Pandas: split list in column into multiple rows [duplicate]

Answer 1

import itertools
dfres = pd.DataFrame([j+(i[2],) for i in df.values for j in itertools.product(*i[0:2])]
        ,columns=df.columns)

   Job position  Job type  id
0             6         1   3
1             2         3   4
2             2         6   4
3             2         5   4
4             6         3   4
5             6         6   4
6             6         5   4
7             1         9  43

Answer 2

Similar to Scott Boston's suggestion, I suggest you explode the columns separately, then merge them together.

For example, for 'Job position':

>>> df['Job position'].apply(pd.Series).reset_index().melt(id_vars='index').dropna()[['index', 'value']].set_index('index')
    value
index   
0   6.0
1   2.0
2   1.0
1   6.0

And, all together:

df = pd.DataFrame({'Job position': [[6], [2, 6], [1]], 'Job type': [[1], [3, 6, 5], [9]], 'id': [3, 4, 43]})
jobs = df['Job position'].apply(pd.Series).reset_index().melt(id_vars='index').dropna()[['index', 'value']].set_index('index')
types = df['Job type'].apply(pd.Series).reset_index().melt(id_vars='index').dropna()[['index', 'value']].set_index('index')
>>> pd.merge(
    pd.merge(
        jobs,
        types,
        left_index=True,
        right_index=True),
    df[['id']],
    left_index=True,
    right_index=True).rename(columns={'value_x': 'Job positions', 'value_y': 'Job type'})
Job positions   Job type    id
0   6.0 1.0 3
1   2.0 3.0 4
1   2.0 6.0 4
1   2.0 5.0 4
1   6.0 3.0 4
1   6.0 6.0 4
1   6.0 5.0 4
2   1.0 9.0 43

Answer 3

Use a comprehension

pd.DataFrame([
    [p, t, i] for P, T, i in df.values
    for p in P for t in T
], columns=df.columns)

   Job position  Job type  id
0             6         1   3
1             2         3   4
2             2         6   4
3             2         5   4
4             6         3   4
5             6         6   4
6             6         5   4
7             1         9  43

---

Alternatives to iterating over values

pd.DataFrame([
    [p, t, i] for P, T, i in df.itertuples(index=False)
    for p in P for t in T
], columns=df.columns)

---

z = zip(df['Job position'], df['Job type'], df['id'])
pd.DataFrame([
    [p, t, i] for P, T, i in z
    for p in P for t in T
], columns=df.columns)

---

To generalize this solution to accommodate any number of columns

pd.DataFrame([
    [p, t] + a for P, T, *a in df.values
    for p in P for t in T
], columns=df.columns)

   Job position  Job type  id
0             6         1   3
1             2         3   4
2             2         6   4
3             2         5   4
4             6         3   4
5             6         6   4
6             6         5   4
7             1         9  43

Answer 4

From data frame constructor

s1=df.Jobposition.str.len()

s2=df.Jobtype.str.len()
pd.DataFrame({'id':df.id.repeat(s1*s2),
  'Jobposition':np.concatenate([np.repeat(x,y) for x,y in zip(df.Jobposition,s2)]),
  'Jobtype':np.concatenate(np.repeat(df.Jobtype,s1).values)})

   Jobposition  Jobtype  id
0            6        1   3
1            2        3   4
1            2        6   4
1            2        5   4
1            6        3   4
1            6        6   4
1            6        5   4
2            1        9  43