blocks|key|3204768|text|是否要迭代？itertools.combinations。常见用法：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3204769|>>>+import+itertools
>>>+itertools.combinations('abcd',2)
<itertools.combinations+object+at+0x01348F30>
>>>+list(itertools.combinations('abcd',2))
[('a',+'b'),+('a',+'c'),+('a',+'d'),+('b',+'c'),+('b',+'d'),+('c',+'d')]
>>>+[''.join(x)+for+x+in+itertools.combinations('abcd',2)]
['ab',+'ac',+'ad',+'bc',+'bd',+'cd']|code-block|syntax|javascript|3204770|如果您只需要计算公式，请使用math.factorial|3204771|import+math

def+nCr(n,r):
++++f+=+math.factorial
++++return+f(n)+/+f(r)+/+f(n-r)

if+__name__+==+'__main__':
++++print+nCr(4,2)|3204772|在Python3中，使用整数除法//而不是/来避免溢出：|style|CODE|3204773|return+f(n)+//+f(r)+//+f(n-r)|3204774|输出|3204775|6|3204776|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html?highlight=itertools.combinations#itertools.combinations|1|http://docs.python.org/library/math.html?highlight=math.factorial#math.factorial^0|6|M|0|0|0|E|E|1|0|0|G|2|L|1|0|0|T|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]]|C|$]]|$1|K|3|L|5|F|7|1F|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1G|8|@$A|1H|B|1I|O|P]|$A|1J|B|1K|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|6|7|1L|8|@$A|1M|B|1N|O|P]]|9|@]|C|$]]|$1|S|3|T|5|6|7|1O|8|@]|9|@]|C|$]]|$1|U|3|V|5|F|7|1P|8|@]|9|@]|C|$G|H]]|$1|W|3|-4|5|6|7|1Q|8|@]|9|@]|C|$]]]|X|$Y|$5|Z|10|11|C|$12|13]]|14|$5|Z|10|11|C|$12|15]]]]

Do you want iteration? <a href="http://docs.python.org/library/itertools.html?highlight=itertools.combinations#itertools.combinations" rel="noreferrer">itertools.combinations</a>. Common usage:
<pre><code>&gt;&gt;&gt; import itertools
&gt;&gt;&gt; itertools.combinations('abcd',2)
&lt;itertools.combinations object at 0x01348F30&gt;
&gt;&gt;&gt; list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
&gt;&gt;&gt; [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
</code></pre>
If you just need to compute the formula, use <a href="http://docs.python.org/library/math.html?highlight=math.factorial#math.factorial" rel="noreferrer">math.factorial</a>:
<pre><code>import math

def nCr(n,r):
 f = math.factorial
 return f(n) / f(r) / f(n-r)

if __name__ == '__main__':
 print nCr(4,2)
</code></pre>
In Python 3, use the integer division <code>//</code> instead of <code>/</code> to avoid overflows:
<code>return f(n) // f(r) // f(n-r)</code>
<h3>Output</h3>
<pre><code>6
</code></pre>

blocks|key|282383|text|下面的程序以高效的方式计算nCr+(与计算阶乘等相比)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|282384|import+operator+as+op
from+functools+import+reduce

def+ncr(n,+r):
++++r+=+min(r,+n-r)
++++numer+=+reduce(op.mul,+range(n,+n-r,+-1),+1)
++++denom+=+reduce(op.mul,+range(1,+r%2B1),+1)
++++return+numer+//+denom++#+or+/+in+Python+2|code-block|syntax|javascript|282385|282386|从Python3.8开始，二项式系数在标准库中以math.comb的形式提供|282387|>>>+from+math+import+comb
>>>+comb(10,3)
120|282388|entityMap^0|D|3|0|0|0|O|9|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|V|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|N|3|O|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|Q|$]]

The following program calculates <code>nCr</code> in an efficient manner (compared to calculating factorials etc.)

<pre><code>import operator as op
from functools import reduce

def ncr(n, r):
 r = min(r, n-r)
 numer = reduce(op.mul, range(n, n-r, -1), 1)
 denom = reduce(op.mul, range(1, r+1), 1)
 return numer // denom # or / in Python 2
</code></pre>

<hr>

As of Python 3.8, binomial coefficients are available in the standard library as <code>math.comb</code>:

<pre><code>&gt;&gt;&gt; from math import comb
&gt;&gt;&gt; comb(10,3)
120
</code></pre>

<blockquote>
 Possible Duplicates: 
 <a href="https://stackoverflow.com/questions/3025162/statistics-combinations-in-python">Statistics: combinations in Python</a> 
 <a href="https://stackoverflow.com/questions/2096573/counting-combinations-and-permutations-efficiently">counting combinations and permutations efficiently</a> 
 <a href="https://stackoverflow.com/questions/3360842/project-euler-problem-in-python-problem-53">Project euler problem in python (problem 53)</a> 
</blockquote>



I'm looking to see if built in with the math library in python is the nCr (n Choose r) function: 

<img src="https://i.stack.imgur.com/OZj2Y.png" alt="enter image description here">

I understand that this can be programmed but I thought that I'd check to see if it's already built in before I do.

Is there a math nCr function in python?

Python

 可能的重复项：我想看看python中的数学库是不是内置了nCr (n Choose r)函数：​​我知道这是可以编程的，但我想我应该先检查一下它是否已经内置。

问python中有数学nCr函数吗？
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