問題描述
我在 Python 中有兩個可迭代對象，我想成對檢查它們:
I have two iterables in Python, and I want to go over them in pairs:
foo = (1, 2, 3)
bar = (4, 5, 6)
for (f, b) in some_iterator(foo, bar):
print("f: ", f, "; b: ", b)
它應該導致:
f: 1; b: 4
f: 2; b: 5
f: 3; b: 6
一種方法是迭代索引:
for i in range(len(foo)):
print("f: ", foo[i], "; b: ", bar[i])
但這對我來說似乎有些不合時宜.有沒有更好的方法?
But that seems somewhat unpythonic to me. Is there a better way to do it?
推薦答案
Python 3
for f, b in zip(foo, bar):
print(f, b)
zip 在 foo 或 bar 中較短者停止時停止.
zip stops when the shorter of foo or bar stops.
在 Python 3 中，zip返回元組的迭代器，如 Python2 中的 itertools.izip.獲取列表元組，使用 list(zip(foo, bar)).并壓縮直到兩個迭代器都筋疲力盡，你會用itertools.zip_longest.
In Python 3, zip
returns an iterator of tuples, like itertools.izip in Python2. To get a list
of tuples, use list(zip(foo, bar)). And to zip until both iterators are
exhausted, you would use
itertools.zip_longest.
在 Python 2 中，zip返回一個元組列表.當 foo 和 bar 不是很大時，這很好.如果它們都是巨大的，那么形成 zip(foo,bar) 是不必要的巨大臨時變量，應替換為 itertools.izip 或itertools.izip_longest，它返回一個迭代器而不是一個列表.
In Python 2, zip
returns a list of tuples. This is fine when foo and bar are not massive. If they are both massive then forming zip(foo,bar) is an unnecessarily massive
temporary variable, and should be replaced by itertools.izip or
itertools.izip_longest, which returns an iterator instead of a list.
import itertools
for f,b in itertools.izip(foo,bar):
print(f,b)
for f,b in itertools.izip_longest(foo,bar):
print(f,b)
izip 在 foo 或 bar 用盡時停止.當 foo 和 bar 都用盡時，izip_longest 停止.當較短的迭代器用盡時，izip_longest 會產生一個元組，其中 None 在對應于該迭代器的位置.您還可以根據需要設置除 None 之外的其他 fillvalue.查看全文.
izip stops when either foo or bar is exhausted.
izip_longest stops when both foo and bar are exhausted.
When the shorter iterator(s) are exhausted, izip_longest yields a tuple with None in the position corresponding to that iterator. You can also set a different fillvalue besides None if you wish. See here for the full story.
還要注意 zip 及其類似 zip 的 brethen 可以接受任意數量的可迭代對象作為參數.例如，
Note also that zip and its zip-like brethen can accept an arbitrary number of iterables as arguments. For example,
for num, cheese, color in zip([1,2,3], ['manchego', 'stilton', 'brie'],
['red', 'blue', 'green']):
print('{} {} {}'.format(num, color, cheese))
打印
1 red manchego
2 blue stilton
3 green brie
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