Etydhv

I have a DF which looks like this.

name id apps
john 1 [[app1, v1], [app2, v2], [app3,v3]]
smith 2 [[app1, v1], [app4, v4]]

I want to expand the apps column such that it looks like this.

name id app_name app_version
john 1 app1 v1
john 1 app2 v2
john 1 app3 v3
smith 2 app1 v1
smith 2 app4 v4

Any help is appreciated

You can .apply(pd.Series) twice to get what you need as an intermediate step, then merge back to the original dataframe.

import pandas as pd

df = pd.DataFrame(
 'name': ['john', 'smith'],
 'id': [1, 2],
 'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']], 
 [['app1', 'v1'], ['app4', 'v4']]]
)

dftmp = df.apps.apply(pd.Series).T.melt().dropna()
dfapp = (dftmp.value
 .apply(pd.Series)
 .set_index(dftmp.variable)
 .rename(columns=0:'app_name', 1:'app_version')
 )

df[['name', 'id']].merge(dfapp, left_index=True, right_index=True)
# returns:
 name id app_name app_version
0 john 1 app1 v1
0 john 1 app2 v2
0 john 1 app3 v3
1 smith 2 app1 v1
1 smith 2 app4 v4

You can always have a brute force solution. Something like:

name, id, app_name, app_version = [], [], [], []
for i in range(len(df)):
 for v in df.loc[i,'apps']:
 app_name.append(v[0])
 app_version.append(v[1])
 name.append(df.loc[i, 'name'])
 id.append(df.loc[i, 'id'])
df = pd.DataFrame('name': name, 'id': id, 'app_name': app_name, 'app_version': app_version)

will do the work.

Note that I assumed df['apps'] is lists of strings if df['apps'] is strings then you need: eval(df.loc[i,'apps']) instead of df.loc[i,'apps']

My suggestion (there may be easier ways) is using DataFrame.apply alongside pd.concat:

import pandas as pd

def expand_row(row):
 apps = row.apps
 apps_names = [app[0] for app in apps]
 apps_versions = [app[1] for app in apps]
 return pd.DataFrame(
 'name': row.name,
 'id': row.id,
 'app_name': apps_names,
 'app_version': apps_versions
 )

df = pd.DataFrame(
 'name': ['john', 'smith'],
 'id': [1, 2],
 'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']], 
 [['app1', 'v1'], ['app4', 'v4']]]
)
temp_dfs = df.apply(expand_row, axis=1).tolist()
expanded = pd.concat(temp_dfs)

print(expanded)

# name id app_name app_version
# 0 0 1 app1 v1
# 1 0 1 app2 v2
# 2 0 1 app3 v3
# 0 1 2 app1 v1
# 1 1 2 app4 v4