pandas 熊猫子集并根据列值删除行
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pandas subset and drop rows based on column value
提问by ramesh
my df:
我的 df：
dframe = pd.DataFrame({"A":list("aaaabbbbccc"), "C":range(1,12)}, index=range(1,12))
Out[9]:
A C
1 a 1
2 a 2
3 a 3
4 a 4
5 b 5
6 b 6
7 b 7
8 b 8
9 c 9
10 c 10
11 c 11
to subset based on column value:
根据列值进行子集化：
In[11]: first = dframe.loc[dframe["A"] == 'a']
In[12]: first
Out[12]:
A C
1 a 1
2 a 2
3 a 3
4 a 4
To drop based on column value:
根据列值删除：
In[16]: dframe = dframe[dframe["A"] != 'a']
In[17]: dframe
Out[16]:
A C
5 b 5
6 b 6
7 b 7
8 b 8
9 c 9
10 c 10
11 c 11
Is there any way to do both in one shot? Like subsetting rows based on a column value and deleting same rows in the original df.
有没有办法一举两得？就像根据列值对行进行子集化并删除原始 df 中的相同行。
回答by chrisb
It's not really in one shot, but typically the way to do this is reuse a boolean mask, like this:
这不是一次拍摄，但通常这样做的方法是重用布尔掩码，如下所示：
In [28]: mask = dframe['A'] == 'a'
In [29]: first, dframe = dframe[mask], dframe[~mask]
In [30]: first
Out[30]:
A C
1 a 1
2 a 2
3 a 3
4 a 4
In [31]: dframe
Out[31]:
A C
5 b 5
6 b 6
7 b 7
8 b 8
9 c 9
10 c 10
11 c 11
回答by Joe T. Boka
You can also use drop()
你也可以使用drop()
dframe = dframe.drop(dframe.index[dframe.A == 'a'])
Output:
输出：
A C
5 b 5
6 b 6
7 b 7
8 b 8
9 c 9
10 c 10
11 c 11
If you want to fix the index, you can do this.
如果你想修复index，你可以这样做。
dframe.index = range(len(dframe))
Output:
输出：
A C
0 b 5
1 b 6
2 b 7
3 b 8
4 c 9
5 c 10
6 c 11
回答by piRSquared
An alternate way to think about it.
另一种思考方式。
gb = dframe.groupby(dframe.A == 'a')
isa, nota = gb.get_group(True), gb.get_group(False)
