在python中混合两个列表的循环方法

Question

如果输入是

round_robin(range(5), "hello")

我需要输出为

[0, 'h', 1, 'e', 2, 'l', 3, 'l', 4, 'o']

我试过了

def round_robin(*seqs):
list1=[]
length=len(seqs)
list1= cycle(iter(items).__name__ for items in seqs)
while length:
    try:
        for x in list1:
            yield x
    except StopIteration:
        length -= 1

pass

但它给出了错误

AttributeError: 'listiterator' object has no attribute '__name__'

如何修改代码以获得想要的输出？

Answer 1

您可以使用zip 函数，然后使用列表理解将结果展平，就像这样

def round_robin(first, second):
    return[item for items in zip(first, second) for item in items]
print round_robin(range(5), "hello")

输出

[0, 'h', 1, 'e', 2, 'l', 3, 'l', 4, 'o']

zip 函数将两个可迭代对象的值分组，如下所示

print zip(range(5), "hello") # [(0, 'h'), (1, 'e'), (2, 'l'), (3, 'l'), (4, 'o')]

我们取出每一个元组并通过列表理解将其展平。

但正如@Ashwini Chaudhary 建议的那样，使用roundrobin receipe from the docs

from itertools import cycle
from itertools import islice
def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    pending = len(iterables)
    nexts = cycle(iter(it).next for it in iterables)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

print list(roundrobin(range(5), "hello"))

Answer 2

您可以在此处找到一系列迭代配方：http://docs.python.org/2.7/library/itertools.html#recipes

from itertools import islice, cycle


def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    pending = len(iterables)
    nexts = cycle(iter(it).next for it in iterables)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))


print list(roundrobin(range(5), "hello"))

编辑：Python 3

https://docs.python.org/3/library/itertools.html#itertools-recipes

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    num_active = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while num_active:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            num_active -= 1
            nexts = cycle(islice(nexts, num_active))

print list(roundrobin(range(5), "hello"))

Answer 3

您可以利用itertools.chain（解开元组）和itertools.izip（转置元素以创建交错模式）来创建结果

>>> from itertools import izip, chain
>>> list(chain.from_iterable(izip(range(5), "hello")))
[0, 'h', 1, 'e', 2, 'l', 3, 'l', 4, 'o']

如果字符串长度不等，请使用 izip_longest 和填充值（最好是空字符串）

Answer 4

Python 2 和 Python 3 的两个 itertools roundrobin 配方的混合如下所示：

from itertools import islice, cycle

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    num_active = len(iterables)
    try:
        iter([]).__next__  # test attribute
        nexts = cycle(iter(it).__next__ for it in iterables)
    except AttributeError:  # Python 2 behavior
        nexts = cycle(iter(it).next for it in iterables)
    while num_active:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            # Remove the iterator we just exhausted from the cycle.
            num_active -= 1
            nexts = cycle(islice(nexts, num_active))

print(list(roundrobin(range(5), "hello")))

Answer 5

list(roundrobin('ABC', 'D', 'EF'))

输出： ['A', 'D', 'E', 'B', 'F', 'C']

def roundrobin(*iterables):
    sentinel = object()
    from itertools import chain
    try:
        from itertools import izip_longest as zip_longest
    except:
        from itertools import zip_longest 
    return (x for x in chain(*zip_longest(fillvalue=sentinel, *iterables)) if x is not sentinel)  

Answer 6

对于寻找 Python 3 的任何人，请使用此

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    num_active = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while num_active:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            # Remove the iterator we just exhausted from the cycle.
            num_active -= 1
            nexts = cycle(islice(nexts, num_active))

不同之处在于 Python 3 的迭代器有 __next__() 而不是 next()。 https://docs.python.org/3/library/itertools.html#recipes

Answer 7

从 itertools 导入周期

A = [[1,2,3],[4,5,6],[7]]

B = [[8],[9,10,11],[12,13]]

对于 A 中的 p：

max1 = len(p) if  max1 <len(p) else max1

对于 B 中的 p：

max1 = len(p) if  max1 <len(p) else max1

i = len(A)

j = 0

C = []

list_num = cycle(k for k in range(i))

对于 list_num 中的 x：

j += 1

if j == i*3:

    break


if A[x]:

    C.append(A[x].pop(0))

if B[x]:

    C.append(B[x].pop(0)) 

输出=====> [1, 8, 4, 9, 7, 12, 2, 5, 10, 13, 3, 6, 11]