[英]How can random.shuffle result in a KeyError?
I just got the error: 我刚收到错误:
Traceback (most recent call last):
File "./download_documents.py", line 153, in <module>
paragraphs, used_pages = find_pages(lang, to_extract)
File "./download_documents.py", line 67, in find_pages
random.shuffle(page_titles_queue)
File "/usr/lib/python2.7/random.py", line 291, in shuffle
x[i], x[j] = x[j], x[i]
KeyError: 1
Which confuses me quite a bit. 这让我很困惑。
random.shuffle
seems to work on zero-element lists and on one-element lists. random.shuffle
似乎适用于零元素列表和单元素列表。 page_titles_queue
is a list of tuples. page_titles_queue
是一个元组列表。 random.shuffle(page_titles_queue)
, there is page_titles_queue.pop()
, but that should not affect the shuffle. random.shuffle(page_titles_queue)
之后的两行,有page_titles_queue.pop()
,但这不应该影响shuffle。 Right? So what are possible reasons for the KeyError? 那么KeyError的可能原因是什么?
I use Python 2.7.12
on Ubuntu 16.04. 我在Ubuntu 16.04上使用
Python 2.7.12
。
random.shuffle
just exchanges items, the line where the exception happened makes this perfectly clear: random.shuffle
只是交换项目,发生异常的random.shuffle
这一点非常清楚:
x[i], x[j] = x[j], x[i]
Where x
is the "sequence" that was passed in. In this case i
and j
will be values in the range range(0, len(x))
and if any of these i
or j
isn't present in the "sequence" it will throw an Exception
. 其中
x
是传入的“序列”。在这种情况下, i
和j
将是范围range(0, len(x))
并且如果“序列”中不存在这些i
或j
任何一个它会抛出Exception
。 In your case it's very likely given that it throws a KeyError
: 在你的情况下,它很可能会抛出一个
KeyError
:
>>> import random
>>> d = {i: i for i in range(7, 10)}
>>> random.shuffle(d)
KeyError: 3
However it works by exchanging the values in case the dictionary contains exactly the keys that make up the range(0, len(x))
: 但是,如果字典包含组成
range(0, len(x))
的键range(0, len(x))
它的工作方式是交换值:
>>> d = {i: i for i in range(10)}
>>> random.shuffle(d)
>>> d
{0: 7, 1: 9, 2: 3, 3: 4, 4: 0, 5: 2, 6: 1, 7: 6, 8: 8, 9: 5}
If one or multiple keys are missing it could work or it could throw an Exception
. 如果缺少一个或多个密钥,它可能会起作用,或者它可能会抛出
Exception
。 That depends on which random numbers will be drawn: 这取决于将绘制哪些随机数:
d = {i: i for i in range(1, 10)}
random.shuffle(d) # works sometimes, but sometimes it throws the KeyError
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