Question

我有一个看起来像这样的DF。

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

我想扩展“应用程序”列，使其看起来像这样。

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

感谢您的帮助

Answer 1

您可以两次var p = Point(x:0,y:0) p = Point(x:p.x+1,y:p.y)来获得所需的中间步骤，然后再合并回原始数据框。

.apply(pd.Series)

Answer 2

您始终可以拥有蛮力解决方案。像这样：

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})

将完成工作。

请注意，如果df ['apps']是字符串，我假设df ['apps']是字符串列表，那么您需要：eval(df.loc[i,'apps'])而不是df.loc[i,'apps']

Answer 3

另一种方法是（也应该相当快）：

#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 4

pd.Series的链很容易理解，如果您想了解更多方法，请检查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

方法二稍微修改我写的功能

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')

然后

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

或

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 5

我的建议（可能有更简单的方法）是将DataFrame.apply与pd.concat一起使用：

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

此外，这是仅使用python的解决方案，如果我的直觉是正确的，则应该很快：

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

Python Pandas将列表列表的列扩展为两个新列

5 个答案: