问题描述
我在 Python 中有两个可迭代对象,我想成对检查它们:
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)
它应该导致:
f: 1; b: 4
f: 2; b: 5
f: 3; b: 6
一种方法是迭代索引:
for i in range(len(foo)):
print("f: ", foo[i], "; b: ", bar[i])
但这对我来说似乎有些不合时宜.有没有更好的方法?
But that seems somewhat unpythonic to me. Is there a better way to do it?
推荐答案
Python 3
for f, b in zip(foo, bar):
print(f, b)
zip
在 foo
或 bar
中较短者停止时停止.
zip
stops when the shorter of foo
or bar
stops.
在 Python 3 中,zip代码>返回元组的迭代器,如 Python2 中的
itertools.izip
.获取列表元组,使用 list(zip(foo, bar))
.并压缩直到两个迭代器都筋疲力尽,你会用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代码>返回一个元组列表.当
foo
和 bar
不是很大时,这很好.如果它们都是巨大的,那么形成 zip(foo,bar)
是不必要的巨大临时变量,应替换为 itertools.izip
或itertools.izip_longest
,它返回一个迭代器而不是一个列表.
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
用尽时停止.当 foo
和 bar
都用尽时,izip_longest
停止.当较短的迭代器用尽时,izip_longest
会产生一个元组,其中 None
在对应于该迭代器的位置.您还可以根据需要设置除 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.
还要注意 zip
及其类似 zip
的 brethen 可以接受任意数量的可迭代对象作为参数.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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