假设我有以下DataFrame:
In [1]: df
Out[1]:
apple banana cherry
0 0 3 good
1 1 4 bad
2 2 5 good
这可以按预期工作:
In [2]: df['apple'][df.cherry == 'bad'] = np.nan
In [3]: df
Out[3]:
apple banana cherry
0 0 3 good
1 NaN 4 bad
2 2 5 good
但这不是:
In [2]: df[['apple', 'banana']][df.cherry == 'bad'] = np.nan
In [3]: df
Out[3]:
apple banana cherry
0 0 3 good
1 1 4 bad
2 2 5 good
为什么呢?如何在不必写出两行的情况下实现'apple'和'banana'值的转换,如
In [2]: df['apple'][df.cherry == 'bad'] = np.nan
In [3]: df['banana'][df.cherry == 'bad'] = np.nan
答案 0 :(得分:33)
您应该使用loc并执行不链接:
In [11]: df.loc[df.cherry == 'bad', ['apple', 'banana']] = np.nan
In [12]: df
Out[12]:
apple banana cherry
0 0 3 good
1 NaN NaN bad
2 2 5 good
查看returning a view vs a copy上的文档,如果你链接到了副本(并扔掉了),但如果你在一个地方进行,那么pandas巧妙地意识到你想要分配给原始的
答案 1 :(得分:4)
这是因为df[['apple', 'banana']][df.cherry == 'bad'] = np.nan
分配给DataFrame的副本。试试这个:
df.ix[df.cherry == 'bad', ['apple', 'banana']] = np.nan