知识问答

重命名熊猫中的特定列

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问题：重命名熊猫中的特定列
回答 0
回答 1
回答 2
如何重命名熊猫中的特定列？
rename 与 axis=1
Index.str.replace
传递一个列表，set_axis与axis=1
How do I rename a specific column in pandas?
rename with axis=1
Index.str.replace
Passing a list to set_axis with axis=1
回答 3

问题：重命名熊猫中的特定列

我有一个名为的数据框data。如何重命名唯一的一列标题？例如gdp到log(gdp)？

data =
    y  gdp  cap
0   1    2    5
1   2    3    9
2   8    7    2
3   3    4    7
4   6    7    7
5   4    8    3
6   8    2    8
7   9    9   10
8   6    6    4
9  10   10    7
点击查看英文原文

I’ve got a dataframe called data. How would I rename the only one column header? For example gdp to log(gdp)?

data =
    y  gdp  cap
0   1    2    5
1   2    3    9
2   8    7    2
3   3    4    7
4   6    7    7
5   4    8    3
6   8    2    8
7   9    9   10
8   6    6    4
9  10   10    7

回答 0

data.rename(columns={'gdp':'log(gdp)'}, inplace=True)

在rename它接受一个字典作为一个PARAM演出columns，所以你只是传递一个字典一次入境。

另请参阅相关

点击查看英文原文 
data.rename(columns={'gdp':'log(gdp)'}, inplace=True)

The rename show that it accepts a dict as a param for columns so you just pass a dict with a single entry.

Also see related

回答 1

list-comprehension如果您需要重命名单个列，则将使用更快的实现。

df.columns = ['log(gdp)' if x=='gdp' else x for x in df.columns]

如果需要重命名多个列，请使用以下条件表达式：

df.columns = ['log(gdp)' if x=='gdp' else 'cap_mod' if x=='cap' else x for x in df.columns]

或者，使用a构造映射dictionary并通过将默认值设置为旧名称来list-comprehension对其执行get操作：

col_dict = {'gdp': 'log(gdp)', 'cap': 'cap_mod'}   ## key→old name, value→new name

df.columns = [col_dict.get(x, x) for x in df.columns]

时间：

%%timeit
df.rename(columns={'gdp':'log(gdp)'}, inplace=True)
10000 loops, best of 3: 168 µs per loop

%%timeit
df.columns = ['log(gdp)' if x=='gdp' else x for x in df.columns]
10000 loops, best of 3: 58.5 µs per loop
点击查看英文原文

A much faster implementation would be to use list-comprehension if you need to rename a single column.

df.columns = ['log(gdp)' if x=='gdp' else x for x in df.columns]

If the need arises to rename multiple columns, either use conditional expressions like:

df.columns = ['log(gdp)' if x=='gdp' else 'cap_mod' if x=='cap' else x for x in df.columns]

Or, construct a mapping using a dictionary and perform the list-comprehension with it’s get operation by setting default value as the old name:

col_dict = {'gdp': 'log(gdp)', 'cap': 'cap_mod'}   ## key→old name, value→new name

df.columns = [col_dict.get(x, x) for x in df.columns]

Timings:

%%timeit
df.rename(columns={'gdp':'log(gdp)'}, inplace=True)
10000 loops, best of 3: 168 µs per loop

%%timeit
df.columns = ['log(gdp)' if x=='gdp' else x for x in df.columns]
10000 loops, best of 3: 58.5 µs per loop

回答 2

如何重命名熊猫中的特定列？

从v0.24 +起，要一次重命名一列（或多列），

如果您需要一次重命名所有列，

  • DataFrame.set_axis()的方法axis=1。传递类似列表的序列。选项也可用于就地修改。

rename 与 axis=1

df = pd.DataFrame('x', columns=['y', 'gdp', 'cap'], index=range(5))
df

   y gdp cap
0  x   x   x
1  x   x   x
2  x   x   x
3  x   x   x
4  x   x   x

使用0.21+，您现在可以使用来指定axis参数rename：

df.rename({'gdp':'log(gdp)'}, axis=1)
# df.rename({'gdp':'log(gdp)'}, axis='columns')
    
   y log(gdp) cap
0  x        x   x
1  x        x   x
2  x        x   x
3  x        x   x
4  x        x   x

（请注意，rename默认情况下它不是就地的，因此您需要将结果分配回去。）

进行此添加是为了提高与其他API的一致性。新axis参数类似于该columns参数，它们执行相同的操作。

df.rename(columns={'gdp': 'log(gdp)'})

   y log(gdp) cap
0  x        x   x
1  x        x   x
2  x        x   x
3  x        x   x
4  x        x   x

rename 还接受为每个列调用一次的回调。

df.rename(lambda x: x[0], axis=1)
# df.rename(lambda x: x[0], axis='columns')

   y  g  c
0  x  x  x
1  x  x  x
2  x  x  x
3  x  x  x
4  x  x  x

对于这种特定情况，您可能要使用

df.rename(lambda x: 'log(gdp)' if x == 'gdp' else x, axis=1)

Index.str.replace

与replacepython中的字符串方法类似，pandas Index和Series（仅对象dtype）定义了一种（“矢量化”）str.replace方法，用于基于字符串和正则表达式的替换。

df.columns = df.columns.str.replace('gdp', 'log(gdp)')
df
 
   y log(gdp) cap
0  x        x   x
1  x        x   x
2  x        x   x
3  x        x   x
4  x        x   x

与其他方法相比，此方法的优点是str.replace支持正则表达式（默认情况下启用）。有关更多信息，请参阅文档。

传递一个列表，set_axis与axis=1

set_axis用标题列表进行调用。该列表的长度必须等于列/索引的大小。set_axis默认情况下会更改原始DataFrame，但您可以指定inplace=False返回修改后的副本。

df.set_axis(['cap', 'log(gdp)', 'y'], axis=1, inplace=False)
# df.set_axis(['cap', 'log(gdp)', 'y'], axis='columns', inplace=False)

  cap log(gdp)  y
0   x        x  x
1   x        x  x
2   x        x  x
3   x        x  x
4   x        x  x

注意：在将来的版本中，inplace默认为True。

方法链接
为什么选择set_axis已经有一种有效的方式分配列的方式df.columns = ...？如Ted Petrou在[此答案]中所示，（https://stackoverflow.com/a/46912050/4909087）set_axis在尝试链接方法时很有用。

比较

# new for pandas 0.21+
df.some_method1()
  .some_method2()
  .set_axis()
  .some_method3()

与

# old way
df1 = df.some_method1()
        .some_method2()
df1.columns = columns
df1.some_method3()

前者是更自然和自由流动的语法。

点击查看英文原文

How do I rename a specific column in pandas?

From v0.24+, to rename one (or more) columns at a time,

If you need to rename ALL columns at once,

  • DataFrame.set_axis() method with axis=1. Pass a list-like sequence. Options are available for in-place modification as well.

rename with axis=1

df = pd.DataFrame('x', columns=['y', 'gdp', 'cap'], index=range(5))
df

   y gdp cap
0  x   x   x
1  x   x   x
2  x   x   x
3  x   x   x
4  x   x   x

With 0.21+, you can now specify an axis parameter with rename:

df.rename({'gdp':'log(gdp)'}, axis=1)
# df.rename({'gdp':'log(gdp)'}, axis='columns')
    
   y log(gdp) cap
0  x        x   x
1  x        x   x
2  x        x   x
3  x        x   x
4  x        x   x

(Note that rename is not in-place by default, so you will need to assign the result back.)

This addition has been made to improve consistency with the rest of the API. The new axis argument is analogous to the columns parameter—they do the same thing.

df.rename(columns={'gdp': 'log(gdp)'})

   y log(gdp) cap
0  x        x   x
1  x        x   x
2  x        x   x
3  x        x   x
4  x        x   x

rename also accepts a callback that is called once for each column.

df.rename(lambda x: x[0], axis=1)
# df.rename(lambda x: x[0], axis='columns')

   y  g  c
0  x  x  x
1  x  x  x
2  x  x  x
3  x  x  x
4  x  x  x

For this specific scenario, you would want to use

df.rename(lambda x: 'log(gdp)' if x == 'gdp' else x, axis=1)

Index.str.replace

Similar to replace method of strings in python, pandas Index and Series (object dtype only) define a (“vectorized”) str.replace method for string and regex-based replacement.

df.columns = df.columns.str.replace('gdp', 'log(gdp)')
df
 
   y log(gdp) cap
0  x        x   x
1  x        x   x
2  x        x   x
3  x        x   x
4  x        x   x

The advantage of this over the other methods is that str.replace supports regex (enabled by default). See the docs for more information.

Passing a list to set_axis with axis=1

Call set_axis with a list of header(s). The list must be equal in length to the columns/index size. set_axis mutates the original DataFrame by default, but you can specify inplace=False to return a modified copy.

df.set_axis(['cap', 'log(gdp)', 'y'], axis=1, inplace=False)
# df.set_axis(['cap', 'log(gdp)', 'y'], axis='columns', inplace=False)

  cap log(gdp)  y
0   x        x  x
1   x        x  x
2   x        x  x
3   x        x  x
4   x        x  x

Note: In future releases, inplace will default to True.

Method Chaining
Why choose set_axis when we already have an efficient way of assigning columns with df.columns = ...? As shown by Ted Petrou in [this answer],(https://stackoverflow.com/a/46912050/4909087) set_axis is useful when trying to chain methods.

Compare

# new for pandas 0.21+
df.some_method1()
  .some_method2()
  .set_axis()
  .some_method3()

Versus

# old way
df1 = df.some_method1()
        .some_method2()
df1.columns = columns
df1.some_method3()

The former is more natural and free flowing syntax.

回答 3

至少有五种不同的方法来重命名熊猫中的特定列，我在下面列出了它们以及原始答案的链接。我还对这些方法进行了计时，发现它们执行的效果大致相同（尽管YMMV取决于您的数据集和方案）。下面的试验情况下是列重命名A M N Z以A2 M2 N2 Z2在一个数据帧的列A到Z含有一百万行。

# Import required modules
import numpy as np
import pandas as pd
import timeit

# Create sample data
df = pd.DataFrame(np.random.randint(0,9999,size=(1000000, 26)), columns=list('ABCDEFGHIJKLMNOPQRSTUVWXYZ'))

# Standard way - https://stackoverflow.com/a/19758398/452587
def method_1():
    df_renamed = df.rename(columns={'A': 'A2', 'M': 'M2', 'N': 'N2', 'Z': 'Z2'})

# Lambda function - https://stackoverflow.com/a/16770353/452587
def method_2():
    df_renamed = df.rename(columns=lambda x: x + '2' if x in ['A', 'M', 'N', 'Z'] else x)

# Mapping function - https://stackoverflow.com/a/19758398/452587
def rename_some(x):
    if x=='A' or x=='M' or x=='N' or x=='Z':
        return x + '2'
    return x
def method_3():
    df_renamed = df.rename(columns=rename_some)

# Dictionary comprehension - https://stackoverflow.com/a/58143182/452587
def method_4():
    df_renamed = df.rename(columns={col: col + '2' for col in df.columns[
        np.asarray([i for i, col in enumerate(df.columns) if 'A' in col or 'M' in col or 'N' in col or 'Z' in col])
    ]})

# Dictionary comprehension - https://stackoverflow.com/a/38101084/452587
def method_5():
    df_renamed = df.rename(columns=dict(zip(df[['A', 'M', 'N', 'Z']], ['A2', 'M2', 'N2', 'Z2'])))

print('Method 1:', timeit.timeit(method_1, number=10))
print('Method 2:', timeit.timeit(method_2, number=10))
print('Method 3:', timeit.timeit(method_3, number=10))
print('Method 4:', timeit.timeit(method_4, number=10))
print('Method 5:', timeit.timeit(method_5, number=10))

输出：

Method 1: 3.650640267
Method 2: 3.163998427
Method 3: 2.998530871
Method 4: 2.9918436889999995
Method 5: 3.2436501520000007

使用对您来说最直观，最容易在应用程序中实现的方法。

点击查看英文原文

There are at least five different ways to rename specific columns in pandas, and I have listed them below along with links to the original answers. I also timed these methods and found them to perform about the same (though YMMV depending on your data set and scenario). The test case below is to rename columns A M N Z to A2 M2 N2 Z2 in a dataframe with columns A to Z containing a million rows.

# Import required modules
import numpy as np
import pandas as pd
import timeit

# Create sample data
df = pd.DataFrame(np.random.randint(0,9999,size=(1000000, 26)), columns=list('ABCDEFGHIJKLMNOPQRSTUVWXYZ'))

# Standard way - https://stackoverflow.com/a/19758398/452587
def method_1():
    df_renamed = df.rename(columns={'A': 'A2', 'M': 'M2', 'N': 'N2', 'Z': 'Z2'})

# Lambda function - https://stackoverflow.com/a/16770353/452587
def method_2():
    df_renamed = df.rename(columns=lambda x: x + '2' if x in ['A', 'M', 'N', 'Z'] else x)

# Mapping function - https://stackoverflow.com/a/19758398/452587
def rename_some(x):
    if x=='A' or x=='M' or x=='N' or x=='Z':
        return x + '2'
    return x
def method_3():
    df_renamed = df.rename(columns=rename_some)

# Dictionary comprehension - https://stackoverflow.com/a/58143182/452587
def method_4():
    df_renamed = df.rename(columns={col: col + '2' for col in df.columns[
        np.asarray([i for i, col in enumerate(df.columns) if 'A' in col or 'M' in col or 'N' in col or 'Z' in col])
    ]})

# Dictionary comprehension - https://stackoverflow.com/a/38101084/452587
def method_5():
    df_renamed = df.rename(columns=dict(zip(df[['A', 'M', 'N', 'Z']], ['A2', 'M2', 'N2', 'Z2'])))

print('Method 1:', timeit.timeit(method_1, number=10))
print('Method 2:', timeit.timeit(method_2, number=10))
print('Method 3:', timeit.timeit(method_3, number=10))
print('Method 4:', timeit.timeit(method_4, number=10))
print('Method 5:', timeit.timeit(method_5, number=10))

Output:

Method 1: 3.650640267
Method 2: 3.163998427
Method 3: 2.998530871
Method 4: 2.9918436889999995
Method 5: 3.2436501520000007

Use the method that is most intuitive to you and easiest for you to implement in your application.