How to create column based on string position of other column in python?

Question

df:

How to create a column based on string position, for example in column_A I need to check 0 position create a column B.

Output;

    Col_A     Col_B
0   011011   pos1,pos4
1   000111   pos1,pos2,pos3
2   011000   pos1,pos4,pos5,pos6
3   011111   pos1
4   011010   pos1,pos4,pos6

Please reformat your question. The question is hard to understand and unclear. — Harshit
– Harshit, Commented Jul 1, 2020 at 7:55

jezrael · Accepted Answer · 2020-07-01 08:00:39Z

2

First convert strings to DataFrame and add columns names by function in rename:

f = lambda x: f'pos{x+1}'
df1 = pd.DataFrame([list(x) for x in df['Col_A']], index=df.index).rename(columns=f)
print (df1)
  pos1 pos2 pos3 pos4 pos5 pos6
0    0    1    1    0    1    1
1    0    0    0    1    1    1
2    0    1    1    0    0    0
3    0    1    1    1    1    1
4    0    1    1    0    1    0

Then compare '0' values by DataFrame.eq and for new column use matrix multiplication by DataFrame.dot with remove separator by Series.str.rstrip:

df['Col_B'] = df1.eq('0').dot(df1.columns + ',').str.rstrip(',')
print (df)

    Col_A                Col_B
0  011011            pos1,pos4
1  000111       pos1,pos2,pos3
2  011000  pos1,pos4,pos5,pos6
3  011111                 pos1
4  011010       pos1,pos4,pos6

answered Jul 1, 2020 at 8:00

jezrael

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1 Comment

Ch3steR Over a year ago

Alternate to avoid extra .str.rstrip: [','.join(df1.columns[mask]) for mask in df1.eq('0').to_numpy()]

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