从列表组合生成Python字典

use*_*543 6 python combinations dictionary list pandas

我试图解决我的问题,但我做不到。

我有三个Python清单:

atr = ['a','b','c']
m = ['h','i','j']
func = ['x','y','z']
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我的问题是根据这三个列表的组合生成Python字典:

所需的输出:

py_dict = {
    'a': [('x','h'), ('x','i'), ('x','j'), ('y','h'), ('y','i'), ('y','j'),('z','h'), ('z','i'), ('z','j')],
    'b': [('x','h'), ('x','i'), ('x','j'), ('y','h'), ('y','i'), ('y','j'),('z','h'), ('z','i'), ('z','j')],
    'c': [('x','h'), ('x','i'), ('x','j'), ('y','h'), ('y','i'), ('y','j'),('z','h'), ('z','i'), ('z','j')]
}
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Aja*_*234 7

您可以使用itertools.product

import itertools
atr = ['a','b','c']
m = ['h','i','j']
func = ['x','y','z']
prod = list(itertools.product(func, m))
result = {i:prod for i in atr}
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输出:

{'a': [('x', 'h'), ('x', 'i'), ('x', 'j'), ('y', 'h'), ('y', 'i'), ('y', 'j'), ('z', 'h'), ('z', 'i'), ('z', 'j')], 'b': [('x', 'h'), ('x', 'i'), ('x', 'j'), ('y', 'h'), ('y', 'i'), ('y', 'j'), ('z', 'h'), ('z', 'i'), ('z', 'j')], 'c': [('x', 'h'), ('x', 'i'), ('x', 'j'), ('y', 'h'), ('y', 'i'), ('y', 'j'), ('z', 'h'), ('z', 'i'), ('z', 'j')]}
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