在pandas DataFrame中设置新列的正确方法是为了避免SettingWithCopyWarning

Question

试图在netc df中创建一个新列,但我收到了警告

netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

C:\Anaconda\lib\site-packages\ipykernel\__main__.py:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

在新版本的Pandas中创建一个字段的正确方法是什么,以避免收到警告？

pd.__version__
Out[45]:
u'0.19.2+0.g825876c.dirty'

Answer 1

正如错误中所述,尝试使用.loc[row_indexer,col_indexer]创建新列.

netc.loc[:,"DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM.

笔记

通过Pandas Indexing Docs,您的代码应该可以运行.

netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

被翻译成

netc.__setitem__('DeltaAMPP', netc.LOAD_AM - netc.VPP12_AM)

哪个应该有可预测的行为.在SettingWithCopyWarning仅存在链式分配期间,警告的意外行为的用户(这是你做的不是).但是,如文档中所述,

有时SettingWithCopy,当没有明显的链式索引时,会出现警告.这些SettingWithCopy是旨在捕获的错误!熊猫可能会试图警告你,你已经这样做了:

然后,文档继续给出一个示例,说明何时可能会出现错误,即使它不是预期的.所以我不知道为什么没有更多的背景会发生这种情况.

Answer 2

在将数据分配给通过索引构造的SettingWithCopyWarningDataFrame 时，我遇到了问题。df两个命令

• df['new_column'] = something
• df.loc[:, 'new_column'] = something

没有警告就无法工作。一旦复制df（DataFrame.copy()）一切都很好。

在下面的代码中，比较df0 = df_test[df_test['a']>3]和df1 = df_test[df_test['a']>3].copy()。对于df0这两个作业都会抛出警告。两者df1都工作得很好。

>>> df_test
      a     b     c     d  e
0   0.0   1.0   2.0   3.0  0
1   4.0   5.0   6.0   7.0  1
2   8.0   9.0  10.0  11.0  2
3  12.0  13.0  14.0  15.0  3
4  16.0  17.0  18.0  19.0  4
>>> df0 = df_test[df_test['a']>3]
>>> df1 = df_test[df_test['a']>3].copy()
>>> df0['e'] = np.arange(4)
__main__:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
>>> df1['e'] = np.arange(4)
>>> df0.loc[2, 'a'] = 77
/opt/anaconda3/lib/python3.7/site-packages/pandas/core/indexing.py:1719: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
  self._setitem_single_column(loc, value, pi)
>>> df1.loc[2, 'a'] = 77
>>> df0
      a     b     c     d  e
1   4.0   5.0   6.0   7.0  0
2  77.0   9.0  10.0  11.0  1
3  12.0  13.0  14.0  15.0  2
4  16.0  17.0  18.0  19.0  3
>>> df1
      a     b     c     d  e
1   4.0   5.0   6.0   7.0  0
2  77.0   9.0  10.0  11.0  1
3  12.0  13.0  14.0  15.0  2
4  16.0  17.0  18.0  19.0  3

顺便说一句：建议阅读有关此问题的文档（警告中的链接）

Answer 3

Your example is incomplete, as it doesn't show where netc comes from. It is likely that netc itself is the product of slicing, and as such Pandas cannot make guarantees that it isn't a view or a copy.

For example, if you're doing this:

netc = netb[netb["DeltaAMPP"] == 0]
netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

then Pandas wouldn't know if netc is a view or a copy. If it were a one-liner, it would effectively be like this:

netb[netb["DeltaAMPP"] == 0]["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

where you can see the double indexing more clearly.

If you want to make netc separate from netb, one possible remedy might be to force a copy in the first line (the loc is to make sure we're not copying two times), like:

netc = netb.loc[netb["DeltaAMPP"] == 0].copy()

If, on the other hand, you want to have netb modified with the new column, you may do:

netb.loc[netb["DeltaAMPP"] == 0, "DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

Answer 4

您需要在创建列时重置索引，特别是如果您对特定值进行了过滤......那么您不需要使用 .loc[row_indexer,col_indexer]

netc.reset_index(drop=True, inplace=True)
netc["DeltaAMPP"] = netc.LOAD_AM - netc.VPP12_AM

然后它应该工作:)