Given a list
给出一个清单
l = [1, 7, 3, 5]
I want to iterate over all pairs of consecutive list items (1,7), (7,3), (3,5)
, i.e.
我想迭代所有连续列表项(1,7),(7,3),(3,5),即
for i in xrange(len(l) - 1):
x = l[i]
y = l[i + 1]
# do something
I would like to do this in a more compact way, like
我想以更紧凑的方式做到这一点,比如
for x, y in someiterator(l): ...
Is there a way to do do this using builtin Python iterators? I'm sure the itertools
module should have a solution, but I just can't figure it out.
有没有办法使用内置的Python迭代器来做到这一点?我确定itertools模块应该有一个解决方案,但我无法弄明白。
5 个解决方案
#1
69
Just use zip
只需使用拉链
>>> l = [1, 7, 3, 5]
>>> for first, second in zip(l, l[1:]):
... print first, second
...
1 7
7 3
3 5
As suggested you might consider using the izip
function in itertools
for very long lists where you don't want to create a new list.
正如所建议的那样,您可以考虑在itertools中使用izip函数,用于您不想创建新列表的很长列表。
import itertools
for first, second in itertools.izip(l, l[1:]):
...
#2
26
Look at pairwise
at itertools recipes: http://docs.python.org/2/library/itertools.html#recipes
在itertools食谱中看看pairwise:http://docs.python.org/2/library/itertools.html#recipes
Quoting from there:
引自那里:
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return izip(a, b)
A General Version
一般版本
A general version, that yields tuples of any given positive natural size, may look like that:
产生任何给定正自然大小的元组的通用版本可能如下所示:
def nwise(iterable, n=2):
iters = tee(iterable, n)
for i, it in enumerate(iters):
next(islice(it, i, i), None)
return izip(*iters)
#3
7
I would create a generic grouper
generator, like this
我会像这样创建一个通用的石斑鱼发生器
def grouper(input_list, n = 2):
for i in xrange(len(input_list) - (n - 1)):
yield input_list[i:i+n]
Sample run 1
样品运行1
for first, second in grouper([1, 7, 3, 5, 6, 8], 2):
print first, second
Output
产量
1 7
7 3
3 5
5 6
6 8
Sample run 1
样品运行1
for first, second, third in grouper([1, 7, 3, 5, 6, 8], 3):
print first, second, third
Output
产量
1 7 3
7 3 5
3 5 6
5 6 8
#4
0
You could use a zip
.
你可以使用拉链。
>>> list(zip(range(5), range(2, 6)))
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
Just like a zipper, it creates pairs. So, to to mix your two lists, you get:
就像拉链一样,它会产生成对。因此,要混合两个列表,您将得到:
>>> l = [1,7,3,5]
>>> list(zip(l[:-1], l[1:]))
[(1, 7), (7, 3), (3, 5)]
Then iterating goes like
然后迭代就像
for x, y in zip(l[:-1], l[1:]):
pass
#5
-1
What's below is very simple/readable and does this job, also likely the most efficient.
下面的内容非常简单/可读并且完成这项工作,也可能是最有效的。
Convert list into generator (or better start with an iterator to star with):
将列表转换为生成器(或者更好地从迭代器开始加注星标):
gen = (x for x in l)
Convert it into pairs:
将其转换为成对:
[(x, gen.next()) for x in gen]
That's all you need.
这就是你所需要的一切。
Of course, better make it a generator too and read from it as you need:
当然,最好把它变成一个发电机并根据需要从中读取:
( (x, gen.next()) for x in gen)
#1
69
Just use zip
只需使用拉链
>>> l = [1, 7, 3, 5]
>>> for first, second in zip(l, l[1:]):
... print first, second
...
1 7
7 3
3 5
As suggested you might consider using the izip
function in itertools
for very long lists where you don't want to create a new list.
正如所建议的那样,您可以考虑在itertools中使用izip函数,用于您不想创建新列表的很长列表。
import itertools
for first, second in itertools.izip(l, l[1:]):
...
#2
26
Look at pairwise
at itertools recipes: http://docs.python.org/2/library/itertools.html#recipes
在itertools食谱中看看pairwise:http://docs.python.org/2/library/itertools.html#recipes
Quoting from there:
引自那里:
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return izip(a, b)
A General Version
一般版本
A general version, that yields tuples of any given positive natural size, may look like that:
产生任何给定正自然大小的元组的通用版本可能如下所示:
def nwise(iterable, n=2):
iters = tee(iterable, n)
for i, it in enumerate(iters):
next(islice(it, i, i), None)
return izip(*iters)
#3
7
I would create a generic grouper
generator, like this
我会像这样创建一个通用的石斑鱼发生器
def grouper(input_list, n = 2):
for i in xrange(len(input_list) - (n - 1)):
yield input_list[i:i+n]
Sample run 1
样品运行1
for first, second in grouper([1, 7, 3, 5, 6, 8], 2):
print first, second
Output
产量
1 7
7 3
3 5
5 6
6 8
Sample run 1
样品运行1
for first, second, third in grouper([1, 7, 3, 5, 6, 8], 3):
print first, second, third
Output
产量
1 7 3
7 3 5
3 5 6
5 6 8
#4
0
You could use a zip
.
你可以使用拉链。
>>> list(zip(range(5), range(2, 6)))
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
Just like a zipper, it creates pairs. So, to to mix your two lists, you get:
就像拉链一样,它会产生成对。因此,要混合两个列表,您将得到:
>>> l = [1,7,3,5]
>>> list(zip(l[:-1], l[1:]))
[(1, 7), (7, 3), (3, 5)]
Then iterating goes like
然后迭代就像
for x, y in zip(l[:-1], l[1:]):
pass
#5
-1
What's below is very simple/readable and does this job, also likely the most efficient.
下面的内容非常简单/可读并且完成这项工作,也可能是最有效的。
Convert list into generator (or better start with an iterator to star with):
将列表转换为生成器(或者更好地从迭代器开始加注星标):
gen = (x for x in l)
Convert it into pairs:
将其转换为成对:
[(x, gen.next()) for x in gen]
That's all you need.
这就是你所需要的一切。
Of course, better make it a generator too and read from it as you need:
当然,最好把它变成一个发电机并根据需要从中读取:
( (x, gen.next()) for x in gen)