它顯示了在要組合一列和額外一列的情況下如何組合行。我現在正在為有很多列的情況尋找解決方案，并且我仍然想基于一列組合行。我希望如何處理是：首先列出一種類型的所有列，然后以與第一次相同的順序列出另一種類型的列。這是一個最小的例子data = [['tom', 'ca', 2], ['ni2ck', 'ma', 2], ['j3uli', 'ny', 4] , ['nic4k', 'ma', 4], ['jul5i', 'ny', 4] , ['nic6k', 'ma', 7], ['ju7li', 'ny', 7] , ['nic8k', 'ma', 7], ['ju9li', 'ny', 7] , ['nic1k', 'ma', 8], ['car', 'ny', 8]]df = pd.DataFrame(data, columns = ['Name', 'Location', 'Age']) df 結果是Name Location Age0 tom ca 21 ni2ck ma 22 j3uli ny 43 nic4k ma 44 jul5i ny 45 nic6k ma 76 ju7li ny 77 nic8k ma 78 ju9li ny 79 nic1k ma 810 car ny 8這將是期望的結果 Name Name Location Location Age0 tom ni2ck ca ma 21 nic1k car ma ny 8Name Name Name Location Location Location Age0 j3uli nic4k jul5i ny ma ny 4Name Name Name Name Location Location Location Location Age0 nic6k ju7li nic8k ju9li ma ny ma ny 7重要的是，正確的位置與相應名稱的順序相同。
1 回答
翻翻過去那場雪
TA貢獻2065條經驗 獲得超14個贊
從@Wen 解決方案開發。而不是pivot，使用pivot_table
df['New']=df.groupby('Age').cumcount()
s= df.pivot_table(index='Age',columns='New',
values=['Name', 'Location'],
aggfunc='first').reindex(['Name', 'Location'], axis=1, level=0)
s.columns = s.columns.map('{0[0]}{0[1]}'.format)
l=[y.dropna(1).reset_index() for _ , y in s.groupby(s.isnull().sum(1))]
In [499]: l[0]
Out[499]:
Age Name0 Name1 Name2 Name3 Location0 Location1 Location2 Location3
0 7 nic6k ju7li nic8k ju9li ma ny ma ny
In [500]: l[1]
Out[500]:
Age Name0 Name1 Name2 Location0 Location1 Location2
0 4 j3uli nic4k jul5i ny ma ny
In [501]: l[2]
Out[501]:
Age Name0 Name1 Location0 Location1
0 2 tom ni2ck ca ma
1 8 nic1k car ma ny
如果您想保留多索引列，請跳過列map上的命令
df['New']=df.groupby('Age').cumcount()
s= df.pivot_table(index='Age',columns='New',
values=['Name', 'Location'],
aggfunc='first').reindex(['Name', 'Location'], axis=1, level=0)
l=[y.dropna(1).reset_index() for _ , y in s.groupby(s.isnull().sum(1))]
In [544]: l[0]
Out[544]:
Age Name Location
New 0 1 2 3 0 1 2 3
0 7 nic6k ju7li nic8k ju9li ma ny ma ny
In [545]: l[1]
Out[545]:
Age Name Location
New 0 1 2 0 1 2
0 4 j3uli nic4k jul5i ny ma ny
In [546]: l[2]
Out[546]:
Age Name Location
New 0 1 0 1
0 2 tom ni2ck ca ma
1 8 nic1k car ma ny
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