Python：将列表中的所有元素相乘，但先减去1

Question

What is the most pythonic and elegant way to multiply the values of all elements of a list minus 1 together in Python?

Ie, calculate (p[0]-1) * (p[1]-1) * ... * (p[n-1]-1) where n is the size of the list p.

Answer 1

Use functools.reduce in python3 along with AJ's example:

>>> l = [5, 8, 7, 2, 3]
>>> l = [item-1 for item in l] #subtract 1
>>> functools.reduce(lambda x, y: x*y, l) #multiply each item
336
>>> 

Answer 2

Using the numpy package

import numpy as np
np.prod(np.array(p)-1)

Or using a python built-in one, eg operator

reduce(operator.mul,\ 
       map(lambda el:el-1,\
           p),\
       1)

Answer 3

>>> l = [5, 8, 7, 2, 3]
>>> reduce(lambda x, y: x*(y-1), l, 1) #multiply each item by its subtracted value
336
>>> 

Thanks to @AChampion for another improvement

Answer 4

There are many ways to multiply all the elements of an iterable.

So, given a list a you can, for example:

Use numpy:

prod([x-1 for x in a])

Use lambda and functools

functools.reduce(lambda x, y: x*y, [x-1 for x in a])

Do it manually

result=1
for x in [x-1 for x in a]: result*=a 

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Note that I used the list comprehension: [x-1 for x in a] but it can also be achieved with numpy: array(a)-1

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Related question: What's the Python function like sum() but for multiplication? product()?

Answer 5

result = 1
for x in p:
    result *= x - 1