Python 如何压缩两个不同大小的列表？
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How to zip two differently sized lists?
提问by user2131116
I want to zip two list with different length
我想压缩两个不同长度的列表
for example
例如
A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
and I expect this
我期待这个
[(1, 'A'), (2, 'B'), (3, 'C'), (4, 'A'), (5, 'B'), (6, 'C'), (7, 'A'), (8, 'B'), (9, 'C')]
But the build-in zipwon't repeat to pair with the list with larger size .
Does there exist any build-in way can achieve this?
thanks
但是内置zip不会重复与更大 size 的列表配对。是否存在任何内置方式可以实现这一目标？谢谢
here is my code
这是我的代码
idx = 0
zip_list = []
for value in larger:
zip_list.append((value,smaller[idx]))
idx += 1
if idx == len(smaller):
idx = 0
采纳答案by sloth
You can use itertools.cycle:
您可以使用itertools.cycle：
Make an iterator returning elements from the iterable and saving a copy of each. When the iterable is exhausted, return elements from the saved copy. Repeats indefinitely.
使迭代器从可迭代对象返回元素并保存每个元素的副本。当迭代用完时，从保存的副本中返回元素。无限重复。
Example:
例子：
A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
from itertools import cycle
zip_list = zip(A, cycle(B)) if len(A) > len(B) else zip(cycle(A), B)
回答by Eser Aygün
Try this.
尝试这个。
A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
Z = []
for i, a in enumerate(A):
Z.append((a, B[i % len(B)]))
Just make sure that the larger list is in A.
只需确保较大的列表在A.
回答by James Robinson
There is probably a better way, but you could make a function that repeats your list to whatever length you want.
可能有更好的方法，但您可以制作一个函数，将您的列表重复到您想要的任何长度。
def repeatlist(l,i):
'''give a list and a total length'''
while len(l) < i:
l += l
while len(l) > i:
l.pop()
Then do
然后做
repeatlist(B,len(A))
zip_list = zip(A,B)
回答by f5r5e5d
symmetric, no conditionals one liner
对称，无条件单行
[*zip(A*(len(B)//len(A) + 1), B*(len(A)//len(B) + 1))]
which strictly answers 'How to zip two differentlysized lists?'
严格回答“如何压缩两个不同大小的列表？”
needs a patch for equal sized lists to be general:
需要一个相同大小列表的补丁才能通用：
[*(zip(A, B) if len(A) == len(B)
else zip(A*(len(B)//len(A) + 1),
B*(len(A)//len(B) + 1)))]
回答by dimitris_ps
And nowadays with list comprehentions
现在有了列表理解
[(i, B[i % 3 - 1]) for i in A]
Or if the elements of Aare not sequential and not worrying about list lengths
或者如果 的元素A不是顺序的并且不担心列表长度
[(j, B[i % len(B)]) for i, j in enumerate(A)] if len(A) >= len(B) else \
[(A[i % len(A)], j) for i, j in enumerate(B)]
回答by Solomon Ucko
For a version that works with any finite number of potentially infinite iterables in any order:
对于以任何顺序处理任何有限数量的潜在无限迭代的版本：
from itertools import cycle, tee, zip_longest
def cyclical_zip(*iterables):
iterables_1, iterables_2 = zip(*map(tee, iterables)) # Allow proper iteration of iterators
for _, x in zip(
zip_longest(*iterables_1), # Limit by the length of the longest iterable
zip(*map(cycle, iterables_2))): # the cycling
yield x
assert list(cyclical_zip([1, 2, 3], 'abcd', 'xy')) == [(1, 'a', 'x'), (2, 'b', 'y'), (3, 'c', 'x'), (1, 'd', 'y')] # An example and test case
回答by Dane White
Solution for an arbitrary number of iterables, and you don't know which one is longest (also allowing a default for any empty iterables):
任意数量的迭代的解决方案，你不知道哪个最长（也允许任何空迭代的默认值）：
from itertools import cycle, zip_longest
def zip_cycle(*iterables, empty_default=None):
cycles = [cycle(i) for i in iterables]
for _ in zip_longest(*iterables):
yield tuple(next(i, empty_default) for i in cycles)
for i in zip_cycle(range(2), range(5), ['a', 'b', 'c'], []):
print(i)
Outputs:
输出：
(0, 0, 'a', None)
(1, 1, 'b', None)
(0, 2, 'c', None)
(1, 3, 'a', None)
(0, 4, 'b', None)
回答by Gajendra D Ambi
d1=['one','two','three']
d2=[1,2,3,4,5]
Zip
压缩
zip(d1,d2)
<zip object at 0x05E494B8>
list of zip
邮编清单
list(zip(d1,d2))
dictionary of list of zip
zip 目录字典
{'one': 1, 'two': 2, 'three': 3}
Note: Python 3.7+
注意：Python 3.7+
