blocks|key|2133345|text|您有一个嵌套的列表，您可以使用itertools.chain.from_iterable将其展平。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2133346|除了这种复杂性之外，问题归结为一个简单的collections.Counter应用程序，因为Counter在计算元组方面没有问题。|2133347|>>>+from+collections+import+Counter
>>>+from+itertools+import+chain
>>>+
>>>+bi_grams=+[[('ABC',+'Memorial'),+('Memorial',+'Hospital')],+[('ABC',+'Memorial'),+('Memorial',+'Clinic')]]
>>>+Counter(chain.from_iterable(bi_grams))
>>>+
Counter({('ABC',+'Memorial'):+2,
+++++++++('Memorial',+'Clinic'):+1,
+++++++++('Memorial',+'Hospital'):+1})|code-block|syntax|javascript|2133348|还有一个非常简单的解决方案，使用for循环：|2133349|>>>+c+=+Counter()
>>>+for+x+in+bi_grams:
...:++++c.update(x)
...:++++
>>>+c
>>>+
Counter({('ABC',+'Memorial'):+2,
+++++++++('Memorial',+'Clinic'):+1,
+++++++++('Memorial',+'Hospital'):+1})|2133350|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.7/library/itertools.html#itertools.chain.from_iterable|1|https://docs.python.org/3.7/library/collections.html#collections.Counter^0|F|T|F|T|0|0|K|J|1A|7|K|J|1|0|0|G|3|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|H|3|I|5|J|7|1E|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1I|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]]]

You have a nested list which you can flatten with <a href="https://docs.python.org/3.7/library/itertools.html#itertools.chain.from_iterable" rel="nofollow noreferrer"><code>itertools.chain.from_iterable</code></a>.

Apart from that complication the problem boils down to a simple application of <a href="https://docs.python.org/3.7/library/collections.html#collections.Counter" rel="nofollow noreferrer"><code>collections.Counter</code></a> because a <code>Counter</code> has no problem with counting tuples.

<pre><code>&gt;&gt;&gt; from collections import Counter
&gt;&gt;&gt; from itertools import chain
&gt;&gt;&gt; 
&gt;&gt;&gt; bi_grams= [[('ABC', 'Memorial'), ('Memorial', 'Hospital')], [('ABC', 'Memorial'), ('Memorial', 'Clinic')]]
&gt;&gt;&gt; Counter(chain.from_iterable(bi_grams))
&gt;&gt;&gt; 
Counter({('ABC', 'Memorial'): 2,
 ('Memorial', 'Clinic'): 1,
 ('Memorial', 'Hospital'): 1})
</code></pre>

There's also a pretty straight forward solution with a <code>for</code> loop:

<pre><code>&gt;&gt;&gt; c = Counter()
&gt;&gt;&gt; for x in bi_grams:
...: c.update(x)
...: 
&gt;&gt;&gt; c
&gt;&gt;&gt; 
Counter({('ABC', 'Memorial'): 2,
 ('Memorial', 'Clinic'): 1,
 ('Memorial', 'Hospital'): 1})
</code></pre>

blocks|key|2133359|text|@timgeb建议的chain.from_iterable可能是可行的方法，但您也可以通过列表理解来扁平化列表，然后应用Counter|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2133360|from+collections+import+Counter

bi_grams=+[[('ABC',+'Memorial'),+('Memorial',+'Hospital')],+[('ABC',+'Memorial'),+('Memorial',+'Clinic')]]

>>>+Counter(i+for+x+in+bi_grams+for+i+in+x)
Counter({('ABC',+'Memorial'):+2,+('Memorial',+'Hospital'):+1,+('Memorial',+'Clinic'):+1})|code-block|syntax|javascript|2133361|entityMap^0|A|J|1O|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

<code>chain.from_iterable</code> as suggested by @timgeb is probably the way to go, but you could also flatten your list via a list comprehension, and then apply <code>Counter</code>:

<pre><code>from collections import Counter

bi_grams= [[('ABC', 'Memorial'), ('Memorial', 'Hospital')], [('ABC', 'Memorial'), ('Memorial', 'Clinic')]]

&gt;&gt;&gt; Counter(i for x in bi_grams for i in x)
Counter({('ABC', 'Memorial'): 2, ('Memorial', 'Hospital'): 1, ('Memorial', 'Clinic'): 1})
</code></pre>

blocks|key|2133386|text|你也可以使用chain(*iterable)的习惯用法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2133387|>>>+from+itertools+import+chain++++++++++++++++++++++++++++++++++++++++++++++++++++>>>+from+collections+import+Counter
>>>+Counter(chain(*bi_grams))
Counter({('ABC',+'Memorial'):+2,+('Memorial',+'Hospital'):+1,+('Memorial',+'Clinic'):+1})|code-block|syntax|javascript|2133388|2133389|使用chain(*iterable)将元组列表展平为内部列表，例如|2133390|>>>+x+=+[[(1,2),+(3,4)],+[(5,6)],+[(7,8)]]

>>>+list(chain(*x))
[(1,+2),+(3,+4),+(5,+6),+(7,+8)]|2133391|Counter简单地计算扁平化列表中的内容：|2133392|>>>+x+=+[[(1,2),+(3,4)],+[(5,6)],+[(7,8)]]

>>>+Counter(chain(*x))
Counter({(1,+2):+1,+(3,+4):+1,+(5,+6):+1,+(7,+8):+1})|2133393|entityMap^0|6|G|0|0|0|2|G|0|0|0|7|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|N|3|O|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|P|3|Q|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|R|3|S|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|T|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|U|$]]

You can use the <code>chain(*iterable)</code> idiom too:

<pre><code>&gt;&gt;&gt; from itertools import chain &gt;&gt;&gt; from collections import Counter
&gt;&gt;&gt; Counter(chain(*bi_grams))
Counter({('ABC', 'Memorial'): 2, ('Memorial', 'Hospital'): 1, ('Memorial', 'Clinic'): 1})
</code></pre>

<hr>

Using <code>chain(*iterable)</code>, flattens the list of list of tuples to the inner list, e.g. 

<pre><code>&gt;&gt;&gt; x = [[(1,2), (3,4)], [(5,6)], [(7,8)]]

&gt;&gt;&gt; list(chain(*x))
[(1, 2), (3, 4), (5, 6), (7, 8)]
</code></pre>

<code>Counter</code> simply counts what's in the flatten list:

<pre><code>&gt;&gt;&gt; x = [[(1,2), (3,4)], [(5,6)], [(7,8)]]

&gt;&gt;&gt; Counter(chain(*x))
Counter({(1, 2): 1, (3, 4): 1, (5, 6): 1, (7, 8): 1})
</code></pre>

I have a list of lists called bi_grams and I want the frequency of each bigram. The length of bi_grams is 23087 so I might need a loop (?)

<pre><code>bi_grams= [[('ABC', 'Memorial'), ('Memorial', 'Hospital')], [('ABC', 'Memorial'), ('Memorial', 'Clinic')]]
</code></pre>

Expected output

<pre><code>dict_items([(('ABC', 'Memorial'), 2), 
(('Memorial', 'Hospital'), 1), (('Memorial', 'Clinic'),1])
</code></pre>

Ive tried <code>Counter(map(tuple, bi_grams[0])).items()</code>
but it only gives me the bigram frequency of the first list. 

I've also tried <code>Counter(map(tuple, bi_grams[0:2])).items()</code> but it is only giving me the count of each list, not the bigrams within the list.

Would really appreciate some help.

frequency of items within list of lists

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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我有一个名为bi_grams的列表，我想要每个二元语法的频率。bi_grams的长度是23087，所以我可能需要一个循环(?)bi_grams= [[('ABC', 'Memorial'), ('Memorial', 'Hospital')], [('ABC', 'Memorial'), ('Memorial', 'C...

问项目在列表列表中的频率
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问项目在列表列表中的频率EN