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[英]Split a list of dictionaries into multiple lists based on uniqueness of one of the key/value pairs in Python
[英]Put multiple lists as value with one list as key
我有 3 個lists
和一個這樣的string
:
a = ['a', 'b', 'c']
b = [1, 2, 3, 4, 5]
c = [10, 11, 14, 15, 16]
string = 'buy'
我需要一個這樣的字典列表:
[{'a':1 , 'b':10 , 'c':'buy'},{'a':2 , 'b':11 , 'c':'buy'},{'a':3 , 'b':14 , 'c':'buy'},...]
result = [ dict(zip(a,[*v, 'buy'])) for v in zip(b,c) ]
您可以只使用 zip b
和c
並使用它來制作字典:
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)]
結果:
[{'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'}]
如果你想動態地a
密鑰,你可以 zip 使用類似 itertools zip_longest
的東西來獲取 static 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)]
結果相同:
[{'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'}]
另一種方法是使用itertools.repeat()
在壓縮元組中包含 static 字符串:
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'))]
這應該可以滿足您的需求:
[{a[0]:vb, a[1]:vc, a[2]:string} for vb, vc in zip(b, c)]
它產生:
[{'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'}]
它從 list a
獲取 dict 鍵,而'buy'
來自string
。
嘗試這個:
a = ['a', 'b', 'c']
b = [1, 2, 3, 4, 5]
c = [10, 11, 14, 15, 16]
print([{ a[0]:i, a[1]:j, a[2]:'buy'} for i,j in zip(b,c)])
或者
print([dict(zip(a, [i, j, "buy"])) for i,j in zip(b,c)])
輸出:
[{'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'}]
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