entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://docs.python.org/2/library/sets.html|blocks|key|foksq|text|您可以使用python的set+datatype来删除这些重复项，因为集合将只包含唯一的组合：|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|4s2mk|import+itertools+as+it
x+=+[-1,+0,+1,+2,+-1,+-4]

permutations+=+[p+for+p+in+set(it.combinations(x,+r=3))]
print(permutations)|code-block|syntax|javascript|2g3od|输出：|1i3e4|[(0,+1,+2),
+(-1,+1,+-1),
+(-1,+2,+-1),
+(0,+-1,+-4),
+(-1,+-1,+-4),
+(-1,+1,+-4),
+(-1,+2,+-4),
+(2,+-1,+-4),
+(1,+2,+-4),
+(-1,+0,+1),
+(1,+2,+-1),
+(-1,+0,+-4),
+(-1,+0,+2),
+(-1,+0,+-1),
+(-1,+1,+2),
+(0,+2,+-4),
+(0,+2,+-1),
+(0,+1,+-4),
+(1,+-1,+-4),
+(0,+1,+-1)]|42bee|然后，use可以使用下面这行代码：|ff9h9|unique_permutations+=+set(tuple(sorted(t))+for+t+in+permutations)|5n27s|9udbu|{(-4,+-1,+-1),
+(-4,+-1,+0),
+(-4,+-1,+1),
+(-4,+-1,+2),
+(-4,+0,+1),
+(-4,+0,+2),
+(-4,+1,+2),
+(-1,+-1,+0),
+(-1,+-1,+1),
+(-1,+-1,+2),
+(-1,+0,+1),
+(-1,+0,+2),
+(-1,+1,+2),
+(0,+1,+2)}|dldd4^0|C|C|0|0|0|0|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|11|G|@]|H|@$I|12|J|13|A|14]]|6|$]]|$A|K|C|L|2|M|F|15|G|@]|H|@]|6|$N|O]]|$A|P|C|Q|2|E|F|16|G|@]|H|@]|6|$]]|$A|R|C|S|2|M|F|17|G|@]|H|@]|6|$N|O]]|$A|T|C|U|2|E|F|18|G|@]|H|@]|6|$]]|$A|V|C|W|2|M|F|19|G|@]|H|@]|6|$N|O]]|$A|X|C|Q|2|E|F|1A|G|@]|H|@]|6|$]]|$A|Y|C|Z|2|M|F|1B|G|@]|H|@]|6|$N|O]]|$A|10|C|-4|2|E|F|1C|G|@]|H|@]|6|$]]]]

You can use python's <a href="https://docs.python.org/2/library/sets.html" rel="nofollow noreferrer">set datatype</a> to remove those duplicates since sets will only contain the unique combinations:

<pre><code>import itertools as it
x = [-1, 0, 1, 2, -1, -4]

permutations = [p for p in set(it.combinations(x, r=3))]
print(permutations)
</code></pre>

Output:

<pre><code>[(0, 1, 2),
 (-1, 1, -1),
 (-1, 2, -1),
 (0, -1, -4),
 (-1, -1, -4),
 (-1, 1, -4),
 (-1, 2, -4),
 (2, -1, -4),
 (1, 2, -4),
 (-1, 0, 1),
 (1, 2, -1),
 (-1, 0, -4),
 (-1, 0, 2),
 (-1, 0, -1),
 (-1, 1, 2),
 (0, 2, -4),
 (0, 2, -1),
 (0, 1, -4),
 (1, -1, -4),
 (0, 1, -1)]
</code></pre>

Then use can use the following line:

<pre><code>unique_permutations = set(tuple(sorted(t)) for t in permutations)
</code></pre>

Output:

<pre><code>{(-4, -1, -1),
 (-4, -1, 0),
 (-4, -1, 1),
 (-4, -1, 2),
 (-4, 0, 1),
 (-4, 0, 2),
 (-4, 1, 2),
 (-1, -1, 0),
 (-1, -1, 1),
 (-1, -1, 2),
 (-1, 0, 1),
 (-1, 0, 2),
 (-1, 1, 2),
 (0, 1, 2)}
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://www.programiz.com/python-programming/methods/built-in/frozenset|blocks|key|2tfqf|text|如何获得组合，然后通过在字典中键入组合的frozenset结果来只获取唯一的组合呢？在创建dictionary之前，这将只使用生成器。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|3u6j6|combs1,+combs2+=+itertools.tee(itertools.combinations(x,+r=3))
res+=+list(dict(zip(map(frozenset,+combs1),+combs2)).values())|code-block|syntax|javascript|8gui^0|K|9|19|A|K|9|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|S|G|@$H|T|I|U|J|K]|$H|V|I|W|J|K]]|L|@$H|X|I|Y|A|Z]]|6|$]]|$A|M|C|N|2|O|F|10|G|@]|L|@]|6|$P|Q]]|$A|R|C|-4|2|E|F|11|G|@]|L|@]|6|$]]]]

What about getting the combinations and then taking only the unique ones by keying a dictionary with the <a href="https://www.programiz.com/python-programming/methods/built-in/frozenset" rel="nofollow noreferrer"><code>frozenset</code></a> result of the combinations. This will only use generators until creating the <code>dictionary</code>.

<pre><code>combs1, combs2 = itertools.tee(itertools.combinations(x, r=3))
res = list(dict(zip(map(frozenset, combs1), combs2)).values())
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://en.wikipedia.org/wiki/Multiset|blocks|key|9ee84|text|它们被称为multisets，我们可以很容易地通过sympy模块获得它们的组合。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|99pjb|from+sympy.utilities.iterables+import+multiset_combinations

list(multiset_combinations([1,+2,+4,+1],+3))
[[1,+1,+2],+[1,+1,+4],+[1,+2,+4]]|code-block|syntax|javascript|feopp|下面是@EdedkiOkoh的例子：|1qc70|x+=+[-1,+0,+1,+2,+-1,+-4]
list(multiset_combinations(x,+3))
[[-4,+-1,+-1],
++++[-4,+-1,+0],
++++[-4,+-1,+1],
++++[-4,+-1,+2],
++++[-4,+0,+1],
++++[-4,+0,+2],
++++[-4,+1,+2],
++++[-1,+-1,+0],
++++[-1,+-1,+1],
++++[-1,+-1,+2],
++++[-1,+0,+1],
++++[-1,+0,+2],
++++[-1,+1,+2],
++++[0,+1,+2]]|dpqha^0|P|5|5|9|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|W|G|@$H|X|I|Y|J|K]]|L|@$H|Z|I|10|A|11]]|6|$]]|$A|M|C|N|2|O|F|12|G|@]|L|@]|6|$P|Q]]|$A|R|C|S|2|E|F|13|G|@]|L|@]|6|$]]|$A|T|C|U|2|O|F|14|G|@]|L|@]|6|$P|Q]]|$A|V|C|-4|2|E|F|15|G|@]|L|@]|6|$]]]]

These are called <a href="https://en.wikipedia.org/wiki/Multiset" rel="nofollow noreferrer">multisets</a>, and we can easily obtain the combinations of these with the <code>sympy</code> module.

<pre><code>from sympy.utilities.iterables import multiset_combinations

list(multiset_combinations([1, 2, 4, 1], 3))
[[1, 1, 2], [1, 1, 4], [1, 2, 4]]
</code></pre>

And here is the example by @EdedkiOkoh:

<pre><code>x = [-1, 0, 1, 2, -1, -4]
list(multiset_combinations(x, 3))
[[-4, -1, -1],
 [-4, -1, 0],
 [-4, -1, 1],
 [-4, -1, 2],
 [-4, 0, 1],
 [-4, 0, 2],
 [-4, 1, 2],
 [-1, -1, 0],
 [-1, -1, 1],
 [-1, -1, 2],
 [-1, 0, 1],
 [-1, 0, 2],
 [-1, 1, 2],
 [0, 1, 2]]
</code></pre>

I have a set of numbers <code>[1, 2, 4, 1]</code>. Now, I want to generate all possible combinations from this set of size k (example k = 3). All the generated output sets must not be duplicate 

Example : <code>[1, 2, 1]</code> and <code>[2, 1, 1]</code> are the same sets but they should not be selected. Only one of them should appear. Is it possible using combinations from itertools in Python?

<pre><code>import itertools
x = [1, 2, 1]
print([p for p in itertools.product(x, repeat=3)])
</code></pre>

I have tried using itertools.product but it doesnt work and 
using combinations from itertools is getting duplicates

I have tried using itertools.combinations
<code>print([p for p in set(itertools.combinations(x, r=3))])</code>

If I give the following input

<pre><code> x = [-1, 0, 1, 2, -1, -4]
</code></pre>

The ouput generated for r = 3 is 

<pre><code>[(0, -1, -4), (-1, -1, -4), (-1, 1, -4), (0, 2, -1), (-1, 0, 2), (-1, 2, -4), (0, 1, 2), (2, -1, -4), (-1, 0, -1), (0, 1, -4), (1, 2, -4), (-1, 0, 1), (-1, 1, 2), (0, 2, -4), (-1, 1, -1), (-1, 2, -1), (1, 2, -1), (0, 1, -1), (-1, 0, -4), (1, -1, -4)]
</code></pre>

<code>(-1, 0, 1)</code> and <code>(0, 1, -1)</code> are duplicate sets with same combinations. I am not sure how to overcome this.

How to generate all distinct combinations (where input elements are repeated) in Python(using Itertools)?

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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 我有一组数字[1, 2, 4, 1]。现在，我想从这个大小为k(例如k= 3)的集合中生成所有可能的组合。所有生成的输出集不能重复 示例：[1, 2, 1]和[2, 1, 1]是相同的集合，但不应选择它们。它们中只有一个应该出现。在Python中使用itertools的组合是可能的吗？ import itertool...

问如何在Python(使用Itertools)中生成所有不同的组合(其中输入元素重复)？
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问如何在Python(使用Itertools)中生成所有不同的组合(其中输入元素重复)？EN