Question

我的数据结构是：

ds = [{
    "name": "groupA",
    "subGroups": [123,456]
},
{
    "name": "groupB",
    "subGroups": ['aaa', 'bbb' , 'ccc']
}]

这给出了以下数据框

df = pd.DataFrame(ds)

    name    subGroups
0   groupA  [123, 456]
1   groupB  [aaa, bbb, ccc]

我想：

    name    subGroupsFlattend
0   groupA  123
1   groupA  456
2   groupB  aaa
3   groupB  bbb
4   groupB  ccc

有什么想法吗？

Answer 1

使用explode：

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Answer 2

您可以使用SO question：

from pandas.io.json import json_normalize

df = json_normalize(ds,  ['subGroups'], 'name').rename(columns={0:'subGroupsFlattend'})
print (df)
  subGroupsFlattend    name
0               123  groupA
1               456  groupA
2               aaa  groupB
3               bbb  groupB
4               ccc  groupB

扁平化词典的替代解决方案：

L = [y for x in ds for y in zip(x["subGroups"], [x["name"]] * len(x["subGroups"]))]
print (L)
[(123, 'groupA'), (456, 'groupA'), ('aaa', 'groupB'), ('bbb', 'groupB'), ('ccc', 'groupB')]

df = pd.DataFrame(L, columns=['subGroupsFlattend','name'])
print (df)
  subGroupsFlattend    name
0               123  groupA
1               456  groupA
2               aaa  groupB
3               bbb  groupB
4               ccc  groupB

编辑：

from itertools import chain
df = pd.DataFrame(ds)

df1 = pd.DataFrame({
    'subGroups' : list(chain.from_iterable(df['subGroups'].tolist())), 
    'name' : df['name'].values.repeat(df['subGroups'].str.len())
})
print (df1)
     name subGroups
0  groupA       123
1  groupA       456
2  groupB       aaa
3  groupB       bbb
4  groupB       ccc

Answer 3

您可以通过以下方式修复输出：

pd.DataFrame({'name':df.name.repeat(df.subGroups.str.len()),'subGroup':df.subGroups.sum()})
Out[364]: 
     name subGroup
0  groupA      123
0  groupA      456
1  groupB      aaa
1  groupB      bbb
1  groupB      ccc

Answer 4

YOBEN_S解决方案，但对于大数据帧而言效率更高。

from itertools import chain
pd.DataFrame({'name':df.name.repeat(df.subGroups.str.len()),
              'subGroup':list(chain.from_iterable(df.subGroups.to_list()))})

在Pandas DataFrame中展平元素列表

4 个答案: