我应该使用 sympy.utilities.iterables.variations() 来获取 permutations_with_replacement 吗? [英] Should I use sympy.utilities.iterables.variations() to get permutations_with_replacement?

问题描述

我正在试验 sympy 的排列没有替换

I'm experimenting with sympy's permutations without replacement

from sympy.functions.combinatorial.numbers import nP
from sympy.utilities.iterables import permutations

nP('abc', 2)
# >>> 6

list(permutations('abc', 2))
# >>> [('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'c'), ('c', 'a'), ('c', 'b')]

接下来，我不想尝试置换替换.似乎没有类似于 combinations_with_replacement() 方法的 permuations_with_replacement() 方法，但是有一个 variations() 方法:

Next, I wan't to try permutations with replacement. It seems that there isn't a permuations_with_replacement() method similar to the combinations_with_replacement() method, but there is a variations() method:

from sympy.utilities.iterables import variations

nP('abc', 2, replacement=True)
# >>> 9

list(variations('abc', 2, repetition=True))
# >>>
    [('a', 'a'),
     ('a', 'b'),
     ('a', 'c'),
     ('b', 'a'),
     ('b', 'b'),
     ('b', 'c'),
     ('c', 'a'),
     ('c', 'b'),
     ('c', 'c')]

variations() 方法是否执行与我期望的 permutations_with_replacement() 相同的功能?

Does the variations() method perform the same function as I am expecting with permutations_with_replacement() to do?

另见:sympy.utilities.iterables.combinations() 替换?

推荐答案

variations 方法完全按照您的想法执行，即计算笛卡尔积，恰当地命名为 product，包的方法.

The variations method does exactly what you think it does, which is to calculate the Cartesian product, aptly named product, method of the package.

这意味着 list(sympy.utilities.iterables.product('abc', repeat=2) 将产生相同的结果.使用 repetition=False，variations 等于 permutations.

This means that list(sympy.utilities.iterables.product('abc', repeat=2) will yield the same results. With repetition=False, variations is equal to permutations instead.

这也可以从variations的内部代码看出:

This can also be seen from the internal code of variations:

if not repetition:
    seq = tuple(seq)
    if len(seq) < n:
        return
    for i in permutations(seq, n):
        yield i
else:
    if n == 0:
        yield ()
    else:
        for i in product(seq, repeat=n):
            yield i

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