自定义您使用的函数,
apply以便为每个组返回一个列表列表:
df.groupby('Column1')[['Column2', 'Column3']].apply(lambda g: g.values.tolist()).to_dict()# {0: [[23, 1]], # 1: [[5, 2], [2, 3], [19, 5]], # 2: [[56, 1], [22, 2]], # 3: [[2, 4], [14, 5]], # 4: [[59, 1]], # 5: [[44, 1], [1, 2], [87, 3]]}
如果您需要显式的元组列表,请使用
list(map(tuple, ...))进行转换:
df.groupby('Column1')[['Column2', 'Column3']].apply(lambda g: list(map(tuple, g.values.tolist()))).to_dict()# {0: [(23, 1)], # 1: [(5, 2), (2, 3), (19, 5)], # 2: [(56, 1), (22, 2)], # 3: [(2, 4), (14, 5)], # 4: [(59, 1)], # 5: [(44, 1), (1, 2), (87, 3)]}
欢迎分享,转载请注明来源:内存溢出
评论列表(0条)