无库列表中整数的乘积

你可以迭代每一行,得到每一行的 n -第个元素,并将每个元素相乘:

>>> from functools import reduce
>>> 
>>> def mul_lst(lst):
    return reduce(lambda x, y: x * y, lst)

>>> 
>>> bigList = [[1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5]]
>>> 
>>> [mul_lst([row[i] for row in bigList]) for i in range(len(bigList))]
[1, 32, 243, 1024, 3125]

任何 图书馆,包括 functools mul_lst

>>> def mul_lst(lst):
    product = lst[0]
    for el in lst[1:]:
        product *= el
    return product

>>> mul_lst([3, 3])
9
>>> mul_lst([2, 2, 2, 2, 2])
32

为什么不简单:

productList = []
for i in range(len(bigList[0]):
    p = 1
    for row in bigList:
        p *= row[i]
    productList.append(p)

或者,对您的解决方案进行一点改进:

productList = [1]* len(bigList[0])
for row in bigList:
    for i, c in enumerate(row):
        productList[i] *= c

转置 嵌套列表,然后使用 reduce

>>> [reduce(lambda a,b: a*b, i) for i in map(list, zip(*bigList))]
[1, 32, 243, 1024, 3125]

def prod(x):
  """ Recursive approach with behavior of np.prod over axis 0 """
  if len(x) is 1:
      return x
  for i, a_ in enumerate(x.pop()):
      x[0][i] *= a_
  return prod(x)