在python中每n个项目拆分一个生成器/可迭代(splitEvery) [英] split a generator/iterable every n items in python (splitEvery)

问题描述

我正在尝试用 Python 编写 Haskell 函数splitEvery".这是它的定义:

splitEvery :: Int ->[e] ->[[e]]@'splitEvery' n@ 将列表分成长度为 n 的部分.最后如果@n@ 不均分长度，则块会更短列表.

这个的基本版本工作正常，但我想要一个可以处理生成器表达式、列表和迭代器的版本.并且，如果有一个生成器作为输入，它应该返回一个生成器作为输出！

测试

# 不应使用生成器或列表进入无限循环splitEvery(itertools.count(), 10)splitEvery(范围(1000)，10)# 如果 n 不均分，最后一块必须更短断言 splitEvery(5, range(9)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]]# 应该给出与生成器相同的正确结果tmp = itertools.islice(itertools.count(), 10)断言列表(splitEvery(5, tmp)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

当前实施

这是我目前拥有的代码，但它不适用于简单列表.

def splitEvery_1(n, iterable):res = list(itertools.islice(iterable, n))而 len(res) != 0:产量资源res = list(itertools.islice(iterable, n))

这个不适用于生成器表达式(感谢 jellybean 修复它):

def splitEvery_2(n, iterable):返回 [iterable[i:i+n] for i in range(0, len(iterable), n)]

必须有一段简单的代码来进行拆分.我知道我可以有不同的功能，但似乎应该很容易做到.我可能被一个不重要的问题困住了，但这真的让我很烦恼.

---

它类似于来自 http://docs.python.org 的石斑鱼/library/itertools.html#itertools.groupby 但我不希望它填充额外的值.

def grouper(n, iterable, fillvalue=None):"石斑鱼(3, 'ABCDEFG', 'x') -->ABC DEF Gxx"args = [iter(iterable)] * n返回 izip_longest(fillvalue=fillvalue, *args)

它确实提到了一种截断最后一个值的方法.这也不是我想要的.

<块引用>

可迭代对象的从左到右的评估顺序是有保证的.这使得使用 izip(*[iter(s)]*n) 将数据系列聚类为 n 个长度的组成为可能.

list(izip(*[iter(range(9))]*5)) == [[0, 1, 2, 3, 4]]# 应该是 [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

解决方案

from itertools import islicedef split_every(n, iterable):我 = 迭代(可迭代)片 = 列表(islice(i，n))而片:屈服件片 = 列表(islice(i，n))

一些测试:

<预><代码>>>>列表(split_every(5，范围(9)))[[0, 1, 2, 3, 4], [5, 6, 7, 8]]>>>list(split_every(3, (x**2 for x in range(20))))[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81, 100, 121], [144, 169, 196], [225, 256, 289], [324, 361]]>>>[''.join(s) for s in split_every(6, 'Hello world')]['你好世界']>>>列表(split_every(100，[]))[]

I'm trying to write the Haskell function 'splitEvery' in Python. Here is it's definition:

splitEvery :: Int -> [e] -> [[e]]
    @'splitEvery' n@ splits a list into length-n pieces.  The last
    piece will be shorter if @n@ does not evenly divide the length of
    the list.

The basic version of this works fine, but I want a version that works with generator expressions, lists, and iterators. And, if there is a generator as an input it should return a generator as an output!

Tests

# should not enter infinite loop with generators or lists
splitEvery(itertools.count(), 10)
splitEvery(range(1000), 10)

# last piece must be shorter if n does not evenly divide
assert splitEvery(5, range(9)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

# should give same correct results with generators
tmp = itertools.islice(itertools.count(), 10)
assert list(splitEvery(5, tmp)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

Current Implementation

Here is the code I currently have but it doesn't work with a simple list.

def splitEvery_1(n, iterable):
    res = list(itertools.islice(iterable, n))
    while len(res) != 0:
        yield res
        res = list(itertools.islice(iterable, n))

This one doesn't work with a generator expression (thanks to jellybean for fixing it):

def splitEvery_2(n, iterable): 
    return [iterable[i:i+n] for i in range(0, len(iterable), n)]

There has to be a simple piece of code that does the splitting. I know I could just have different functions but it seems like it should be and easy thing to do. I'm probably getting stuck on an unimportant problem but it's really bugging me.

---

It is similar to grouper from http://docs.python.org/library/itertools.html#itertools.groupby but I don't want it to fill extra values.

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

It does mention a method that truncates the last value. This isn't what I want either.

The left-to-right evaluation order of the iterables is guaranteed. This makes possible an idiom for clustering a data series into n-length groups using izip(*[iter(s)]*n).

list(izip(*[iter(range(9))]*5)) == [[0, 1, 2, 3, 4]]
# should be [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

解决方案

from itertools import islice

def split_every(n, iterable):
    i = iter(iterable)
    piece = list(islice(i, n))
    while piece:
        yield piece
        piece = list(islice(i, n))

Some tests:

>>> list(split_every(5, range(9)))
[[0, 1, 2, 3, 4], [5, 6, 7, 8]]

>>> list(split_every(3, (x**2 for x in range(20))))
[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81, 100, 121], [144, 169, 196], [225, 256, 289], [324, 361]]

>>> [''.join(s) for s in split_every(6, 'Hello world')]
['Hello ', 'world']

>>> list(split_every(100, []))
[]

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