问题:在Python Pandas中向现有DataFrame添加新列
我有以下索引的DataFrame,其中的命名列和行不是连续数字:
a b c d
2 0.671399 0.101208 -0.181532 0.241273
3 0.446172 -0.243316 0.051767 1.577318
5 0.614758 0.075793 -0.451460 -0.012493
我想'e'
在现有数据框架中添加一个新列,并且不想更改数据框架中的任何内容(即,新列始终与DataFrame具有相同的长度)。
0 -0.335485
1 -1.166658
2 -0.385571
dtype: float64
如何e
在上述示例中添加列?
I have the following indexed DataFrame with named columns and rows not- continuous numbers:
a b c d
2 0.671399 0.101208 -0.181532 0.241273
3 0.446172 -0.243316 0.051767 1.577318
5 0.614758 0.075793 -0.451460 -0.012493
I would like to add a new column, 'e'
, to the existing data frame and do not want to change anything in the data frame (i.e., the new column always has the same length as the DataFrame).
0 -0.335485
1 -1.166658
2 -0.385571
dtype: float64
How can I add column e
to the above example?
回答 0
使用原始的df1索引创建系列:
df1['e'] = pd.Series(np.random.randn(sLength), index=df1.index)
编辑2015年
有人报告SettingWithCopyWarning
使用此代码。
但是,该代码仍可以在当前的熊猫0.10.1版本中完美运行。
>>> sLength = len(df1['a'])
>>> df1
a b c d
6 -0.269221 -0.026476 0.997517 1.294385
8 0.917438 0.847941 0.034235 -0.448948
>>> df1['e'] = pd.Series(np.random.randn(sLength), index=df1.index)
>>> df1
a b c d e
6 -0.269221 -0.026476 0.997517 1.294385 1.757167
8 0.917438 0.847941 0.034235 -0.448948 2.228131
>>> p.version.short_version
'0.16.1'
该SettingWithCopyWarning
目标对数据帧的副本通知可能无效转让的。它不一定表示您做错了(它可能会触发误报),但从0.13.0起,它会让您知道有更多适合同一目的的方法。然后,如果收到警告,请遵循其建议:尝试使用.loc [row_index,col_indexer] = value代替
>>> df1.loc[:,'f'] = pd.Series(np.random.randn(sLength), index=df1.index)
>>> df1
a b c d e f
6 -0.269221 -0.026476 0.997517 1.294385 1.757167 -0.050927
8 0.917438 0.847941 0.034235 -0.448948 2.228131 0.006109
>>>
实际上,这是目前熊猫文档中描述的更有效的方法
编辑2017
如评论和@Alexander所示,当前最好将Series的值添加为DataFrame的新列的最佳方法是使用assign
:
df1 = df1.assign(e=pd.Series(np.random.randn(sLength)).values)
Use the original df1 indexes to create the series:
df1['e'] = pd.Series(np.random.randn(sLength), index=df1.index)
Edit 2015
Some reported getting the SettingWithCopyWarning
with this code.
However, the code still runs perfectly with the current pandas version 0.16.1.
>>> sLength = len(df1['a'])
>>> df1
a b c d
6 -0.269221 -0.026476 0.997517 1.294385
8 0.917438 0.847941 0.034235 -0.448948
>>> df1['e'] = pd.Series(np.random.randn(sLength), index=df1.index)
>>> df1
a b c d e
6 -0.269221 -0.026476 0.997517 1.294385 1.757167
8 0.917438 0.847941 0.034235 -0.448948 2.228131
>>> p.version.short_version
'0.16.1'
The SettingWithCopyWarning
aims to inform of a possibly invalid assignment on a copy of the Dataframe. It doesn’t necessarily say you did it wrong (it can trigger false positives) but from 0.13.0 it let you know there are more adequate methods for the same purpose. Then, if you get the warning, just follow its advise: Try using .loc[row_index,col_indexer] = value instead
>>> df1.loc[:,'f'] = pd.Series(np.random.randn(sLength), index=df1.index)
>>> df1
a b c d e f
6 -0.269221 -0.026476 0.997517 1.294385 1.757167 -0.050927
8 0.917438 0.847941 0.034235 -0.448948 2.228131 0.006109
>>>
In fact, this is currently the more efficient method as described in pandas docs
Edit 2017
As indicated in the comments and by @Alexander, currently the best method to add the values of a Series as a new column of a DataFrame could be using assign
:
df1 = df1.assign(e=pd.Series(np.random.randn(sLength)).values)
回答 1
This is the simple way of adding a new column: df['e'] = e
回答 2
我想在现有数据框中添加新列’e’,并且不更改数据框中的任何内容。(该系列的长度总是与数据帧相同。)
我假设中的索引值e
与中的索引值匹配df1
。
初始化名为的新列e
并为其分配系列中的值的最简单方法e
:
df['e'] = e.values
分配(熊猫0.16.0+)
从Pandas 0.16.0开始,您还可以使用assign
,它为DataFrame分配新列,并返回一个新对象(副本),该对象除包含新列外还包含所有原始列。
df1 = df1.assign(e=e.values)
按照此示例(还包括assign
函数的源代码),您还可以包括多个列:
df = pd.DataFrame({'a': [1, 2], 'b': [3, 4]})
>>> df.assign(mean_a=df.a.mean(), mean_b=df.b.mean())
a b mean_a mean_b
0 1 3 1.5 3.5
1 2 4 1.5 3.5
在您的示例中:
np.random.seed(0)
df1 = pd.DataFrame(np.random.randn(10, 4), columns=['a', 'b', 'c', 'd'])
mask = df1.applymap(lambda x: x <-0.7)
df1 = df1[-mask.any(axis=1)]
sLength = len(df1['a'])
e = pd.Series(np.random.randn(sLength))
>>> df1
a b c d
0 1.764052 0.400157 0.978738 2.240893
2 -0.103219 0.410599 0.144044 1.454274
3 0.761038 0.121675 0.443863 0.333674
7 1.532779 1.469359 0.154947 0.378163
9 1.230291 1.202380 -0.387327 -0.302303
>>> e
0 -1.048553
1 -1.420018
2 -1.706270
3 1.950775
4 -0.509652
dtype: float64
df1 = df1.assign(e=e.values)
>>> df1
a b c d e
0 1.764052 0.400157 0.978738 2.240893 -1.048553
2 -0.103219 0.410599 0.144044 1.454274 -1.420018
3 0.761038 0.121675 0.443863 0.333674 -1.706270
7 1.532779 1.469359 0.154947 0.378163 1.950775
9 1.230291 1.202380 -0.387327 -0.302303 -0.509652
首次引入此新功能时,可以在此处找到说明。
I would like to add a new column, ‘e’, to the existing data frame and do not change anything in the data frame. (The series always got the same length as a dataframe.)
I assume that the index values in e
match those in df1
.
The easiest way to initiate a new column named e
, and assign it the values from your series e
:
df['e'] = e.values
assign (Pandas 0.16.0+)
As of Pandas 0.16.0, you can also use assign
, which assigns new columns to a DataFrame and returns a new object (a copy) with all the original columns in addition to the new ones.
df1 = df1.assign(e=e.values)
As per this example (which also includes the source code of the assign
function), you can also include more than one column:
df = pd.DataFrame({'a': [1, 2], 'b': [3, 4]})
>>> df.assign(mean_a=df.a.mean(), mean_b=df.b.mean())
a b mean_a mean_b
0 1 3 1.5 3.5
1 2 4 1.5 3.5
In context with your example:
np.random.seed(0)
df1 = pd.DataFrame(np.random.randn(10, 4), columns=['a', 'b', 'c', 'd'])
mask = df1.applymap(lambda x: x <-0.7)
df1 = df1[-mask.any(axis=1)]
sLength = len(df1['a'])
e = pd.Series(np.random.randn(sLength))
>>> df1
a b c d
0 1.764052 0.400157 0.978738 2.240893
2 -0.103219 0.410599 0.144044 1.454274
3 0.761038 0.121675 0.443863 0.333674
7 1.532779 1.469359 0.154947 0.378163
9 1.230291 1.202380 -0.387327 -0.302303
>>> e
0 -1.048553
1 -1.420018
2 -1.706270
3 1.950775
4 -0.509652
dtype: float64
df1 = df1.assign(e=e.values)
>>> df1
a b c d e
0 1.764052 0.400157 0.978738 2.240893 -1.048553
2 -0.103219 0.410599 0.144044 1.454274 -1.420018
3 0.761038 0.121675 0.443863 0.333674 -1.706270
7 1.532779 1.469359 0.154947 0.378163 1.950775
9 1.230291 1.202380 -0.387327 -0.302303 -0.509652
The description of this new feature when it was first introduced can be found here.
回答 3
似乎在最新的Pandas版本中,可行的方法是使用df.assign:
df1 = df1.assign(e=np.random.randn(sLength))
它不会产生SettingWithCopyWarning
。
It seems that in recent Pandas versions the way to go is to use df.assign:
df1 = df1.assign(e=np.random.randn(sLength))
It doesn’t produce SettingWithCopyWarning
.
回答 4
通过NumPy直接执行此操作将是最有效的:
df1['e'] = np.random.randn(sLength)
请注意,我最初的建议(很旧)是使用map
(慢得多):
df1['e'] = df1['a'].map(lambda x: np.random.random())
Doing this directly via NumPy will be the most efficient:
df1['e'] = np.random.randn(sLength)
Note my original (very old) suggestion was to use map
(which is much slower):
df1['e'] = df1['a'].map(lambda x: np.random.random())
回答 5
超简单的列分配
将熊猫数据框实现为列的有序字典。
这意味着__getitem__
[]
不仅可以用于获取特定列,__setitem__
[] =
还可以用于分配新列。
例如,只需使用[]
访问器,就可以向该数据框添加一列
size name color
0 big rose red
1 small violet blue
2 small tulip red
3 small harebell blue
df['protected'] = ['no', 'no', 'no', 'yes']
size name color protected
0 big rose red no
1 small violet blue no
2 small tulip red no
3 small harebell blue yes
请注意,即使数据框的索引已关闭,此操作也有效。
df.index = [3,2,1,0]
df['protected'] = ['no', 'no', 'no', 'yes']
size name color protected
3 big rose red no
2 small violet blue no
1 small tulip red no
0 small harebell blue yes
[] =是要走的路,但要当心!
但是,如果您有一个pd.Series
并尝试将其分配给索引关闭的数据帧,则会遇到麻烦。参见示例:
df['protected'] = pd.Series(['no', 'no', 'no', 'yes'])
size name color protected
3 big rose red yes
2 small violet blue no
1 small tulip red no
0 small harebell blue no
这是因为pd.Series
默认情况下,a的索引从0枚举到n。而熊猫[] =
方法试图 变得“聪明”
实际发生了什么。
使用[] =
方法时,pandas使用左手数据帧的索引和右手序列的索引安静地执行外部联接或外部合并。df['column'] = series
边注
这很快就会引起认知失调,因为该[]=
方法试图根据输入来做很多不同的事情,除非您只知道熊猫是如何工作的,否则无法预测结果。因此,我建议不要使用[]=
in代码库,但是在笔记本中浏览数据时可以使用。
解决问题
如果您有一个pd.Series
并且希望从上到下分配它,或者您正在编码生产性代码并且不确定索引顺序,那么为此类问题提供保护是值得的。
您可以将转换pd.Series
为a np.ndarray
或a list
,这可以解决问题。
df['protected'] = pd.Series(['no', 'no', 'no', 'yes']).values
要么
df['protected'] = list(pd.Series(['no', 'no', 'no', 'yes']))
但这不是很明确。
某些编码器可能会说:“嘿,这看起来很多余,我将对其进行优化”。
显式方式
设置的索引pd.Series
是的索引df
是明确的。
df['protected'] = pd.Series(['no', 'no', 'no', 'yes'], index=df.index)
或更现实的说,您可能pd.Series
已经有空了。
protected_series = pd.Series(['no', 'no', 'no', 'yes'])
protected_series.index = df.index
3 no
2 no
1 no
0 yes
现在可以分配
df['protected'] = protected_series
size name color protected
3 big rose red no
2 small violet blue no
1 small tulip red no
0 small harebell blue yes
另一种方式 df.reset_index()
由于索引不一致是问题所在,因此,如果您认为数据框的索引不应该指示事物,则可以简单地删除索引,这应该更快,但是它不是很干净,因为您的函数现在可能做两件事。
df.reset_index(drop=True)
protected_series.reset_index(drop=True)
df['protected'] = protected_series
size name color protected
0 big rose red no
1 small violet blue no
2 small tulip red no
3 small harebell blue yes
注意 df.assign
尽管df.assign
让您更清楚地知道自己在做什么,但实际上却存在与上述相同的所有问题[]=
df.assign(protected=pd.Series(['no', 'no', 'no', 'yes']))
size name color protected
3 big rose red yes
2 small violet blue no
1 small tulip red no
0 small harebell blue no
请注意df.assign
,您的专栏没有被调用self
。会导致错误。这很df.assign
臭,因为函数中存在这些伪像。
df.assign(self=pd.Series(['no', 'no', 'no', 'yes'])
TypeError: assign() got multiple values for keyword argument 'self'
您可能会说,“好吧,那我就不使用了self
”。但是谁知道这个函数将来会如何变化以支持新的论点。也许您的列名将成为熊猫新更新中的一个参数,从而导致升级问题。
Super simple column assignment
A pandas dataframe is implemented as an ordered dict of columns.
This means that the __getitem__
[]
can not only be used to get a certain column, but __setitem__
[] =
can be used to assign a new column.
For example, this dataframe can have a column added to it by simply using the []
accessor
size name color
0 big rose red
1 small violet blue
2 small tulip red
3 small harebell blue
df['protected'] = ['no', 'no', 'no', 'yes']
size name color protected
0 big rose red no
1 small violet blue no
2 small tulip red no
3 small harebell blue yes
Note that this works even if the index of the dataframe is off.
df.index = [3,2,1,0]
df['protected'] = ['no', 'no', 'no', 'yes']
size name color protected
3 big rose red no
2 small violet blue no
1 small tulip red no
0 small harebell blue yes
[]= is the way to go, but watch out!
However, if you have a pd.Series
and try to assign it to a dataframe where the indexes are off, you will run in to trouble. See example:
df['protected'] = pd.Series(['no', 'no', 'no', 'yes'])
size name color protected
3 big rose red yes
2 small violet blue no
1 small tulip red no
0 small harebell blue no
This is because a pd.Series
by default has an index enumerated from 0 to n. And the pandas [] =
method tries to be “smart”
What actually is going on.
When you use the [] =
method pandas is quietly performing an outer join or outer merge using the index of the left hand dataframe and the index of the right hand series. df['column'] = series
Side note
This quickly causes cognitive dissonance, since the []=
method is trying to do a lot of different things depending on the input, and the outcome cannot be predicted unless you just know how pandas works. I would therefore advice against the []=
in code bases, but when exploring data in a notebook, it is fine.
Going around the problem
If you have a pd.Series
and want it assigned from top to bottom, or if you are coding productive code and you are not sure of the index order, it is worth it to safeguard for this kind of issue.
You could downcast the pd.Series
to a np.ndarray
or a list
, this will do the trick.
df['protected'] = pd.Series(['no', 'no', 'no', 'yes']).values
or
df['protected'] = list(pd.Series(['no', 'no', 'no', 'yes']))
But this is not very explicit.
Some coder may come along and say “Hey, this looks redundant, I’ll just optimize this away”.
Explicit way
Setting the index of the pd.Series
to be the index of the df
is explicit.
df['protected'] = pd.Series(['no', 'no', 'no', 'yes'], index=df.index)
Or more realistically, you probably have a pd.Series
already available.
protected_series = pd.Series(['no', 'no', 'no', 'yes'])
protected_series.index = df.index
3 no
2 no
1 no
0 yes
Can now be assigned
df['protected'] = protected_series
size name color protected
3 big rose red no
2 small violet blue no
1 small tulip red no
0 small harebell blue yes
Alternative way with df.reset_index()
Since the index dissonance is the problem, if you feel that the index of the dataframe should not dictate things, you can simply drop the index, this should be faster, but it is not very clean, since your function now probably does two things.
df.reset_index(drop=True)
protected_series.reset_index(drop=True)
df['protected'] = protected_series
size name color protected
0 big rose red no
1 small violet blue no
2 small tulip red no
3 small harebell blue yes
Note on df.assign
While df.assign
make it more explicit what you are doing, it actually has all the same problems as the above []=
df.assign(protected=pd.Series(['no', 'no', 'no', 'yes']))
size name color protected
3 big rose red yes
2 small violet blue no
1 small tulip red no
0 small harebell blue no
Just watch out with df.assign
that your column is not called self
. It will cause errors. This makes df.assign
smelly, since there are these kind of artifacts in the function.
df.assign(self=pd.Series(['no', 'no', 'no', 'yes'])
TypeError: assign() got multiple values for keyword argument 'self'
You may say, “Well, I’ll just not use self
then”. But who knows how this function changes in the future to support new arguments. Maybe your column name will be an argument in a new update of pandas, causing problems with upgrading.
回答 6
最简单的方法:
data['new_col'] = list_of_values
data.loc[ : , 'new_col'] = list_of_values
这样,您可以在熊猫对象中设置新值时避免所谓的链接索引。单击此处以进一步阅读。
Easiest ways:-
data['new_col'] = list_of_values
data.loc[ : , 'new_col'] = list_of_values
This way you avoid what is called chained indexing when setting new values in a pandas object. Click here to read further.
回答 7
如果您要将整个新列设置为初始基值(例如None
),则可以执行以下操作:df1['e'] = None
实际上,这将为单元分配“对象”类型。因此,稍后您可以将复杂的数据类型(如列表)放到单个单元格中。
If you want to set the whole new column to an initial base value (e.g. None
), you can do this: df1['e'] = None
This actually would assign “object” type to the cell. So later you’re free to put complex data types, like list, into individual cells.
回答 8
I got the dreaded SettingWithCopyWarning
, and it wasn’t fixed by using the iloc syntax. My DataFrame was created by read_sql from an ODBC source. Using a suggestion by lowtech above, the following worked for me:
df.insert(len(df.columns), 'e', pd.Series(np.random.randn(sLength), index=df.index))
This worked fine to insert the column at the end. I don’t know if it is the most efficient, but I don’t like warning messages. I think there is a better solution, but I can’t find it, and I think it depends on some aspect of the index.
Note. That this only works once and will give an error message if trying to overwrite and existing column.
Note As above and from 0.16.0 assign is the best solution. See documentation http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.assign.html#pandas.DataFrame.assign
Works well for data flow type where you don’t overwrite your intermediate values.
回答 9
- 首先创建一个
list_of_e
具有相关数据的python 。
- 用这个:
df['e'] = list_of_e
- First create a python’s
list_of_e
that has relevant data.
- Use this:
df['e'] = list_of_e
回答 10
如果您要添加的列是一个系列变量,则只需:
df["new_columns_name"]=series_variable_name #this will do it for you
即使您要替换现有的列,此方法也能很好地工作,只需键入与要替换的列相同的new_columns_name,它将用新的系列数据覆盖现有的列数据。
If the column you are trying to add is a series variable then just :
df["new_columns_name"]=series_variable_name #this will do it for you
This works well even if you are replacing an existing column.just type the new_columns_name same as the column you want to replace.It will just overwrite the existing column data with the new series data.
回答 11
如果数据框和Series对象具有相同的index,则pandas.concat
也可以在这里工作:
import pandas as pd
df
# a b c d
#0 0.671399 0.101208 -0.181532 0.241273
#1 0.446172 -0.243316 0.051767 1.577318
#2 0.614758 0.075793 -0.451460 -0.012493
e = pd.Series([-0.335485, -1.166658, -0.385571])
e
#0 -0.335485
#1 -1.166658
#2 -0.385571
#dtype: float64
# here we need to give the series object a name which converts to the new column name
# in the result
df = pd.concat([df, e.rename("e")], axis=1)
df
# a b c d e
#0 0.671399 0.101208 -0.181532 0.241273 -0.335485
#1 0.446172 -0.243316 0.051767 1.577318 -1.166658
#2 0.614758 0.075793 -0.451460 -0.012493 -0.385571
如果它们没有相同的索引:
e.index = df.index
df = pd.concat([df, e.rename("e")], axis=1)
If the data frame and Series object have the same index, pandas.concat
also works here:
import pandas as pd
df
# a b c d
#0 0.671399 0.101208 -0.181532 0.241273
#1 0.446172 -0.243316 0.051767 1.577318
#2 0.614758 0.075793 -0.451460 -0.012493
e = pd.Series([-0.335485, -1.166658, -0.385571])
e
#0 -0.335485
#1 -1.166658
#2 -0.385571
#dtype: float64
# here we need to give the series object a name which converts to the new column name
# in the result
df = pd.concat([df, e.rename("e")], axis=1)
df
# a b c d e
#0 0.671399 0.101208 -0.181532 0.241273 -0.335485
#1 0.446172 -0.243316 0.051767 1.577318 -1.166658
#2 0.614758 0.075793 -0.451460 -0.012493 -0.385571
In case they don’t have the same index:
e.index = df.index
df = pd.concat([df, e.rename("e")], axis=1)
回答 12
万无一失:
df.loc[:, 'NewCol'] = 'New_Val'
例:
df = pd.DataFrame(data=np.random.randn(20, 4), columns=['A', 'B', 'C', 'D'])
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
3 -0.147354 0.778707 0.479145 2.284143
4 -0.529529 0.000571 0.913779 1.395894
5 2.592400 0.637253 1.441096 -0.631468
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
8 0.606985 -2.232903 -1.358107 -2.855494
9 -0.692013 0.671866 1.179466 -1.180351
10 -1.093707 -0.530600 0.182926 -1.296494
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
18 0.693458 0.144327 0.329500 -0.655045
19 0.104425 0.037412 0.450598 -0.923387
df.drop([3, 5, 8, 10, 18], inplace=True)
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
4 -0.529529 0.000571 0.913779 1.395894
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
9 -0.692013 0.671866 1.179466 -1.180351
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
19 0.104425 0.037412 0.450598 -0.923387
df.loc[:, 'NewCol'] = 0
df
A B C D NewCol
0 -0.761269 0.477348 1.170614 0.752714 0
1 1.217250 -0.930860 -0.769324 -0.408642 0
2 -0.619679 -1.227659 -0.259135 1.700294 0
4 -0.529529 0.000571 0.913779 1.395894 0
6 0.757178 0.240012 -0.553820 1.177202 0
7 -0.986128 -1.313843 0.788589 -0.707836 0
9 -0.692013 0.671866 1.179466 -1.180351 0
11 -0.143273 -0.503199 -1.328728 0.610552 0
12 -0.923110 -1.365890 -1.366202 -1.185999 0
13 -2.026832 0.273593 -0.440426 -0.627423 0
14 -0.054503 -0.788866 -0.228088 -0.404783 0
15 0.955298 -1.430019 1.434071 -0.088215 0
16 -0.227946 0.047462 0.373573 -0.111675 0
17 1.627912 0.043611 1.743403 -0.012714 0
19 0.104425 0.037412 0.450598 -0.923387 0
Foolproof:
df.loc[:, 'NewCol'] = 'New_Val'
Example:
df = pd.DataFrame(data=np.random.randn(20, 4), columns=['A', 'B', 'C', 'D'])
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
3 -0.147354 0.778707 0.479145 2.284143
4 -0.529529 0.000571 0.913779 1.395894
5 2.592400 0.637253 1.441096 -0.631468
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
8 0.606985 -2.232903 -1.358107 -2.855494
9 -0.692013 0.671866 1.179466 -1.180351
10 -1.093707 -0.530600 0.182926 -1.296494
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
18 0.693458 0.144327 0.329500 -0.655045
19 0.104425 0.037412 0.450598 -0.923387
df.drop([3, 5, 8, 10, 18], inplace=True)
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
4 -0.529529 0.000571 0.913779 1.395894
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
9 -0.692013 0.671866 1.179466 -1.180351
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
19 0.104425 0.037412 0.450598 -0.923387
df.loc[:, 'NewCol'] = 0
df
A B C D NewCol
0 -0.761269 0.477348 1.170614 0.752714 0
1 1.217250 -0.930860 -0.769324 -0.408642 0
2 -0.619679 -1.227659 -0.259135 1.700294 0
4 -0.529529 0.000571 0.913779 1.395894 0
6 0.757178 0.240012 -0.553820 1.177202 0
7 -0.986128 -1.313843 0.788589 -0.707836 0
9 -0.692013 0.671866 1.179466 -1.180351 0
11 -0.143273 -0.503199 -1.328728 0.610552 0
12 -0.923110 -1.365890 -1.366202 -1.185999 0
13 -2.026832 0.273593 -0.440426 -0.627423 0
14 -0.054503 -0.788866 -0.228088 -0.404783 0
15 0.955298 -1.430019 1.434071 -0.088215 0
16 -0.227946 0.047462 0.373573 -0.111675 0
17 1.627912 0.043611 1.743403 -0.012714 0
19 0.104425 0.037412 0.450598 -0.923387 0
回答 13
让我补充一点,就像hum3一样,.loc
没有解决SettingWithCopyWarning
,我不得不求助于df.insert()
。在我的情况下,“假”链索引产生了误报 dict['a']['e']
,其中'e'
是新列,并且dict['a']
是来自字典的DataFrame。
另请注意,如果您知道自己在做什么,则可以使用pd.options.mode.chained_assignment = None
,而可以使用此处提供的其他解决方案之一来切换警告
。
Let me just add that, just like for hum3, .loc
didn’t solve the SettingWithCopyWarning
and I had to resort to df.insert()
. In my case false positive was generated by “fake” chain indexing dict['a']['e']
, where 'e'
is the new column, and dict['a']
is a DataFrame coming from dictionary.
Also note that if you know what you are doing, you can switch of the warning using
pd.options.mode.chained_assignment = None
and than use one of the other solutions given here.
回答 14
to insert a new column at a given location (0 <= loc <= amount of columns) in a data frame, just use Dataframe.insert:
DataFrame.insert(loc, column, value)
Therefore, if you want to add the column e at the end of a data frame called df, you can use:
e = [-0.335485, -1.166658, -0.385571]
DataFrame.insert(loc=len(df.columns), column='e', value=e)
value can be a Series, an integer (in which case all cells get filled with this one value), or an array-like structure
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.insert.html
回答 15
在分配新列之前,如果您已建立索引数据,则需要对索引进行排序。至少就我而言,我必须:
data.set_index(['index_column'], inplace=True)
"if index is unsorted, assignment of a new column will fail"
data.sort_index(inplace = True)
data.loc['index_value1', 'column_y'] = np.random.randn(data.loc['index_value1', 'column_x'].shape[0])
Before assigning a new column, if you have indexed data, you need to sort the index. At least in my case I had to:
data.set_index(['index_column'], inplace=True)
"if index is unsorted, assignment of a new column will fail"
data.sort_index(inplace = True)
data.loc['index_value1', 'column_y'] = np.random.randn(data.loc['index_value1', 'column_x'].shape[0])
回答 16
但是要注意的一件事是,如果您这样做
df1['e'] = Series(np.random.randn(sLength), index=df1.index)
这实际上是df1.index上的左连接。因此,如果要产生外部联接效果,我可能不完善的解决方案是创建一个具有索引值的数据框,该索引值覆盖数据的整个范围,然后使用上面的代码。例如,
data = pd.DataFrame(index=all_possible_values)
df1['e'] = Series(np.random.randn(sLength), index=df1.index)
One thing to note, though, is that if you do
df1['e'] = Series(np.random.randn(sLength), index=df1.index)
this will effectively be a left join on the df1.index. So if you want to have an outer join effect, my probably imperfect solution is to create a dataframe with index values covering the universe of your data, and then use the code above. For example,
data = pd.DataFrame(index=all_possible_values)
df1['e'] = Series(np.random.randn(sLength), index=df1.index)
回答 17
我一直在寻找一种通用方法,将numpy.nan
s 的列添加到数据框而不会变得愚蠢SettingWithCopyWarning
。
从以下内容:
- 答案在这里
- 关于将变量作为关键字参数传递的问题
- 这种在线生成
numpy
NaN数组的方法
我想出了这个:
col = 'column_name'
df = df.assign(**{col:numpy.full(len(df), numpy.nan)})
I was looking for a general way of adding a column of numpy.nan
s to a dataframe without getting the dumb SettingWithCopyWarning
.
From the following:
- the answers here
- this question about passing a variable as a keyword argument
- this method for generating a
numpy
array of NaNs in-line
I came up with this:
col = 'column_name'
df = df.assign(**{col:numpy.full(len(df), numpy.nan)})
回答 18
要将新列“ e”添加到现有数据框中
df1.loc[:,'e'] = Series(np.random.randn(sLength))
To add a new column, ‘e’, to the existing data frame
df1.loc[:,'e'] = Series(np.random.randn(sLength))
回答 19
为了完整性-使用DataFrame.eval()方法的另一种解决方案:
数据:
In [44]: e
Out[44]:
0 1.225506
1 -1.033944
2 -0.498953
3 -0.373332
4 0.615030
5 -0.622436
dtype: float64
In [45]: df1
Out[45]:
a b c d
0 -0.634222 -0.103264 0.745069 0.801288
4 0.782387 -0.090279 0.757662 -0.602408
5 -0.117456 2.124496 1.057301 0.765466
7 0.767532 0.104304 -0.586850 1.051297
8 -0.103272 0.958334 1.163092 1.182315
9 -0.616254 0.296678 -0.112027 0.679112
解:
In [46]: df1.eval("e = @e.values", inplace=True)
In [47]: df1
Out[47]:
a b c d e
0 -0.634222 -0.103264 0.745069 0.801288 1.225506
4 0.782387 -0.090279 0.757662 -0.602408 -1.033944
5 -0.117456 2.124496 1.057301 0.765466 -0.498953
7 0.767532 0.104304 -0.586850 1.051297 -0.373332
8 -0.103272 0.958334 1.163092 1.182315 0.615030
9 -0.616254 0.296678 -0.112027 0.679112 -0.622436
For the sake of completeness – yet another solution using DataFrame.eval() method:
Data:
In [44]: e
Out[44]:
0 1.225506
1 -1.033944
2 -0.498953
3 -0.373332
4 0.615030
5 -0.622436
dtype: float64
In [45]: df1
Out[45]:
a b c d
0 -0.634222 -0.103264 0.745069 0.801288
4 0.782387 -0.090279 0.757662 -0.602408
5 -0.117456 2.124496 1.057301 0.765466
7 0.767532 0.104304 -0.586850 1.051297
8 -0.103272 0.958334 1.163092 1.182315
9 -0.616254 0.296678 -0.112027 0.679112
Solution:
In [46]: df1.eval("e = @e.values", inplace=True)
In [47]: df1
Out[47]:
a b c d e
0 -0.634222 -0.103264 0.745069 0.801288 1.225506
4 0.782387 -0.090279 0.757662 -0.602408 -1.033944
5 -0.117456 2.124496 1.057301 0.765466 -0.498953
7 0.767532 0.104304 -0.586850 1.051297 -0.373332
8 -0.103272 0.958334 1.163092 1.182315 0.615030
9 -0.616254 0.296678 -0.112027 0.679112 -0.622436
回答 20
To create an empty column
df['i'] = None
回答 21
以下是我的工作…但是,我对熊猫和Python真的很陌生,所以没有什么承诺。
df = pd.DataFrame([[1, 2], [3, 4], [5,6]], columns=list('AB'))
newCol = [3,5,7]
newName = 'C'
values = np.insert(df.values,df.shape[1],newCol,axis=1)
header = df.columns.values.tolist()
header.append(newName)
df = pd.DataFrame(values,columns=header)
The following is what I did… But I’m pretty new to pandas and really Python in general, so no promises.
df = pd.DataFrame([[1, 2], [3, 4], [5,6]], columns=list('AB'))
newCol = [3,5,7]
newName = 'C'
values = np.insert(df.values,df.shape[1],newCol,axis=1)
header = df.columns.values.tolist()
header.append(newName)
df = pd.DataFrame(values,columns=header)
回答 22
如果得到SettingWithCopyWarning
,一个简单的解决方法是复制您要向其中添加列的DataFrame。
df = df.copy()
df['col_name'] = values
If you get the SettingWithCopyWarning
, an easy fix is to copy the DataFrame you are trying to add a column to.
df = df.copy()
df['col_name'] = values