Pandas: split list in column into multiple rows [duplicate]
Solution 1:
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
Solution 2:
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
Solution 3:
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