遍历列表中的每两个元素

问题:遍历列表中的每两个元素

如何进行for循环或列表理解,以便每次迭代都给我两个元素?

l = [1,2,3,4,5,6]

for i,k in ???:
    print str(i), '+', str(k), '=', str(i+k)

输出:

1+2=3
3+4=7
5+6=11

How do I make a for loop or a list comprehension so that every iteration gives me two elements?

l = [1,2,3,4,5,6]

for i,k in ???:
    print str(i), '+', str(k), '=', str(i+k)

Output:

1+2=3
3+4=7
5+6=11

回答 0

您需要一个pairwise()(或grouped())实施。

对于Python 2:

from itertools import izip

def pairwise(iterable):
    "s -> (s0, s1), (s2, s3), (s4, s5), ..."
    a = iter(iterable)
    return izip(a, a)

for x, y in pairwise(l):
   print "%d + %d = %d" % (x, y, x + y)

或更笼统地说:

from itertools import izip

def grouped(iterable, n):
    "s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), (s2n,s2n+1,s2n+2,...s3n-1), ..."
    return izip(*[iter(iterable)]*n)

for x, y in grouped(l, 2):
   print "%d + %d = %d" % (x, y, x + y)

在Python 3中,您可以替换izip为内置zip()函数,然后删除import

所有信贷蒂诺他的回答我的问题,我发现这是非常有效的,因为它只是在列表上循环一次,并在此过程中不会产生任何不必要的名单。

注意:不要将其与Python自己的文档中pairwise配方混淆,后者由@lazyr在评论中指出。itertoolss -> (s0, s1), (s1, s2), (s2, s3), ...

对于想要在Python 3上使用mypy进行类型检查的用户而言,几乎没有什么附加的

from typing import Iterable, Tuple, TypeVar

T = TypeVar("T")

def grouped(iterable: Iterable[T], n=2) -> Iterable[Tuple[T, ...]]:
    """s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), ..."""
    return zip(*[iter(iterable)] * n)

You need a pairwise() (or grouped()) implementation.

For Python 2:

from itertools import izip

def pairwise(iterable):
    "s -> (s0, s1), (s2, s3), (s4, s5), ..."
    a = iter(iterable)
    return izip(a, a)

for x, y in pairwise(l):
   print "%d + %d = %d" % (x, y, x + y)

Or, more generally:

from itertools import izip

def grouped(iterable, n):
    "s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), (s2n,s2n+1,s2n+2,...s3n-1), ..."
    return izip(*[iter(iterable)]*n)

for x, y in grouped(l, 2):
   print "%d + %d = %d" % (x, y, x + y)

In Python 3, you can replace izip with the built-in zip() function, and drop the import.

All credit to martineau for his answer to my question, I have found this to be very efficient as it only iterates once over the list and does not create any unnecessary lists in the process.

N.B: This should not be confused with the pairwise recipe in Python’s own itertools documentation, which yields s -> (s0, s1), (s1, s2), (s2, s3), ..., as pointed out by @lazyr in the comments.

Little addition for those who would like to do type checking with mypy on Python 3:

from typing import Iterable, Tuple, TypeVar

T = TypeVar("T")

def grouped(iterable: Iterable[T], n=2) -> Iterable[Tuple[T, ...]]:
    """s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), ..."""
    return zip(*[iter(iterable)] * n)

回答 1

那么你需要两个元素的元组,所以

data = [1,2,3,4,5,6]
for i,k in zip(data[0::2], data[1::2]):
    print str(i), '+', str(k), '=', str(i+k)

哪里:

  • data[0::2] 意味着创建元素的子集, (index % 2 == 0)
  • zip(x,y) 根据x和y集合的相同索引元素创建一个元组集合。

Well you need tuple of 2 elements, so

data = [1,2,3,4,5,6]
for i,k in zip(data[0::2], data[1::2]):
    print str(i), '+', str(k), '=', str(i+k)

Where:

  • data[0::2] means create subset collection of elements that (index % 2 == 0)
  • zip(x,y) creates a tuple collection from x and y collections same index elements.

回答 2

>>> l = [1,2,3,4,5,6]

>>> zip(l,l[1:])
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]

>>> zip(l,l[1:])[::2]
[(1, 2), (3, 4), (5, 6)]

>>> [a+b for a,b in zip(l,l[1:])[::2]]
[3, 7, 11]

>>> ["%d + %d = %d" % (a,b,a+b) for a,b in zip(l,l[1:])[::2]]
['1 + 2 = 3', '3 + 4 = 7', '5 + 6 = 11']
>>> l = [1,2,3,4,5,6]

>>> zip(l,l[1:])
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]

>>> zip(l,l[1:])[::2]
[(1, 2), (3, 4), (5, 6)]

>>> [a+b for a,b in zip(l,l[1:])[::2]]
[3, 7, 11]

>>> ["%d + %d = %d" % (a,b,a+b) for a,b in zip(l,l[1:])[::2]]
['1 + 2 = 3', '3 + 4 = 7', '5 + 6 = 11']

回答 3

一个简单的解决方案。

l = [1、2、3、4、5、6]

对于范围(0,len(l),2)中的i:
    打印str(l [i]),'+',str(l [i + 1]),'=',str(l [i] + l [i + 1])

A simple solution.

l = [1, 2, 3, 4, 5, 6]

for i in range(0, len(l), 2):
    print str(l[i]), '+', str(l[i + 1]), '=', str(l[i] + l[i + 1])

回答 4

尽管使用的所有答案zip都是正确的,但我发现自己实现功能会导致代码更具可读性:

def pairwise(it):
    it = iter(it)
    while True:
        try:
            yield next(it), next(it)
        except StopIteration:
            # no more elements in the iterator
            return

it = iter(it)部分确保它it实际上是一个迭代器,而不仅仅是一个迭代器。如果it已经是迭代器,则此行为无操作。

用法:

for a, b in pairwise([0, 1, 2, 3, 4, 5]):
    print(a + b)

While all the answers using zip are correct, I find that implementing the functionality yourself leads to more readable code:

def pairwise(it):
    it = iter(it)
    while True:
        try:
            yield next(it), next(it)
        except StopIteration:
            # no more elements in the iterator
            return

The it = iter(it) part ensures that it is actually an iterator, not just an iterable. If it already is an iterator, this line is a no-op.

Usage:

for a, b in pairwise([0, 1, 2, 3, 4, 5]):
    print(a + b)

回答 5

我希望这将是更优雅的方法。

a = [1,2,3,4,5,6]
zip(a[::2], a[1::2])

[(1, 2), (3, 4), (5, 6)]

I hope this will be even more elegant way of doing it.

a = [1,2,3,4,5,6]
zip(a[::2], a[1::2])

[(1, 2), (3, 4), (5, 6)]

回答 6

如果您对性能感兴趣,我做了一个小型基准测试(使用我的library simple_benchmark)来比较解决方案的性能,并且我从一个软件包中添加了一个函数:iteration_utilities.grouper

from iteration_utilities import grouper
import matplotlib as mpl
from simple_benchmark import BenchmarkBuilder

bench = BenchmarkBuilder()

@bench.add_function()
def Johnsyweb(l):
    def pairwise(iterable):
        "s -> (s0, s1), (s2, s3), (s4, s5), ..."
        a = iter(iterable)
        return zip(a, a)

    for x, y in pairwise(l):
        pass

@bench.add_function()
def Margus(data):
    for i, k in zip(data[0::2], data[1::2]):
        pass

@bench.add_function()
def pyanon(l):
    list(zip(l,l[1:]))[::2]

@bench.add_function()
def taskinoor(l):
    for i in range(0, len(l), 2):
        l[i], l[i+1]

@bench.add_function()
def mic_e(it):
    def pairwise(it):
        it = iter(it)
        while True:
            try:
                yield next(it), next(it)
            except StopIteration:
                return

    for a, b in pairwise(it):
        pass

@bench.add_function()
def MSeifert(it):
    for item1, item2 in grouper(it, 2):
        pass

bench.use_random_lists_as_arguments(sizes=[2**i for i in range(1, 20)])
benchmark_result = bench.run()
mpl.rcParams['figure.figsize'] = (8, 10)
benchmark_result.plot_both(relative_to=MSeifert)

因此,如果您想要没有外部依赖项的最快解决方案,则可能应该使用Johnysweb给出的方法(在撰写本文时,它是最受支持和接受的答案)。

如果您不介意其他依赖项,那么grouperfrom iteration_utilities可能会更快。

其他想法

其中一些方法有一些限制,这里没有讨论。

例如,一些解决方案仅适用于序列(即列表,字符串等),例如Margus / pyanon / taskinoor解决方案使用索引,而其他解决方案可用于任何可迭代的(即序列生成器,迭代器),例如Johnysweb / mic_e /我的解决方案。

然后,Johnysweb还提供了一种解决方案,该解决方案适用于除2以外的其他大小,而其他答案则不行(好的,iteration_utilities.grouper还允许将元素数设置为“ group”)。

还有一个问题是,如果列表中的元素数量奇数,应该怎么办。剩余的物品应否开除?是否应该填充列表以使其大小均匀?剩余的物品是否应作为单一物品退还?另一个答案不能直接解决这一点,但是,如果我没有忽略任何内容,那么他​​们都会遵循将剩余项目排除的方法(taskinoor答案除外-这实际上会引发异常)。

有了grouper你可以决定你想要做什么:

>>> from iteration_utilities import grouper

>>> list(grouper([1, 2, 3], 2))  # as single
[(1, 2), (3,)]

>>> list(grouper([1, 2, 3], 2, truncate=True))  # ignored
[(1, 2)]

>>> list(grouper([1, 2, 3], 2, fillvalue=None))  # padded
[(1, 2), (3, None)]

In case you’re interested in the performance, I did a small benchmark (using my library simple_benchmark) to compare the performance of the solutions and I included a function from one of my packages: iteration_utilities.grouper

from iteration_utilities import grouper
import matplotlib as mpl
from simple_benchmark import BenchmarkBuilder

bench = BenchmarkBuilder()

@bench.add_function()
def Johnsyweb(l):
    def pairwise(iterable):
        "s -> (s0, s1), (s2, s3), (s4, s5), ..."
        a = iter(iterable)
        return zip(a, a)

    for x, y in pairwise(l):
        pass

@bench.add_function()
def Margus(data):
    for i, k in zip(data[0::2], data[1::2]):
        pass

@bench.add_function()
def pyanon(l):
    list(zip(l,l[1:]))[::2]

@bench.add_function()
def taskinoor(l):
    for i in range(0, len(l), 2):
        l[i], l[i+1]

@bench.add_function()
def mic_e(it):
    def pairwise(it):
        it = iter(it)
        while True:
            try:
                yield next(it), next(it)
            except StopIteration:
                return

    for a, b in pairwise(it):
        pass

@bench.add_function()
def MSeifert(it):
    for item1, item2 in grouper(it, 2):
        pass

bench.use_random_lists_as_arguments(sizes=[2**i for i in range(1, 20)])
benchmark_result = bench.run()
mpl.rcParams['figure.figsize'] = (8, 10)
benchmark_result.plot_both(relative_to=MSeifert)

So if you want the fastest solution without external dependencies you probably should just use the approach given by Johnysweb (at the time of writing it’s the most upvoted and accepted answer).

If you don’t mind the additional dependency then the grouper from iteration_utilities will probably be a bit faster.

Additional thoughts

Some of the approaches have some restrictions, that haven’t been discussed here.

For example a few solutions only work for sequences (that is lists, strings, etc.), for example Margus/pyanon/taskinoor solutions which uses indexing while other solutions work on any iterable (that is sequences and generators, iterators) like Johnysweb/mic_e/my solutions.

Then Johnysweb also provided a solution that works for other sizes than 2 while the other answers don’t (okay, the iteration_utilities.grouper also allows setting the number of elements to “group”).

Then there is also the question about what should happen if there is an odd number of elements in the list. Should the remaining item be dismissed? Should the list be padded to make it even sized? Should the remaining item be returned as single? The other answer don’t address this point directly, however if I haven’t overlooked anything they all follow the approach that the remaining item should be dismissed (except for taskinoors answer – that will actually raise an Exception).

With grouper you can decide what you want to do:

>>> from iteration_utilities import grouper

>>> list(grouper([1, 2, 3], 2))  # as single
[(1, 2), (3,)]

>>> list(grouper([1, 2, 3], 2, truncate=True))  # ignored
[(1, 2)]

>>> list(grouper([1, 2, 3], 2, fillvalue=None))  # padded
[(1, 2), (3, None)]

回答 7

使用zipiter一起命令:

我发现此解决方案使用iter起来非常优雅:

it = iter(l)
list(zip(it, it))
# [(1, 2), (3, 4), (5, 6)]

我在Python 3 zip文档中找到了该文件

it = iter(l)
print(*(f'{u} + {v} = {u+v}' for u, v in zip(it, it)), sep='\n')

# 1 + 2 = 3
# 3 + 4 = 7
# 5 + 6 = 11

一次概括到N元素:

N = 2
list(zip(*([iter(l)] * N)))
# [(1, 2), (3, 4), (5, 6)]

Use the zip and iter commands together:

I find this solution using iter to be quite elegant:

it = iter(l)
list(zip(it, it))
# [(1, 2), (3, 4), (5, 6)]

Which I found in the Python 3 zip documentation.

it = iter(l)
print(*(f'{u} + {v} = {u+v}' for u, v in zip(it, it)), sep='\n')

# 1 + 2 = 3
# 3 + 4 = 7
# 5 + 6 = 11

To generalise to N elements at a time:

N = 2
list(zip(*([iter(l)] * N)))
# [(1, 2), (3, 4), (5, 6)]

回答 8

for (i, k) in zip(l[::2], l[1::2]):
    print i, "+", k, "=", i+k

zip(*iterable) 返回一个元组以及每个可迭代的下一个元素。

l[::2] 返回列表的1、3、5等元素:第一个冒号表示切片从开头开始,因为后面没有数字,仅当您要在切片中使用“ step”时,才需要第二个冒号’(在这种情况下为2)。

l[1::2]做同样的事情,但从列表的第二个元素开始,因此它返回原始列表的第二,第四,第六等元素。

for (i, k) in zip(l[::2], l[1::2]):
    print i, "+", k, "=", i+k

zip(*iterable) returns a tuple with the next element of each iterable.

l[::2] returns the 1st, the 3rd, the 5th, etc. element of the list: the first colon indicates that the slice starts at the beginning because there’s no number behind it, the second colon is only needed if you want a ‘step in the slice’ (in this case 2).

l[1::2] does the same thing but starts in the second element of the lists so it returns the 2nd, the 4th, 6th, etc. element of the original list.


回答 9

带包装:

l = [1,2,3,4,5,6]
while l:
    i, k, *l = l
    print(str(i), '+', str(k), '=', str(i+k))

With unpacking:

l = [1,2,3,4,5,6]
while l:
    i, k, *l = l
    print(str(i), '+', str(k), '=', str(i+k))

回答 10

对于任何人都可能有帮助,这是解决类似问题的解决方案,但对具有重叠对(而不是互斥对)。

从Python itertools文档中

from itertools import izip

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

或更笼统地说:

from itertools import izip

def groupwise(iterable, n=2):
    "s -> (s0,s1,...,sn-1), (s1,s2,...,sn), (s2,s3,...,sn+1), ..."
    t = tee(iterable, n)
    for i in range(1, n):
        for j in range(0, i):
            next(t[i], None)
    return izip(*t)

For anyone it might help, here is a solution to a similar problem but with overlapping pairs (instead of mutually exclusive pairs).

From the Python itertools documentation:

from itertools import izip

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

Or, more generally:

from itertools import izip

def groupwise(iterable, n=2):
    "s -> (s0,s1,...,sn-1), (s1,s2,...,sn), (s2,s3,...,sn+1), ..."
    t = tee(iterable, n)
    for i in range(1, n):
        for j in range(0, i):
            next(t[i], None)
    return izip(*t)

回答 11

您可以使用more_itertools软件包。

import more_itertools

lst = range(1, 7)
for i, j in more_itertools.chunked(lst, 2):
    print(f'{i} + {j} = {i+j}')

you can use more_itertools package.

import more_itertools

lst = range(1, 7)
for i, j in more_itertools.chunked(lst, 2):
    print(f'{i} + {j} = {i+j}')

回答 12

我需要将列表除以数字并固定不变。

l = [1,2,3,4,5,6]

def divideByN(data, n):
        return [data[i*n : (i+1)*n] for i in range(len(data)//n)]  

>>> print(divideByN(l,2))
[[1, 2], [3, 4], [5, 6]]

>>> print(divideByN(l,3))
[[1, 2, 3], [4, 5, 6]]

I need to divide a list by a number and fixed like this.

l = [1,2,3,4,5,6]

def divideByN(data, n):
        return [data[i*n : (i+1)*n] for i in range(len(data)//n)]  

>>> print(divideByN(l,2))
[[1, 2], [3, 4], [5, 6]]

>>> print(divideByN(l,3))
[[1, 2, 3], [4, 5, 6]]

回答 13

有很多方法可以做到这一点。例如:

lst = [1,2,3,4,5,6]
[(lst[i], lst[i+1]) for i,_ in enumerate(lst[:-1])]    
>>>[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]

[i for i in zip(*[iter(lst)]*2)]    
>>>[(1, 2), (3, 4), (5, 6)]

There are many ways to do that. For example:

lst = [1,2,3,4,5,6]
[(lst[i], lst[i+1]) for i,_ in enumerate(lst[:-1])]    
>>>[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]

[i for i in zip(*[iter(lst)]*2)]    
>>>[(1, 2), (3, 4), (5, 6)]

回答 14

认为这是分享我对n> 2的概括的一个好地方,这只是可迭代项上的滑动窗口:

def sliding_window(iterable, n):
    its = [ itertools.islice(iter, i, None) 
            for i, iter
            in enumerate(itertools.tee(iterable, n)) ]                               

    return itertools.izip(*its)

Thought that this is a good place to share my generalization of this for n>2, which is just a sliding window over an iterable:

def sliding_window(iterable, n):
    its = [ itertools.islice(iter, i, None) 
            for i, iter
            in enumerate(itertools.tee(iterable, n)) ]                               

    return itertools.izip(*its)

回答 15

这个问题的标题具有误导性,您似乎正在寻找连续对,但如果要遍历所有可能对的集合,则可以使用以下方法:

for i,v in enumerate(items[:-1]):
        for u in items[i+1:]:

The title of this question is misleading, you seem to be looking for consecutive pairs, but if you want to iterate over the set of all possible pairs than this will work :

for i,v in enumerate(items[:-1]):
        for u in items[i+1:]:

回答 16

使用输入法,因此您可以使用mypy静态分析工具验证数据:

from typing import Iterator, Any, Iterable, TypeVar, Tuple

T_ = TypeVar('T_')
Pairs_Iter = Iterator[Tuple[T_, T_]]

def legs(iterable: Iterator[T_]) -> Pairs_Iter:
    begin = next(iterable)
    for end in iterable:
        yield begin, end
        begin = end

Using typing so you can verify data using mypy static analysis tool:

from typing import Iterator, Any, Iterable, TypeVar, Tuple

T_ = TypeVar('T_')
Pairs_Iter = Iterator[Tuple[T_, T_]]

def legs(iterable: Iterator[T_]) -> Pairs_Iter:
    begin = next(iterable)
    for end in iterable:
        yield begin, end
        begin = end

回答 17

一种简单的方法:

[(a[i],a[i+1]) for i in range(0,len(a),2)]

如果您的数组是a并且您想成对迭代它,这将很有用。要遍历三胞胎或更多,只需更改“ range”步骤命令,例如:

[(a[i],a[i+1],a[i+2]) for i in range(0,len(a),3)]

(如果数组长度和步长不合适,则必须处理多余的值)

A simplistic approach:

[(a[i],a[i+1]) for i in range(0,len(a),2)]

this is useful if your array is a and you want to iterate on it by pairs. To iterate on triplets or more just change the “range” step command, for example:

[(a[i],a[i+1],a[i+2]) for i in range(0,len(a),3)]

(you have to deal with excess values if your array length and the step do not fit)


回答 18

在这里,我们可以找到alt_elem适合您的for循环的方法。

def alt_elem(list, index=2):
    for i, elem in enumerate(list, start=1):
        if not i % index:
           yield tuple(list[i-index:i])


a = range(10)
for index in [2, 3, 4]:
    print("With index: {0}".format(index))
    for i in alt_elem(a, index):
       print(i)

输出:

With index: 2
(0, 1)
(2, 3)
(4, 5)
(6, 7)
(8, 9)
With index: 3
(0, 1, 2)
(3, 4, 5)
(6, 7, 8)
With index: 4
(0, 1, 2, 3)
(4, 5, 6, 7)

注意:考虑到func中执行的操作,上述解决方案可能不是有效的。

Here we can have alt_elem method which can fit in your for loop.

def alt_elem(list, index=2):
    for i, elem in enumerate(list, start=1):
        if not i % index:
           yield tuple(list[i-index:i])


a = range(10)
for index in [2, 3, 4]:
    print("With index: {0}".format(index))
    for i in alt_elem(a, index):
       print(i)

Output:

With index: 2
(0, 1)
(2, 3)
(4, 5)
(6, 7)
(8, 9)
With index: 3
(0, 1, 2)
(3, 4, 5)
(6, 7, 8)
With index: 4
(0, 1, 2, 3)
(4, 5, 6, 7)

Note: Above solution might not be efficient considering operations performed in func.


回答 19

a_list = [1,2,3,4,5,6]
empty_list = [] 
for i in range(0,len(a_list),2):
   empty_list.append(a_list[i]+a_list[i+1])   
print(empty_list)
a_list = [1,2,3,4,5,6]
empty_list = [] 
for i in range(0,len(a_list),2):
   empty_list.append(a_list[i]+a_list[i+1])   
print(empty_list)