dfc = pd.DataFrame({'A': ['aaa', np.nan, 'ccc'], 'B': [1, 2, 3], 'C': [np.nan, np.nan, np.nan], 'D': ['mm', np.nan, 10.0]})
A B C D
0 aaa 1 NaN mm
1 NaN 2 NaN NaN
2 ccc 3 NaN 10
dfc.loc[0][['A']]= 100
SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrameit turns out that assigning to the product of chained indexing has inherently unpredictable results
dfc.loc[0,'A'] = 100
A B C D
0 100 1 NaN mm
1 NaN 2 NaN NaN
2 ccc 3 NaN 10
cc2 = dfc[['A','B']].fillna(0.0,inplace=True)
# SettingWithCopyWarning:
# A value is trying to be set on a copy of a slice from a DataFrame
cc2 = dfc[['A','B']].fillna(0.0) #works, only on copy
cc1 = dfc.loc[:,['A','B']].fillna(0.0,inplace=True)
# return: None. using .loc indexed to a list of columns won't support inplace operations
dfc.fillna({'A':0, 'C':0}, inplace=True) # works
dfc.fillna({x:0 for x in ['A','C']}, inplace=True) # also works
dfc.fillna(0.0,inplace=True) # works
参考链接：http://pandas.pydata.org/pandas-docs/stable/indexing.html#returning-a-view-versus-a-copy
转载于:https://www.cnblogs.com/lyy-totoro/p/6946266.html
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