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假设我有一个素数列表 [ 2,3,5 ]，我想获得所有 N^3 个此类产品的列表（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 )

这样做的pythonic方法是什么？（在长度为 L 的列表的情况下）

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2 回答 2

1

我希望我说对了。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，一个衬里可以是：

productlist = [reduce(mul,t) for t in product(*[[i**j for j in range(n)] for i in primes])]
于 2013-08-18T12:33:52.480 回答
1

这应该这样做：

from itertools import product
from operator import mul

def all_products(l, k):
    for t in product(*[[(p, e) for e in range(k)] for p in l]):
        yield reduce(mul, [x[0] ** x[1] for x in t], 1)

关键是使用itertools.product。

用法：

for prod in all_products([2, 3, 5], 3):
    print prod

这将为您提供形式2^a0 * 3^a1 * 5^a2where的所有产品0 <= aj < 3。

于 2013-08-18T12:39:12.750 回答