Python Pandas Expand a Column of List of Lists to Two New Column

Question

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

Answer 1

I have taken James' answer and amended as per rafaelc's suggestion about using pd.Dataframe.apps.tolist()) to speed it up:

import pandas as pd

df_melted = pd.DataFrame(df.apps.tolist()).T.melt().dropna()
df_tmp = pd.DataFrame(df_melted.value.tolist(), 
             columns = ['app_name', 'app_version'],
             index = df_melted.variable)

df = df_prices.join(df_tmp)
df.drop(columns=['prices'], inplace = True)

Answer 2

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

def expand_row(row):
    return pd.DataFrame({
        'name': row['name'], # row.name is the name of the series
        'id': row['id'],
        'app_name': [app[0] for app in row.apps],
        'app_version': [app[1] for app in row.apps]
    })

temp_dfs = df.apply(expand_row, axis=1).tolist()
expanded = pd.concat(temp_dfs)
expanded = expanded.reset_index() # put index in the correct order

print(expanded)

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

Also, here is a solution using python only, which, if my intuition is correct, should be fast:

rows = df.values.tolist()
expanded = [[row[0], row[1], app[0], app[1]]
            for row in rows
            for app in row[2]]
df = pd.DataFrame(
    expanded, columns=['name', 'id', 'app_name', 'app_version'])

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

Answer 3

Chain of pd.Series easy to understand, also if you would like know more methods ,check unnesting

df.set_index(['name','id']).apps.apply(pd.Series).\
         stack().apply(pd.Series).\
            reset_index(level=[0,1]).\
                rename(columns={0:'app_name',1:'app_version'})
Out[541]: 
    name  id app_name app_version
0   john   1     app1          v1
1   john   1     app2          v2
2   john   1     app3          v3
0  smith   2     app1          v1
1  smith   2     app4          v4

---

Method two slightly modify the function I write

def unnesting(df, explode):
    idx = df.index.repeat(df[explode[0]].str.len())
    df1 = pd.concat([
        pd.DataFrame({x: sum(df[x].tolist(),[])}) for x in explode], axis=1)
    df1.index = idx
    return df1.join(df.drop(explode, 1), how='left')

---

Then

yourdf=unnesting(df,['apps'])

yourdf['app_name'],yourdf['app_version']=yourdf.apps.str[0],yourdf.apps.str[1]
yourdf
Out[548]: 
         apps  id   name app_name app_version
0  [app1, v1]   1   john     app1          v1
0  [app2, v2]   1   john     app2          v2
0  [app3, v3]   1   john     app3          v3
1  [app1, v1]   2  smith     app1          v1
1  [app4, v4]   2  smith     app4          v4

Or

yourdf=unnesting(df,['apps']).reindex(columns=df.columns.tolist()+['app_name','app_version'])
yourdf[['app_name','app_version']]=yourdf.apps.tolist()
yourdf
Out[567]: 
         apps  id   name app_name app_version
0  [app1, v1]   1   john     app1          v1
0  [app2, v2]   1   john     app2          v2
0  [app3, v3]   1   john     app3          v3
1  [app1, v1]   2  smith     app1          v1
1  [app4, v4]   2  smith     app4          v4

Answer 4

Another approach would be (should be quite fast too):

#Repeat the columns without the list by the str length of the list
m=df.drop('apps',1).loc[df.index.repeat(df.apps.str.len())].reset_index(drop=True)
#creating a df exploding the list to 2 columns
n=pd.DataFrame(np.concatenate(df.apps.values),columns=['app_name','app_version'])
#concat them together
df_new=pd.concat([m,n],axis=1)

---

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

Answer 5

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']

Answer 6

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