您可以压缩b 和c 并使用它来制作字典:
a = ['a', 'b', 'c']
b = [1, 2, 3, 4, 5]
c = [10, 11, 14, 15, 16]
[{'a':a, 'b':b, 'c':'buy'} for a, b, in zip(b,c)]
结果:
[{'a': 1, 'b': 10, 'c': 'buy'},
{'a': 2, 'b': 11, 'c': 'buy'},
{'a': 3, 'b': 14, 'c': 'buy'},
{'a': 4, 'b': 15, 'c': 'buy'},
{'a': 5, 'b': 16, 'c': 'buy'}]
如果您想动态包含来自a 的密钥,您也可以使用类似itertools zip_longest 的东西将它们压缩以获取静态字符串:
from itertools import zip_longest
a = ['a', 'b', 'c']
b = [1, 2, 3, 4, 5]
c = [10, 11, 14, 15, 16]
[dict(zip_longest(a, tup, fillvalue='buy')) for tup in zip(b,c)]
同样的结果:
[{'a': 1, 'b': 10, 'c': 'buy'},
{'a': 2, 'b': 11, 'c': 'buy'},
{'a': 3, 'b': 14, 'c': 'buy'},
{'a': 4, 'b': 15, 'c': 'buy'},
{'a': 5, 'b': 16, 'c': 'buy'}]
还有一种方法是使用itertools.repeat() 将静态字符串包含在压缩元组中:
from itertools import repeat
a = ['a', 'b', 'c']
b = [1, 2, 3, 4, 5]
c = [10, 11, 14, 15, 16]
[dict(zip(a, tup)) for tup in zip(b,c, repeat('buy'))]