問題描述

為了更好地了解 itertools groupby 的感覺，所以我按數(shù)字對元組列表進(jìn)行了分組，并嘗試獲取結(jié)果組的列表.但是，當(dāng)我將 groupby 的結(jié)果轉(zhuǎn)換為列表時，我得到了一個奇怪的結(jié)果:除了最后一組之外，所有的都是空的.這是為什么?我認(rèn)為將迭代器轉(zhuǎn)換為列表效率會降低，但永遠(yuǎn)不會改變行為.我猜列表是空的，因為遍歷了內(nèi)部迭代器，但是什么時候/在哪里發(fā)生?

I was playing around to get a better feeling for itertools groupby, so I grouped a list of tuples by the number and tried to get a list of the resulting groups. When I convert the result of groupby to a list however, I get a strange result: all but the last group are empty. Why is that? I assumed turning an iterator into a list would be less efficient but never change behavior. I guess the lists are empty because the inner iterators are traversed but when/where does that happen?

import itertools

l=list(zip([1,2,2,3,3,3],['a','b','c','d','e','f']))
#[(1, 'a'), (2, 'b'), (2, 'c'), (3, 'd'), (3, 'e'), (3, 'f')]

grouped_l = list(itertools.groupby(l, key=lambda x:x[0]))
#[(1, <itertools._grouper at ...>), (2, <itertools._grouper at ...>), (3, <itertools._grouper at ...>)]

[list(x[1]) for x in grouped_l]
[[], [], [(3, 'f')]]


grouped_i = itertools.groupby(l, key=lambda x:x[0])
#<itertools.groupby at ...>
[list(x[1]) for x in grouped_i]
[[(1, 'a')], [(2, 'b'), (2, 'c')], [(3, 'd'), (3, 'e'), (3, 'f')]]

推薦答案

來自 itertools.groupby() 文檔:

返回的組本身是一個迭代器，它與 groupby() 共享底層迭代器.因為源是共享的，當(dāng)groupby()對象前進(jìn)時，之前的組就不再可見了.

The returned group is itself an iterator that shares the underlying iterable with groupby(). Because the source is shared, when the groupby() object is advanced, the previous group is no longer visible.

將 groupby() 的輸出轉(zhuǎn)換為列表會推進(jìn) groupby() 對象.

Turning the output from groupby() into a list advances the groupby() object.

因此，您不應(yīng)該將 itertools.groupby 對象類型轉(zhuǎn)換為列表.如果您想將值存儲為 list，那么您應(yīng)該執(zhí)行類似 list comprehension 的操作，以便創(chuàng)建 groupby 對象的副本:

Hence, you shouldn't be type-casting itertools.groupby object to list. If you want to store the values as list, then you should be doing something like this list comprehension in order to create copy of groupby object:

grouped_l = [(a, list(b)) for a, b in itertools.groupby(l, key=lambda x:x[0])]

這將允許您多次迭代您的列表(從 groupby 對象轉(zhuǎn)換).但是，如果您對只迭代一次結(jié)果感興趣，那么您在問題中提到的第二個解決方案將滿足您的要求.

This will allow you to iterate your list (transformed from groupby object) multiple times. However, if you are interested in only iterating the result once, then the second solution you mentioned in the question will suffice your requirement.

這篇關(guān)于在 groupby 周圍列出空組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)！