将结果分组到列表字典

你可以

groupby

Column1

Column3

apply(list)

to_dict

In [81]: df.groupby('Column1')['Column3'].apply(list).to_dict()Out[81]: {0: [1], 1: [2, 3, 5], 2: [1, 2], 3: [4, 5], 4: [1], 5: [1, 2, 3]}

或者，做

In [433]: {k: list(v) for k, v in df.groupby('Column1')['Column3']}Out[433]: {0: [1], 1: [2, 3, 5], 2: [1, 2], 3: [4, 5], 4: [1], 5: [1, 2, 3]}