2 回答

TA贡献1777条经验 获得超10个赞
像这样的东西也可以工作:
In [2468]: df.value2 = df['value2'].apply(str)
In [2494]: res = df.groupby('value1')['value2'].apply(lambda x:','.join(x)).reset_index()
In [2498]: res['count'] = df.groupby('value1').size().reset_index()[0]
In [2499]: res
Out[2499]:
value1 value2 count
0 1 1,4,5 3
1 2 3,1 2

TA贡献1827条经验 获得超8个赞
使用aggwithjoin和聚合size,但有必要将列转换为字符串:
tups = [('value2', lambda x: ','.join(x.astype(str))), ('count', 'size')]
df1 = df.groupby('value1')['value2'].agg(tups).reset_index()
print (df1)
value1 value2 count
0 1 1,4,5 3
1 2 3,1 2
选择:
tups = [('value2', ','.join), ('count', 'size')]
df1 = df['value2'].astype(str).groupby(df['value1']).agg(tups).reset_index()
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