blocks|key|3017828|text|您可以将值分离成单独的列表。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3017829|from+bisect+import+bisect

data+=+[(3,+1),+(2,+2),+(5,+6)]
fst,+snd+=+zip(*data)
idx+=+bisect(fst,+2)|code-block|syntax|javascript|3017830|但是请注意，要使bisect工作，您的数据确实应该被排序.|offset|length|style|CODE|3017831|entityMap^0|0|0|8|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can separate out the values into separate lists.

<pre><code>from bisect import bisect

data = [(3, 1), (2, 2), (5, 6)]
fst, snd = zip(*data)
idx = bisect(fst, 2)
</code></pre>

Note however, that for <code>bisect</code> to work, your data really should be ordered...

blocks|key|3017906|text|在某些情况下，只是简单的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3017907|bisect(list_of_tuples,+(3,+None))|code-block|syntax|javascript|3017908|就够了。|3017909|因为None比任何整数都要小，所以这将给出以至少3开头的第一个元组的索引，或者如果它们都小于3时则给出len(list_of_tuples)。注意，list_of_tuples是排序的。|offset|length|style|CODE|3017910|entityMap^0|0|0|0|2|4|1F|J|22|E|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]|$K|W|L|X|M|N]|$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

In some cases just the simple

<pre><code>bisect(list_of_tuples, (3, None))
</code></pre>

will be enough.

Because <code>None</code> will compare less than any integer, this will give you the index of the first tuple starting with at least 3, or <code>len(list_of_tuples)</code> if all of them are smaller than 3. Note that <code>list_of_tuples</code> is sorted.

blocks|key|3046301|text|检查文档的底部：http://docs.python.org/3/library/bisect.html。如果您想要与元素本身相比，您应该创建一个单独的所谓键列表。在您的示例中，只包含一个元组的In列表。使用第二个列表来用二分法计算索引。然后使用它将元素插入原始元素(元组列表)，并将键(元组中的一个)插入到新的键列表(ints列表)中。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3046302|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/3/library/bisect.html^0|8|18|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Check the bottom section of the documentation: <a href="http://docs.python.org/3/library/bisect.html" rel="nofollow">http://docs.python.org/3/library/bisect.html</a>. If you want to compare to something else than the element itself, you should create a separate list of so-called keys. In your case a list of ints containing only [1] of the tuple. Use that second list to compute the index with bisect. Then use that to both insert the element into the original (list of tuples) and the key ([1] of the tuple) into the new list of keys (list of ints).

blocks|key|3017975|text|建议转换输入列表的答案通过将应该是O(log+)的操作转换为O(n)操作，从而违背了均分的目的。更好的解决方案是对输入使用视图：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3017976|class+_list_view:
++++def+__init__(self,+a,+key):
++++++++self.a+=+a
++++++++self.key+=+key

++++def+__getitem__(self,+index):
++++++++return+self.key(self.a[index])


def+bisect_left(a,+x,+lo=0,+hi=None,+key=id):
++++from+bisect+import+bisect_left
++++hi+=+hi+or+len(a)
++++if+key+==+id:
++++++++return+bisect_left(a,+x,+lo,+hi)
++++return+bisect_left(_list_view(a,+key),+x,+lo,+hi)|code-block|syntax|javascript|3017977|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Answers that suggest transforming the input list are defeating the purpose of bisect by converting what should be an O(log n) into an O(n) operation. A better solution is to use a view on the input:
<pre><code>class _list_view:
 def __init__(self, a, key):
 self.a = a
 self.key = key

 def __getitem__(self, index):
 return self.key(self.a[index])


def bisect_left(a, x, lo=0, hi=None, key=id):
 from bisect import bisect_left
 hi = hi or len(a)
 if key == id:
 return bisect_left(a, x, lo, hi)
 return bisect_left(_list_view(a, key), x, lo, hi)
</code></pre>

blocks|key|3017960|text|我也面临着同样的问题。我在存储(file_id,+word_frequency)元组的列表，并希望按照元组中的第二个元素(即word_frequency+)对列表进行排序。我做了一些研究，并了解了python是如何比较元组的，这里描述的是https://howtodoinjava.com/python/compare-tuples/。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3017961|本质上，它看两个元组的第一个元素，并取较小的元组。如果第一个元素是相同的，那么它会比较第二个值a+so。|3017962|所以我所做的就是用元组(word_frequency,+file_id)交换元素。现在，我得到了我的名单，根据词频使用等分排序。|3017963|希望这能有所帮助|3017964|entityMap|0|LINK|mutability|MUTABLE|url|https://howtodoinjava.com/python/compare-tuples/^0|F|P|1Q|E|3B|1C|0|0|0|B|P|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|M|$N|$5|O|P|Q|E|$R|S]]]]

I faced the same issue. I was storing list of <code>(file_id, word_frequency)</code> tuples and wanted to get the list sorted by second element in the tuple which is <code>word_frequency</code>. I did some research and found out how python compares tuples which is describe here <a href="https://howtodoinjava.com/python/compare-tuples/" rel="nofollow noreferrer">https://howtodoinjava.com/python/compare-tuples/</a>.

Essentially it looks at first element of two tuples and takes the smaller one. If the first elements are same then it compares second values an so on.

So what I did was exchanged the elements in tuple <code>(word_frequency, file_id)</code>. Now I got my list sorted according to word frequency using bisect. 

Hope this helps

I'm trying to figure out how to use bisect in a list of tuples
for example

<pre><code>[(3, 1), (2, 2), (5, 6)]
</code></pre>

How can I bisect this list according to the [1] in each tuple?

<pre><code>list_dict [(69, 8), (70, 8), ((65, 67), 6)]
tup1,tup2 (69, 8) (70, 8)
list_dict [((65, 67), 6)]
fst, snd ((65, 67),) (6,)
</code></pre>

And I'm inserting to bisect 

<pre><code>idx = bisect.bisect(fst, tup1[1]+tup2[1])
</code></pre>

Which gives me <code>unorderable types: int() &lt; tuple()</code>

Using bisect in a list of tuples?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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例如，我正在试图找出如何在元组列表中使用二分法[(3, 1), (2, 2), (5, 6)]如何根据每个元组中的1对此列表进行平分？list_dict [(69, 8), (70, 8), ((65, 67), 6)]tup1,tup2 (69, 8) (70, 8)list_dict [((65, 67), 6)]fst, snd ((65, 67),) (6,)我要插入到二分idx = bi

问在元组列表中使用等分？
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