基于列值的熊猫子集和放置行

Question

my 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:

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:

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.

Answer 1

It's not really in one shot, but typically the way to do this is reuse a boolean mask, like this:

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

Answer 2

You can also use drop()

dframe = dframe.drop(dframe.index[dframe.A == 'a'])

Output:

    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.

dframe.index = range(len(dframe))

Output:

    A   C
0   b   5
1   b   6
2   b   7
3   b   8
4   c   9
5   c   10
6   c   11

Answer 3

An alternate way to think about it.

gb = dframe.groupby(dframe.A == 'a')
isa, nota = gb.get_group(True), gb.get_group(False)