如何将列表中的所有项目与Python相乘？

I need to write a function that takes a list of numbers and multiplies them together. Example: [1,2,3,4,5,6] will give me 1*2*3*4*5*6. I could really use your help.

Python 3: use functools.reduce:

>>> from functools import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720

Python 2: use reduce:

>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720

For compatible with 2 and 3 use pip install six, then:

>>> from six.moves import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720

You can use:

import operator
import functools
functools.reduce(operator.mul, [1,2,3,4,5,6], 1)

See reduce and operator.mul documentations for an explanation.

You need the import functools line in Python 3+.

I would use the numpy.prod to perform the task. See below.

import numpy as np
mylist = [1, 2, 3, 4, 5, 6] 
result = np.prod(np.array(mylist))  

If you want to avoid importing anything and avoid more complex areas of Python, you can use a simple for loop

product = 1  # Don't use 0 here, otherwise, you'll get zero 
             # because anything times zero will be zero.
list = [1, 2, 3]
for x in list:
    product *= x

Starting Python 3.8, a .prod function has been included to the math module in the standard library:

math.prod(iterable, *, start=1)

The method returns the product of a start value (default: 1) times an iterable of numbers:

import math
math.prod([1, 2, 3, 4, 5, 6])

>>> 720

If the iterable is empty, this will produce 1 (or the start value, if provided).

Here’s some performance measurements from my machine. Relevant in case this is performed for small inputs in a long-running loop:

import functools, operator, timeit
import numpy as np

def multiply_numpy(iterable):
    return np.prod(np.array(iterable))

def multiply_functools(iterable):
    return functools.reduce(operator.mul, iterable)

def multiply_manual(iterable):
    prod = 1
    for x in iterable:
        prod *= x

    return prod

sizesToTest = [5, 10, 100, 1000, 10000, 100000]

for size in sizesToTest:
    data = [1] * size

    timerNumpy = timeit.Timer(lambda: multiply_numpy(data))
    timerFunctools = timeit.Timer(lambda: multiply_functools(data))
    timerManual = timeit.Timer(lambda: multiply_manual(data))

    repeats = int(5e6 / size)
    resultNumpy = timerNumpy.timeit(repeats)
    resultFunctools = timerFunctools.timeit(repeats)
    resultManual = timerManual.timeit(repeats)
    print(f'Input size: {size:>7d} Repeats: {repeats:>8d}    Numpy: {resultNumpy:.3f}, Functools: {resultFunctools:.3f}, Manual: {resultManual:.3f}')

Results:

Input size:       5 Repeats:  1000000    Numpy: 4.670, Functools: 0.586, Manual: 0.459
Input size:      10 Repeats:   500000    Numpy: 2.443, Functools: 0.401, Manual: 0.321
Input size:     100 Repeats:    50000    Numpy: 0.505, Functools: 0.220, Manual: 0.197
Input size:    1000 Repeats:     5000    Numpy: 0.303, Functools: 0.207, Manual: 0.185
Input size:   10000 Repeats:      500    Numpy: 0.265, Functools: 0.194, Manual: 0.187
Input size:  100000 Repeats:       50    Numpy: 0.266, Functools: 0.198, Manual: 0.185

You can see that Numpy is quite a bit slower on smaller inputs, since it allocates an array before multiplication is performed. Also, watch out for the overflow in Numpy.

I personally like this for a function that multiplies all elements of a generic list together:

def multiply(n):
    total = 1
    for i in range(0, len(n)):
        total *= n[i]
    print total

It’s compact, uses simple things (a variable and a for loop), and feels intuitive to me (it looks like how I’d think of the problem, just take one, multiply it, then multiply by the next, and so on!)

The simple way is:

import numpy as np
np.exp(np.log(your_array).sum())

Numpy has the prod() function that returns the product of a list, or in this case since it’s numpy, it’s the product of an array over a given axis:

import numpy
a = [1,2,3,4,5,6]
b = numpy.prod(a)

…or else you can just import numpy.prod():

from numpy import prod
a = [1,2,3,4,5,6]
b = prod(a)

Found this question today but I noticed that it does not have the case where there are None‘s in the list. So, the complete solution would be:

from functools import reduce

a = [None, 1, 2, 3, None, 4]
print(reduce(lambda x, y: (x if x else 1) * (y if y else 1), a))

In the case of addition, we have:

print(reduce(lambda x, y: (x if x else 0) + (y if y else 0), a))

nums = str(tuple([1,2,3]))
mul_nums = nums.replace(',','*')
print(eval(mul_nums))

I would like this in following way:

    def product_list(p):
          total =1 #critical step works for all list
          for i in p:
             total=total*i # this will ensure that each elements are multiplied by itself
          return total
   print product_list([2,3,4,2]) #should print 48

This is my code:

def product_list(list_of_numbers):
    xxx = 1
    for x in list_of_numbers:
        xxx = xxx*x
    return xxx

print(product_list([1,2,3,4]))

result : (‘1*1*2*3*4’, 24)

How about using recursion?

def multiply(lst):
    if len(lst) > 1:
        return multiply(lst[:-1])* lst[-1]
    else:
        return lst[0]

My solution:

def multiply(numbers):
    a = 1
    for num in numbers:
        a *= num
        return a

  pass

”’the only simple method to understand the logic use for loop”’

Lap=[2,5,7,7,9] x=1 for i in Lap: x=i*x print(x)

It is very simple do not import anything. This is my code. This will define a function that multiplies all the items in a list and returns their product.

def myfunc(lst):
    multi=1
    for product in lst:
        multi*=product
    return product