python实现-身高和体重排序_python 按照身高体重排序

题目描述：
一组学生，每个学生有身高和体重两个特征，现在要对这些学生排序：先按照身高排序，身高相同的按照体重排序

1.用字典来存储 身高和体重的键值对，基于题目描述，身高存在相同的情况，所以体重作为键
不要因为先排序身高，第一次就排序身高，否则第二次排序会将已排序的序列打乱

dic = {80: 180, 70: 180, 60: 190, 90: 190, 56: 180}
ls1 = sorted(dic.items(), key=lambda kv: kv[0])
print(ls1)
ls2 = sorted(ls1, key=lambda kv: kv[1])
print(ls2)

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输出为：

[(56, 180), (60, 190), (70, 180), (80, 180), (90, 190)]
[(56, 180), (70, 180), (80, 180), (60, 190), (90, 190)]
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2.1存在问题：体重也存在相同的情况，作为键值就不合理，换列表存储，每个元素是(weight，height),先后顺序也可以调换

weight_height = [(66, 180), (66, 190), (70, 160), (70, 180), (60, 170), (80, 170)]
ls1 = sorted(weight_height, key=lambda kv: kv[0])
print(ls1)
ls2 = sorted(ls1, key=lambda kv: kv[1])
print(ls2)

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输出为：

[(60, 170), (66, 180), (66, 190), (70, 160), (70, 180), (80, 170)]
[(70, 160), (60, 170), (80, 170), (66, 180), (70, 180), (66, 190)]
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3.变更一下，先按照体重排序，体重相同的按照身高排序
直接修改排序的先后顺序即可

weight_height = [(66, 180), (66, 190), (70, 160), (70, 180), (60, 170), (80, 170)]
ls1 = sorted(weight_height, key=lambda kv: kv[1])
print(ls1)
ls2 = sorted(ls1, key=lambda kv: kv[0])
print(ls2)
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输出为：

[(70, 160), (60, 170), (80, 170), (66, 180), (70, 180), (66, 190)]
[(60, 170), (66, 180), (66, 190), (70, 160), (70, 180), (80, 170)]

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4.题目拓展，学生是有编号的，先按照身高排序，身高相同的按照体重排序，身高和体重相同的按照编号排序
所以，排序的先后顺序就是：编号 体重 身高

weight_height = [(0, 66, 180), (1, 66, 190), (2, 66, 190), (3, 70, 160), (4, 70, 160), (5, 70, 180), (6, 60, 170),
                 (7, 80, 170), (8, 70, 170)]

ls1 = sorted(weight_height, key=lambda kv: kv[0])
print(ls1)
ls2 = sorted(ls1, key=lambda kv: kv[1])
print(ls2)
ls3 = sorted(ls2, key=lambda kv: kv[2])
print(ls3)

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输出为：

[(0, 66, 180), (1, 66, 190), (2, 66, 190), (3, 70, 160), (4, 70, 160), (5, 70, 180), (6, 60, 170), (7, 80, 170), (8, 70, 170)]
[(6, 60, 170), (0, 66, 180), (1, 66, 190), (2, 66, 190), (3, 70, 160), (4, 70, 160), (5, 70, 180), (8, 70, 170), (7, 80, 170)]
[(3, 70, 160), (4, 70, 160), (6, 60, 170), (8, 70, 170), (7, 80, 170), (0, 66, 180), (5, 70, 180), (1, 66, 190), (2, 66, 190)]
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