在 Python 中迭代列表
我试图迭代一个列表并获取列表的每个部分，对其进行编码并在全部完成后将结果连接起来。例如，我有一个字符串，它生成一个列表，每个元素的长度为 16 个字符。
message = (u'sixteen-letters.sixteen-letters.sixteen-letters.sixteen-letters.')
result = split16(message, 16)
msg = ';'.join(encode(result.pop(0)) for i in result)
编码函数采用 16 字节字符串并返回结果。然而，按照它的编写方式，它只编码列表中一半的元素。
如果我尝试理解：
result = [encode(split16(message, 16) for message in list_of_messages)]
result = ''.join(result)
它会导致整个列表立即发送。我需要做的是将每个元素分别发送到编码函数，获取结果然后将它们连接在一起。
有没有一种简单的方法可以实现这一目标？
I am trying to iterate through a list and take each part of the list, encode it and join the result up when it is all done. As an example, I have a string which produces a list with each element being 16 characters in length.
message = (u'sixteen-letters.sixteen-letters.sixteen-letters.sixteen-letters.')
result = split16(message, 16)
msg = ';'.join(encode(result.pop(0)) for i in result)
The encode function takes a 16 byte string and returns the result. However with the way it is written, it only encodes half of the elements in the list.
If I try comprehension:
result = [encode(split16(message, 16) for message in list_of_messages)]
result = ''.join(result)
It results in the whole list being sent at once. What I need to do is send each element to the encode function separately, get the result then join them together.
Is there an easy way of achieving this?
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你想做这样的事情吗？
当然，
如果您的 split16 函数足够复杂，也可能是这样。
Are you trying to do something like this?
of course it could be just
if your
split16function complicated enough.我对你到底想要做什么有点困惑，你发布的代码中缺少一个括号，这让情况更加复杂：
应该是：
或：
我认为第二个会做你想做的事。
这段代码
失败了，因为在每一步你都在迭代
result，但在每一步都用pop缩短它。它应该只是：I am a bit confused about what you are exactly trying to do, which is compounded by a missing paren in the code you posted:
Should that be:
or:
I think the second will do what you want.
This code:
is failing because at every step you are iterating through
result, but shortening it at every step withpop. It should just be:我不太清楚你在追求什么，但是
I'm not quite clear what you are after, but