Count occurrences of items in Series in each row of a DataFrame

Question

I have a pandas.DataFrame that looks like this.

COL1    COL2    COL3
C1      None    None
C1      C2      None
C1      C1      None
C1      C2           


        

Answer 1

You could apply value_counts:

In [11]: df.apply(pd.Series.value_counts, axis=1)
Out[11]: 
   C1  C2  C3  None
0   1 NaN NaN     2
1   1   1 NaN     1
2   2 NaN NaN     1
3   1   1   1   NaN

So you can fill the NaN and applend just the base values you want:

In [12]: df.apply(pd.Series.value_counts, axis=1)[['C1', 'C2', 'C3']].fillna(0)
Out[12]: 
   C1  C2  C3
0   1   0   0
1   1   1   0
2   2   0   0
3   1   1   1

Note: there's an open issue to have a value_counts method directly for a DataFrame (which I think should be introduced by pandas 0.15).

Answer 2

Andy's answer is spot on.

I'm adding this answer, if C1,C2...Cn list is huge and we want to view only subset of them.

dff = df.copy()
dff['C1']=(df == 'C1').T.sum()
dff['C2']=(df == 'C2').T.sum()
dff['C3']=(df == 'C3').T.sum()
dff
  COL1  COL2  COL3  C1  C2  C3
0   C1  None  None   1   0   0
1   C1    C2  None   1   1   0
2   C1    C1  None   2   0   0
3   C1    C2    C3   1   1   1

Answer 3

Usually apply + serise function to whole dataframe will slowing down the whole process , Additional Reading : Link

df.mask(df.eq('None')).stack().str.get_dummies().sum(level=0)
Out[165]: 
   C1  C2  C3
0   1   0   0
1   1   1   0
2   2   0   0
3   1   1   1

---

Or you can do with Counter

from  collections import Counter

pd.DataFrame([ Counter(x) for x in df.values]).drop('None',1)
Out[170]: 
   C1   C2   C3
0   1  NaN  NaN
1   1  1.0  NaN
2   2  NaN  NaN
3   1  1.0  1.0