Pandas 由布尔`loc` 和随后的`iloc` 索引
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Pandas indexing by both boolean `loc` and subsequent `iloc`
提问by tsawallis
I want to index a Pandas dataframe using a boolean mask, then set a value in a subset of the filtered dataframe based on an integer index, and have this value reflected in the dataframe. That is, I would be happy if this worked on a view of the dataframe.
我想使用布尔掩码对 Pandas 数据帧进行索引，然后根据整数索引在过滤后的数据帧的子集中设置一个值，并将该值反映在数据帧中。也就是说，如果这适用于数据框的视图，我会很高兴。
Example:
例子：
In [293]:
df = pd.DataFrame({'a': [0, 1, 2, 3, 4, 5, 6, 7],
'b': [5, 5, 2, 2, 5, 5, 2, 2],
'c': [0, 0, 0, 0, 0, 0, 0, 0]})
mask = (df['a'] < 7) & (df['b'] == 2)
df.loc[mask, 'c']
Out[293]:
2 0
3 0
6 0
Name: c, dtype: int64
Now I would like to set the values of the first two elements returned in the filtered dataframe. Chaining an iloconto the loccall above works to index:
现在我想设置过滤数据帧中返回的前两个元素的值。将 an 链接iloc到loc上面的调用可以索引：
In [294]:
df.loc[mask, 'c'].iloc[0: 2]
Out[294]:
2 0
3 0
Name: c, dtype: int64
But not to assign:
但不分配：
In [295]:
df.loc[mask, 'c'].iloc[0: 2] = 1
print(df)
a b c
0 0 5 0
1 1 5 0
2 2 2 0
3 3 2 0
4 4 5 0
5 5 5 0
6 6 2 0
7 7 2 0
Making the assign value the same length as the slice (i.e. = [1, 1]) also doesn't work. Is there a way to assign these values?
使分配值与切片的长度相同（即= [1, 1]）也不起作用。有没有办法分配这些值？
采纳答案by EdChum
This does work but is a little ugly, basically we use the index generated from the mask and make an additional call to loc:
这确实有效，但有点难看，基本上我们使用从掩码生成的索引并额外调用loc：
In [57]:
df.loc[df.loc[mask,'c'].iloc[0:2].index, 'c'] = 1
df
Out[57]:
a b c
0 0 5 0
1 1 5 0
2 2 2 1
3 3 2 1
4 4 5 0
5 5 5 0
6 6 2 0
7 7 2 0
So breaking the above down:
所以分解上述内容：
In [60]:
# take the index from the mask and iloc
df.loc[mask, 'c'].iloc[0: 2]
Out[60]:
2 0
3 0
Name: c, dtype: int64
In [61]:
# call loc using this index, we can now use this to select column 'c' and set the value
df.loc[df.loc[mask,'c'].iloc[0:2].index]
Out[61]:
a b c
2 2 2 0
3 3 2 0
回答by JoeCondron
How about.
怎么样。
ix = df.index[mask][:2]
df.loc[ix, 'c'] = 1
Same idea as EdChum but more elegant as suggested in the comment.
与 EdChum 相同的想法，但如评论中建议的那样更优雅。
EDIT: Have to be a little bit careful with this one as it may give unwanted results with a non-unique index, since there could be multiple rows indexed by either of the label in ixabove. If the index is non-unique and you only want the first 2 (or n) rows that satisfy the boolean key, it would be safer to use .ilocwith integer indexing with something like
编辑：必须对这个稍微小心一点，因为它可能会使用非唯一索引给出不需要的结果，因为ix上面的任何一个标签都可能索引多行。如果索引是非唯一的，并且您只想要满足布尔键的前 2（或 n）行，则将.iloc整数索引与类似的内容一起使用会更安全
ix = np.where(mask)[0][:2]
df.iloc[ix, 'c'] = 1
回答by JohnE
I don't know if this is any more elegant, but it's a little different:
我不知道这是否更优雅，但它有点不同：
mask = mask & (mask.cumsum() < 3)
df.loc[mask, 'c'] = 1
a b c
0 0 5 0
1 1 5 0
2 2 2 1
3 3 2 1
4 4 5 0
5 5 5 0
6 6 2 0
7 7 2 0
