I have following dataframe named df.

The mapping condition is {'x' : 1, 'z' : 2, 'ELSE' : 0}

my desired output dataframe should look like,

Which means, even any of the letters in column letter is x, then the column map should be 1.

Without iterating through each row of the dataframe, is there any way to do that?

CodePudding user response：

You can use

pure pandas

cond = {'x' : 1, 'z' : 2, 'ELSE' : 0}

df['map'] = (df['letter']
 .str.split(',').explode()
 .map(lambda x: cond.get(x, cond['ELSE']))
 .groupby(level=0).max()
)

In case of multiple values I would get the max.

Alternative for the first valid match:

df['map'] = (df['letter']
 .str.split(',').explode()
 .map(cond)
 .groupby(level=0).first()
 .fillna(cond['ELSE'], downcast='infer')
)

list comprehension

Or using a list comprehension, here the first valid match would be used:

cond = {'x' : 1, 'z' : 2, 'ELSE' : 0}

df['map'] = [next((cond[x] for x in s.split(',') if x in cond),
                  cond['ELSE'])  for s in df['letter']]

   id letter  map
0   1    x,y    1
1   2      z    2
2   3      a    0

CodePudding user response：

use np.select

import numpy as np

cond1 = df['letter'].str.contains('x')
cond2 = df['letter'].str.contains('z')
df.assign(map=np.select([cond1, cond2], [1, 2], 0))

output:

    id  letter  map
0   1   x,y     1
1   2   z       2
2   3   a       0