pandas - 按列名屏蔽数据框
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pandas - mask dataframe by column name
提问by Fabio Lamanna
Starting from this simple dataframe df:
从这个简单的数据框开始df：
col1,col2
1,3
2,1
3,8
I would like to apply a boolean maskin function of the name of the column. I know that it is easy for values:
我想mask在列名称的函数中应用一个布尔值。我知道价值观很容易：
mask = df <= 1
df = df[mask]
which returns:
返回：
mask:
面具：
col1 col2
0 True False
1 False True
2 False False
df:
df：
col1 col2
0 1 NaN
1 NaN 1
2 NaN NaN
as expected. Now I would like to obtain a boolean mask based on the column name, something like:
正如预期的那样。现在我想根据列名获得一个布尔掩码，例如：
mask = df == df['col_1']
which should return:
应该返回：
mask
面具
col1 col2
0 True False
1 True False
2 True False
EDIT:
编辑：
This seems weird, but I need those kind of masks to later filtering by columns seaborn heatmaps.
这看起来很奇怪，但我需要这些掩码，以便以后按列 seaborn 热图进行过滤。
回答by KT.
As noted in the comments, situations where you would need to get a "mask" like that seem rare (and chances are, you not in one of them). Consequently, there is probably no nice "built-in" solution for them in Pandas.
正如评论中所指出的，您需要获得这样的“面具”的情况似乎很少见（而且很有可能，您不在其中之一）。因此，在 Pandas 中可能没有很好的“内置”解决方案。
None the less, you can achieve what you need, using a hack like the following, for example:
尽管如此，您可以使用如下所示的 hack 来实现您需要的功能，例如：
mask = (df == df) & (df.columns == 'col_1')
Update:. As noted in the comments, if your data frame contains nulls, the mask computed this way will always be Falseat the corresponding locations. If this is a problem, the safer option is:
更新：。如评论中所述，如果您的数据框包含空值，则以这种方式计算的掩码将始终False位于相应的位置。如果这是一个问题，更安全的选择是：
mask = ((df == df) | df.isnull()) & (df.columns == 'col_1')
回答by Anton Protopopov
You could transpose your dataframe than compare it with the columns and then transpose back. A bit weird but working example:
您可以转置您的数据框，而不是将其与列进行比较，然后转回。有点奇怪但有效的例子：
import pandas as pd
from io import StringIO
data = """
col1,col2
1,3
2,1
3,8
"""
df = pd.read_csv(StringIO(data))
mask = (df.T == df['col1']).T
In [176]: df
Out[176]:
col1 col2
0 1 3
1 2 1
2 3 8
In [178]: mask
Out[178]:
col1 col2
0 True False
1 True False
2 True False
EDIT
编辑
I found another answer for that, you could use isinmethod:
我找到了另一个答案，你可以使用isin方法：
In [41]: df.isin(df.col1)
Out[41]:
col1 col2
0 True False
1 True False
2 True False
EDIT2
编辑2
As @DSM show in the comment that these two cases not working correctly. So you should use @KT. method. But.. Let's play more with transpose:
正如@DSM 在评论中显示的那样，这两种情况无法正常工作。所以你应该使用@KT。方法。但是..让我们玩转转：
df.col2 = df.col1
In [149]: df
Out[149]:
col1 col2
0 1 1
1 2 2
2 3 3
In [147]: df.isin(df.T[df.columns == 'col1'].T)
Out[147]:
col1 col2
0 True False
1 True False
2 True False
