Question

如果我不指望它，我会收到SettingWithCopyWarning错误：

N.In <38>: # Column B does not exist yet
N.In <39>: df['B'] = df['A']/25
N.In <40>: df['B'] = df['A']/50

/Users/josh/anaconda/envs/py27/lib/python2.7/site-packages/pandas/core/indexing.py:389: SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead
  self.obj[item] = s

和

N.In <41>: df.loc[:,'B'] = df['A']/50

/Users/josh/anaconda/envs/py27/lib/python2.7/site-packages/pandas/core/indexing.py:389: SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead
  self.obj[item] = s

为什么会发生在案例1和案例2中？

Answer 1

在案例1中，df['A']会创建df的副本。正如Pandas documentation所解释的那样，这可能会在链接时导致意外结果，从而引发警告。案例2看起来正确，但可能出现误报：

警告：链式分配警告/异常的目标是告知用户可能无效的分配。可能有假阳性;无意中链式分配的情况报道。

要为单个数据帧关闭SettingWithCopyWarning，请使用

df.is_copy = False

要完全关闭链式分配警告，请使用

options.mode.chained_assignment = None

Answer 2

应该取消警告的另一种解决方案：

df = df.copy()
df['B'] = df['A']/25
df['B'] = df['A']/50

SettingWithCopyWarning，即使使用loc（？）

2 个答案: