假设我有一个素数列表 [ 2,3,5 ]，我想获得所有 N^3 此类乘积的列表(或迭代器):
pow( 2, 0 ) * pow( 3, 0 ) * pow( 5, 0 )
pow( 2, 1 ) * pow( 3, 0 ) * pow( 5, 0 )
pow( 2, 0 ) * pow( 3, 1 ) * pow( 5, 0 )
pow( 2, 0 ) * pow( 3, 0 ) * pow( 5, 1 )
pow( 2, 1 ) * pow( 3, 1 ) * pow( 5, 0 )
pow( 2, 0 ) * pow( 3, 1 ) * pow( 5, 1 )
pow( 2, 1 ) * pow( 3, 0 ) * pow( 5, 1 )
[...]
pow( 2, N-1 ) * pow( 3, N-1 ) * pow( 5, N-1 )
Python 的方法是什么？ (列表长度为L的情况)
最佳答案
希望我的理解是对的。检查这个(N=3):
from itertools import product
from operators import mul
primes = [2,3,5]
n = 3
sets = product(*[[(i,j) for j in range(n)] for i in primes])
# Now 'sets' contains all the combinations you want. If you just wanted pow(i,j), write i**j instead and skip the map in the next enumeration
# list(sets)
#[((2, 0), (3, 0), (5, 0)),
# ((2, 0), (3, 0), (5, 1)),
# ((2, 0), (3, 0), (5, 2)),
# ... ... ...
# ((2, 2), (3, 2), (5, 0)),
# ((2, 2), (3, 2), (5, 1)),
# ((2, 2), (3, 2), (5, 2))]
productlist = []
for t in sets:
productlist.append(reduce(mul,map(lambda tp:tp[0]**tp[1],t)))
# now productlist contains the multiplication of each n(=3) items:
#[1, 5, 25, 3, 15, 75, 9, 45, 225, 2, 10, 50, 6, 30, 150, 18, 90, 450, 4, 20, 100, 12, 60, 300, 36, 180, 900]
# pow( 2, 0 ) * pow( 3, 0 ) * pow( 5, 0 ) = 1
# pow( 2, 0 ) * pow( 3, 0 ) * pow( 5, 1 ) = 5
# pow( 2, 0 ) * pow( 3, 0 ) * pow( 5, 2 ) = 25
# .... ...
或者，单衬可以是:
productlist = [reduce(mul,t) for t in product(*[[i**j for j in range(n)] for i in primes])]
关于计算所有乘积组合的 Pythonic 方法，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/18298953/