blocks|key|3580790|text|您可以使用enumerate|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3580791|original_list+=+[1,2,1,4,4,4,3,4,4,1,4,3,3]
groups+=+{a:[i+for+i,+c+in+enumerate(original_list)+if+c+==+a]+for+a+in+set(original_list)}|code-block|syntax|javascript|3580792|输出：|3580793|{1:+[0,+2,+9],+2:+[1],+3:+[6,+11,+12],+4:+[3,+4,+5,+7,+8,+10]}|3580794|entityMap^0|5|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

You can use <code>enumerate</code>:

<pre><code>original_list = [1,2,1,4,4,4,3,4,4,1,4,3,3]
groups = {a:[i for i, c in enumerate(original_list) if c == a] for a in set(original_list)}
</code></pre>

Output:

<pre><code>{1: [0, 2, 9], 2: [1], 3: [6, 11, 12], 4: [3, 4, 5, 7, 8, 10]}
</code></pre>

blocks|key|5048905|text|这可以通过列表理解来实现。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5048906|>>>+x+=+[1,2,1,4,4,4,3,4,4,1,4,3,3]
>>>+[[i+for+i+in+range(len(x))+if+x[i]==y]+for+y+in+sorted(set(x))]
[[0,+2,+9],+[1],+[6,+11,+12],+[3,+4,+5,+7,+8,+10]]|code-block|syntax|javascript|5048907|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|$]]

This can be achieved with a list comprehension.

<pre><code>&gt;&gt;&gt; x = [1,2,1,4,4,4,3,4,4,1,4,3,3]
&gt;&gt;&gt; [[i for i in range(len(x)) if x[i]==y] for y in sorted(set(x))]
[[0, 2, 9], [1], [6, 11, 12], [3, 4, 5, 7, 8, 10]]
</code></pre>

blocks|key|1264228|text|使用enumerate和defaultdict，您可以使用以下命令构建值到其索引的映射|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1264229|from+collections+import+defaultdict

dd+=+defaultdict(list)
for+index,+value+in+enumerate(original_list):
++++dd[value].append(index)

print(dd)
#+defaultdict(<class+'list'>,+{1:+[0,+2,+9],+2:+[1],+4:+[3,+4,+5,+7,+8,+10],+3:+[6,+11,+12]})|code-block|syntax|javascript|1264230|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/functions.html#enumerate|1|https://docs.python.org/3/library/collections.html#collections.defaultdict^0|2|9|C|B|2|9|0|C|B|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]|$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

Using <a href="https://docs.python.org/3/library/functions.html#enumerate" rel="nofollow noreferrer"><code>enumerate</code></a> and a <a href="https://docs.python.org/3/library/collections.html#collections.defaultdict" rel="nofollow noreferrer"><code>defaultdict</code></a>, you can build a mapping of values to their indices with 

<pre><code>from collections import defaultdict

dd = defaultdict(list)
for index, value in enumerate(original_list):
 dd[value].append(index)

print(dd)
# defaultdict(&lt;class 'list'&gt;, {1: [0, 2, 9], 2: [1], 4: [3, 4, 5, 7, 8, 10], 3: [6, 11, 12]})
</code></pre>

blocks|key|1264248|text|您可以将collections.defaultdict用于一次性解决方案。然后，如果需要的话，可以使用sorted来按值对索引进行排序。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1264249|original_list+=+[1,2,1,4,4,4,3,4,4,1,4,3,3]

from+collections+import+defaultdict
from+operator+import+itemgetter

dd+=+defaultdict(list)

for+idx,+value+in+enumerate(original_list):
++++dd[value].append(idx)

keys,+values+=+zip(*sorted(dd.items(),+key=itemgetter(0)))

print(keys,+values,+sep='\n')

(1,+2,+3,+4)
([0,+2,+9],+[1],+[6,+11,+12],+[3,+4,+5,+7,+8,+10])|code-block|syntax|javascript|1264250|为了便于比较，dd的值在Python+3.6%2B中是按插入顺序排列的(在3.7%2B中是正式的，在3.6中作为CPython实现的详细信息)：|1264251|print(list(dd.values()))

[[0,+2,+9],+[1],+[3,+4,+5,+7,+8,+10],+[6,+11,+12]]|1264252|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.6/library/collections.html#collections.defaultdict^0|4|N|1F|6|4|N|0|0|0|7|2|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

You can use <a href="https://docs.python.org/3.6/library/collections.html#collections.defaultdict" rel="nofollow noreferrer"><code>collections.defaultdict</code></a> for a one-pass solution. Then use <code>sorted</code> if you need, as in your desired result, to sort your indices by value.

<pre><code>original_list = [1,2,1,4,4,4,3,4,4,1,4,3,3]

from collections import defaultdict
from operator import itemgetter

dd = defaultdict(list)

for idx, value in enumerate(original_list):
 dd[value].append(idx)

keys, values = zip(*sorted(dd.items(), key=itemgetter(0)))

print(keys, values, sep='\n')

(1, 2, 3, 4)
([0, 2, 9], [1], [6, 11, 12], [3, 4, 5, 7, 8, 10])
</code></pre>

For comparison, the values of <code>dd</code> are insertion ordered in Python 3.6+ (officially in 3.7+, as a CPython implementation detail in 3.6):

<pre><code>print(list(dd.values()))

[[0, 2, 9], [1], [3, 4, 5, 7, 8, 10], [6, 11, 12]]
</code></pre>

blocks|key|5048988|text|下面是我如何使用numpy，使用我在注释中链接的argsort函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5048989|original+=+[1,2,1,4,4,4,3,4,4,1,4,3,3]
indexes+=+[]
s+=+set()

for+n+in+np.argsort(original):
++++if+original[n]+in+s:
++++++++indexes[-1].append(n)
++++else:
++++++++indexes.append([n])
++++++++s.add(original[n])

print(indexes)|code-block|syntax|javascript|5048990|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|$]]

Here is how I would do it with numpy, using the argsort function I linked in the comments.

<pre><code>original = [1,2,1,4,4,4,3,4,4,1,4,3,3]
indexes = []
s = set()

for n in np.argsort(original):
 if original[n] in s:
 indexes[-1].append(n)
 else:
 indexes.append([n])
 s.add(original[n])

print(indexes)
</code></pre>

I'm trying to split a list of integers into sublists of the the indexes of equal integers. So say I have a list:

<pre><code>original_list = [1,2,1,4,4,4,3,4,4,1,4,3,3]
</code></pre>

The desired output would be:

<pre><code>indexes : [[0,2,9], [1], [6,11,12], [3,4,5,7,8,10]]
# corresponds to sublists: [[1,1,1] [2], [3,3,3], [4,4,4,4,4,4]]
</code></pre>

I can't figure out how to do this though, as most solutions require you to first sort the original list, but in my case, this messes up the indices. Itertools or np.arrays have not helped me for this reason, as they only group sequential equal elements. 

Does anyone know of a solution for this problem? I would love to hear!

Create sublists of indexes of equal values from list

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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我正在尝试将一个整数列表拆分成相等整数的索引的子列表。因此，假设我有一个列表：original_list = [1,2,1,4,4,4,3,4,4,1,4,3,3]期望的输出将是：indexes : [[0,2,9], [1], [6,11,12], [3,4,5,7,8,10]]# corresponds to su...

问从列表创建相等值的索引子列表
EN

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问从列表创建相等值的索引子列表EN