

In [145]: %timeit (pd.DataFrame([ x.split() for x in df['col'].tolist()]))
10 loops, best of 3: 54.1 ms per loop

但是如果列NaN只包含str.splitexpand=True返回的参数一起使用DataFrame值为(document)的,那么它解释了为什么它比较慢:

df = pd.DataFrame(['a b c']*10, columns=['col'])
df.loc[0] = np.nan
print (df.head())
     col
0    NaN
1  a b c
2  a b c
3  a b c
4  a b c

print (df.col.str.split(expand=True))
     0     1     2
0  NaN  None  None
1    a     b     c
2    a     b     c
3    a     b     c
4    a     b     c
5    a     b     c
6    a     b     c
7    a     b     c
8    a     b     c
9    a     b     c
import pandas as pd
import numpy as np

df = pd.DataFrame({'ItemQty': {0: 3, 1: 25}, 
                   'Seatblocks': {0: '2:218:10:4,6', 1: '1:13:36:1,12 1:13:37:1,13'}, 
                   'ItemExt': {0: 60, 1: 300}, 
                   'CustomerName': {0: 'McCartney, Paul', 1: 'Lennon, John'}, 
                   'CustNum': {0: 32363, 1: 31316}, 
                   'Item': {0: 'F04', 1: 'F01'}}, 
                    columns=['CustNum','CustomerName','ItemQty','Item','Seatblocks','ItemExt'])

print (df)
   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0    32363  McCartney, Paul        3  F04               2:218:10:4,6       60
1    31316     Lennon, John       25  F01  1:13:36:1,12 1:13:37:1,13      300

Another similar solution with chaining is use reset_index and rename:

print (df.drop('Seatblocks', axis=1)
             .join
             (
             df.Seatblocks
             .str
             .split(expand=True)
             .stack()
             .reset_index(drop=True, level=1)
             .rename('Seatblocks')           
             ))

   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    32363  McCartney, Paul        3  F04       60  2:218:10:4,6
1    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13

If in column are NOT NaN values, the fastest solution is use list comprehension with DataFrame constructor:

df = pd.DataFrame(['a b c']*100000, columns=['col'])

In [141]: %timeit (pd.DataFrame(dict(zip(range(3), [df['col'].apply(lambda x : x.split(' ')[i]) for i in range(3)]))))
1 loop, best of 3: 211 ms per loop

In [142]: %timeit (pd.DataFrame(df.col.str.split().tolist()))
10 loops, best of 3: 87.8 ms per loop

In [143]: %timeit (pd.DataFrame(list(df.col.str.split())))
10 loops, best of 3: 86.1 ms per loop

In [144]: %timeit (df.col.str.split(expand=True))
10 loops, best of 3: 156 ms per loop

In [145]: %timeit (pd.DataFrame([ x.split() for x in df['col'].tolist()]))
10 loops, best of 3: 54.1 ms per loop

But if column contains NaN only works str.split with parameter expand=True which return DataFrame (documentation), and it explain why it is slowier:

df = pd.DataFrame(['a b c']*10, columns=['col'])
df.loc[0] = np.nan
print (df.head())
     col
0    NaN
1  a b c
2  a b c
3  a b c
4  a b c

print (df.col.str.split(expand=True))
     0     1     2
0  NaN  None  None
1    a     b     c
2    a     b     c
3    a     b     c
4    a     b     c
5    a     b     c
6    a     b     c
7    a     b     c
8    a     b     c
9    a     b     c

回答 3

另一种方法是这样的:

temp = df['Seatblocks'].str.split(' ')
data = data.reindex(data.index.repeat(temp.apply(len)))
data['new_Seatblocks'] = np.hstack(temp)

Another approach would be like this:

temp = df['Seatblocks'].str.split(' ')
data = data.reindex(data.index.repeat(temp.apply(len)))
data['new_Seatblocks'] = np.hstack(temp)

回答 4

也可以使用groupby()而不需要加入和stack()。

使用上面的示例数据:

import pandas as pd
import numpy as np


df = pd.DataFrame({'ItemQty': {0: 3, 1: 25}, 
                   'Seatblocks': {0: '2:218:10:4,6', 1: '1:13:36:1,12 1:13:37:1,13'}, 
                   'ItemExt': {0: 60, 1: 300}, 
                   'CustomerName': {0: 'McCartney, Paul', 1: 'Lennon, John'}, 
                   'CustNum': {0: 32363, 1: 31316}, 
                   'Item': {0: 'F04', 1: 'F01'}}, 
                    columns=['CustNum','CustomerName','ItemQty','Item','Seatblocks','ItemExt']) 
print(df)

   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0  32363    McCartney, Paul  3        F04  2:218:10:4,6               60     
1  31316    Lennon, John     25       F01  1:13:36:1,12 1:13:37:1,13  300  


#first define a function: given a Series of string, split each element into a new series
def split_series(ser,sep):
    return pd.Series(ser.str.cat(sep=sep).split(sep=sep)) 
#test the function, 
split_series(pd.Series(['a b','c']),sep=' ')
0    a
1    b
2    c
dtype: object

df2=(df.groupby(df.columns.drop('Seatblocks').tolist()) #group by all but one column
          ['Seatblocks'] #select the column to be split
          .apply(split_series,sep=' ') # split 'Seatblocks' in each group
         .reset_index(drop=True,level=-1).reset_index()) #remove extra index created

print(df2)
   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13
2    32363  McCartney, Paul        3  F04       60  2:218:10:4,6

Can also use groupby() with no need to join and stack().

Use above example data:

import pandas as pd
import numpy as np


df = pd.DataFrame({'ItemQty': {0: 3, 1: 25}, 
                   'Seatblocks': {0: '2:218:10:4,6', 1: '1:13:36:1,12 1:13:37:1,13'}, 
                   'ItemExt': {0: 60, 1: 300}, 
                   'CustomerName': {0: 'McCartney, Paul', 1: 'Lennon, John'}, 
                   'CustNum': {0: 32363, 1: 31316}, 
                   'Item': {0: 'F04', 1: 'F01'}}, 
                    columns=['CustNum','CustomerName','ItemQty','Item','Seatblocks','ItemExt']) 
print(df)

   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0  32363    McCartney, Paul  3        F04  2:218:10:4,6               60     
1  31316    Lennon, John     25       F01  1:13:36:1,12 1:13:37:1,13  300  


#first define a function: given a Series of string, split each element into a new series
def split_series(ser,sep):
    return pd.Series(ser.str.cat(sep=sep).split(sep=sep)) 
#test the function, 
split_series(pd.Series(['a b','c']),sep=' ')
0    a
1    b
2    c
dtype: object

df2=(df.groupby(df.columns.drop('Seatblocks').tolist()) #group by all but one column
          ['Seatblocks'] #select the column to be split
          .apply(split_series,sep=' ') # split 'Seatblocks' in each group
         .reset_index(drop=True,level=-1).reset_index()) #remove extra index created

print(df2)
   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13
2    32363  McCartney, Paul        3  F04       60  2:218:10:4,6

回答 5

这似乎比该线程其他地方建议的方法容易得多。

在熊猫数据框中拆分行

This seems a far easier method than those suggested elsewhere in this thread.

split rows in pandas dataframe


熊猫唯一值多列

问题:熊猫唯一值多列

df = pd.DataFrame({'Col1': ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
                   'Col2': ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
                   'Col3': np.random.random(5)})

返回“ Col1”和“ Col2”的唯一值的最佳方法是什么?

所需的输出是

'Bob', 'Joe', 'Bill', 'Mary', 'Steve'
df = pd.DataFrame({'Col1': ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
                   'Col2': ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
                   'Col3': np.random.random(5)})

What is the best way to return the unique values of ‘Col1’ and ‘Col2’?

The desired output is

'Bob', 'Joe', 'Bill', 'Mary', 'Steve'

回答 0

pd.unique 从输入数组或DataFrame列或索引返回唯一值。

此函数的输入必须是一维的,因此将需要合并多列。最简单的方法是选择所需的列,然后在展平的NumPy数组中查看值。整个操作如下所示:

>>> pd.unique(df[['Col1', 'Col2']].values.ravel('K'))
array(['Bob', 'Joe', 'Bill', 'Mary', 'Steve'], dtype=object)

请注意,这ravel()是一个数组方法,它返回多维数组的视图(如果可能)。该参数'K'告诉方法按元素在内存中存储的顺序展平数组(熊猫通常以Fortran连续的顺序存储基础数组;列在行之前)。这可能比使用该方法的默认“ C”顺序快得多。


另一种方法是选择列并将其传递给np.unique

>>> np.unique(df[['Col1', 'Col2']].values)
array(['Bill', 'Bob', 'Joe', 'Mary', 'Steve'], dtype=object)

ravel()此处不需要使用该方法,因为该方法可以处理多维数组。即使这样,它也可能比pd.unique使用基于排序的算法而不是哈希表来标识唯一值的方法要慢。

对于较大的DataFrame,速度上的差异非常大(尤其是在只有少数唯一值的情况下):

>>> df1 = pd.concat([df]*100000, ignore_index=True) # DataFrame with 500000 rows
>>> %timeit np.unique(df1[['Col1', 'Col2']].values)
1 loop, best of 3: 1.12 s per loop

>>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel('K'))
10 loops, best of 3: 38.9 ms per loop

>>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel()) # ravel using C order
10 loops, best of 3: 49.9 ms per loop

pd.unique returns the unique values from an input array, or DataFrame column or index.

The input to this function needs to be one-dimensional, so multiple columns will need to be combined. The simplest way is to select the columns you want and then view the values in a flattened NumPy array. The whole operation looks like this:

>>> pd.unique(df[['Col1', 'Col2']].values.ravel('K'))
array(['Bob', 'Joe', 'Bill', 'Mary', 'Steve'], dtype=object)

Note that ravel() is an array method than returns a view (if possible) of a multidimensional array. The argument 'K' tells the method to flatten the array in the order the elements are stored in memory (pandas typically stores underlying arrays in Fortran-contiguous order; columns before rows). This can be significantly faster than using the method’s default ‘C’ order.


An alternative way is to select the columns and pass them to np.unique:

>>> np.unique(df[['Col1', 'Col2']].values)
array(['Bill', 'Bob', 'Joe', 'Mary', 'Steve'], dtype=object)

There is no need to use ravel() here as the method handles multidimensional arrays. Even so, this is likely to be slower than pd.unique as it uses a sort-based algorithm rather than a hashtable to identify unique values.

The difference in speed is significant for larger DataFrames (especially if there are only a handful of unique values):

>>> df1 = pd.concat([df]*100000, ignore_index=True) # DataFrame with 500000 rows
>>> %timeit np.unique(df1[['Col1', 'Col2']].values)
1 loop, best of 3: 1.12 s per loop

>>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel('K'))
10 loops, best of 3: 38.9 ms per loop

>>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel()) # ravel using C order
10 loops, best of 3: 49.9 ms per loop

回答 1

DataFrame在其列中设置了一些简单的字符串:

>>> df
   a  b
0  a  g
1  b  h
2  d  a
3  e  e

您可以连接感兴趣的列并调用unique函数:

>>> pandas.concat([df['a'], df['b']]).unique()
array(['a', 'b', 'd', 'e', 'g', 'h'], dtype=object)

I have setup a DataFrame with a few simple strings in it’s columns:

>>> df
   a  b
0  a  g
1  b  h
2  d  a
3  e  e

You can concatenate the columns you are interested in and call unique function:

>>> pandas.concat([df['a'], df['b']]).unique()
array(['a', 'b', 'd', 'e', 'g', 'h'], dtype=object)

回答 2

In [5]: set(df.Col1).union(set(df.Col2))
Out[5]: {'Bill', 'Bob', 'Joe', 'Mary', 'Steve'}

要么:

set(df.Col1) | set(df.Col2)
In [5]: set(df.Col1).union(set(df.Col2))
Out[5]: {'Bill', 'Bob', 'Joe', 'Mary', 'Steve'}

Or:

set(df.Col1) | set(df.Col2)

回答 3

如果使用多个列,则使用numpy v1.13 +更新的解决方案需要在np.unique中指定轴,否则该数组将隐式展平。

import numpy as np

np.unique(df[['col1', 'col2']], axis=0)

此更改于2016年11月引入:https : //github.com/numpy/numpy/commit/1f764dbff7c496d6636dc0430f083ada9ff4e4be

An updated solution using numpy v1.13+ requires specifying the axis in np.unique if using multiple columns, otherwise the array is implicitly flattened.

import numpy as np

np.unique(df[['col1', 'col2']], axis=0)

This change was introduced Nov 2016: https://github.com/numpy/numpy/commit/1f764dbff7c496d6636dc0430f083ada9ff4e4be


回答 4

pandas解决方案:使用set()。

import pandas as pd
import numpy as np

df = pd.DataFrame({'Col1' : ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
              'Col2' : ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
               'Col3' : np.random.random(5)})

print df

print set(df.Col1.append(df.Col2).values)

输出:

   Col1   Col2      Col3
0   Bob    Joe  0.201079
1   Joe  Steve  0.703279
2  Bill    Bob  0.722724
3  Mary    Bob  0.093912
4   Joe  Steve  0.766027
set(['Steve', 'Bob', 'Bill', 'Joe', 'Mary'])

Non-pandas solution: using set().

import pandas as pd
import numpy as np

df = pd.DataFrame({'Col1' : ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
              'Col2' : ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
               'Col3' : np.random.random(5)})

print df

print set(df.Col1.append(df.Col2).values)

Output:

   Col1   Col2      Col3
0   Bob    Joe  0.201079
1   Joe  Steve  0.703279
2  Bill    Bob  0.722724
3  Mary    Bob  0.093912
4   Joe  Steve  0.766027
set(['Steve', 'Bob', 'Bill', 'Joe', 'Mary'])

回答 5

对于那些喜欢大熊猫的人来说,适用于它们,当然还有lambda函数:

df['Col3'] = df[['Col1', 'Col2']].apply(lambda x: ''.join(x), axis=1)

for those of us that love all things pandas, apply, and of course lambda functions:

df['Col3'] = df[['Col1', 'Col2']].apply(lambda x: ''.join(x), axis=1)

回答 6

这是另一种方式


import numpy as np
set(np.concatenate(df.values))

here’s another way


import numpy as np
set(np.concatenate(df.values))

回答 7

list(set(df[['Col1', 'Col2']].as_matrix().reshape((1,-1)).tolist()[0]))

输出将是[‘Mary’,’Joe’,’Steve’,’Bob’,’Bill’]

list(set(df[['Col1', 'Col2']].as_matrix().reshape((1,-1)).tolist()[0]))

The output will be [‘Mary’, ‘Joe’, ‘Steve’, ‘Bob’, ‘Bill’]