My dataframe has a column pairs that contains a key-pair list. Each key is unique in the list. e.g:

df = pd.DataFrame({
        'id':  ['1', '2', '3'],
        'abc':None,
        'pairs': [ ['abc/123', 'foo/345', 'xyz/789'],  ['abc/456', 'foo/111', 'xyz/789'],  ['xxx/222', 'foo/555', 'xyz/333'] ]
      })

Dataframe is:

  id | abc  | pairs
  ------------------------------------
  1  |None  | [abc/123, foo/345, xyz/789]
  2  |None  | [abc/456, foo/111, xyz/789]
  3  |None  | [xxx/222, foo/555, xyz/333]

The column abc is filled with the value in column pairs if an element (idx=0) split by \ has the value (key) ==’abc’.

Expected df:

  id | abc  | pairs
  ------------------------------------
  1  |123   | [abc/123, foo/345, xyz/789]
  2  |456   | [abc/456, foo/111, xyz/789]
  3  |None  | [xxx/222, foo/555, xyz/333]

I look for something like:

df.loc[df['pairs'].map(lambda x: 'abc' in (l.split('/')[0] for l in x)), 'abc'] = 'FOUND'

my problem is to replace the FOUND by the correct value the l.split('/')[0]

>Solution :

Try this

# data
df = pd.DataFrame({
        'id':  ['1', '2', '3'],
        'abc':None,
        'pairs': [ ['abc/123', 'foo/345', 'xyz/789'],  ['abc/456', 'foo/111', 'xyz/789'],  ['xxx/222', 'foo/555', 'xyz/333'] ]
      })
# construct a dict in loop and get value of abc key
df['abc'] = df['pairs'].apply(lambda x: dict(e.split('/') for e in x).get('abc'))
df