問題描述

我需要一個 2D 循環(huán)，其中第一個循環(huán)使用迭代器，第二個循環(huán)使用生成器，但是這個簡單的函數(shù)無法工作，誰能幫忙檢查一下?

I need a 2D loop of which the first loop uses an iterator and the second uses a generator, but this simple function failed to work, can anyone help to check?

def alphabet(begin, end):
    for number in xrange(ord(begin), ord(end)+1):
        yield chr(number)

def test(a, b):
    for i in a:
        for j in b:
            print i, j

test(xrange(8, 10), alphabet('A', 'C'))

The result shows:
>>> 8 A
>>> 8 B
>>> 8 c

不知道為什么?如果有人可以提供幫助，請?zhí)崆爸轮x.

don't know why? thanks in advance if any one can help.

推薦答案

既然你要求澄清，我就多說一點；但真的 Ignacio 的回答總結得很好:您只能迭代生成器一次.您示例中的代碼嘗試對其進行三次迭代，a 中的每個值一次.

Since you've asked for clarification, I'll say a bit more; but really Ignacio's answer sums it up pretty well: you can only iterate over a generator once. The code in your example tries to iterate over it three times, once for each value in a.

要明白我的意思，請考慮這個簡單的例子:

To see what I mean, consider this simplistic example:

>>> def mygen(x):
...     i = 0
...     while i < x:
...         yield i
...         i += 1
... 
>>> mg = mygen(4)
>>> list(mg)
[0, 1, 2, 3]
>>> list(mg)
[]

當 mygen 被調用時，它會創(chuàng)建一個可以只迭代一次的對象.當您嘗試再次對其進行迭代時，您會得到一個空的可迭代對象.

When mygen is called, it creates an object which can be iterated over exactly once. When you try to iterate over it again, you get an empty iterable.

這意味著你必須重新調用 mygen，每次你想要迭代它`.所以換句話說(使用相當冗長的風格)......

This means you have to call mygen anew, every time you want to iterate over it`. So in other words (using a rather verbose style)...

>>> def make_n_lists(gen, gen_args, n):
...     list_of_lists = []
...     for _ in range(n):
...         list_of_lists.append(list(gen(*gen_args)))
...     return list_of_lists
... 
>>> make_n_lists(mygen, (3,), 3)
[[0, 1, 2], [0, 1, 2], [0, 1, 2]]

如果您想將參數(shù)綁定到生成器并將其作為無參數(shù)函數(shù)傳遞，您可以這樣做(使用更簡潔的樣式):

If you wanted to bind your arguments to your generator and pass that as an argumentless function, you could do this (using a more terse style):

>>> def make_n_lists(gen_func, n):
...     return [list(gen_func()) for _ in range(n)]
... 
>>> make_n_lists(lambda: mygen(3), 3)
[[0, 1, 2], [0, 1, 2], [0, 1, 2]]

lambda 只是定義了一個匿名函數(shù)；以上與此相同:

The lambda just defines an anonymous function; the above is identical to this:

>>> def call_mygen_with_3():
...     return mygen(3)
... 
>>> make_n_lists(call_mygen_with_3, 3)
[[0, 1, 2], [0, 1, 2], [0, 1, 2]]

這篇關于使用生成器和迭代器時 Python 多循環(huán)失敗的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！