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pythonlambdareduceenumerate
Szakul1
Szakul1

Reputation: 147

Reduce with enumerate Python

I'm trying to get product of list where every element is multiplied by its index + 1. I tried with reduce and enumerate

value = reduce(lambda a, b: (a[0]+1)*a[1]*b[1], enumerate(list))

but it gives TypeError: 'int' object is not subscriptable. Is it possible to do it in one line?

Editv2

example of list [1,2,3]

desire output list[0]*1 * list[1]*2 * list[2]*3 = 36

Upvotes: 0

Views: 736

Answers (3)

Sanjay Subramaniam
Sanjay Subramaniam

Reputation: 98

The trick is to make a new list and multiply through it.

Create a new list where a number at index i is i*list[i], indexing starting with 1:

>>> new_list = [a*b for a, b in enumerate(list, 1)]
>>> new_list
[1, 4, 9]

and multiply over your new list:

>>> reduce((lambda x, y: x*y), new_list)
36

In one line:

>>> reduce((lambda x, y: x*y), [a*b for a, b in enumerate(list, 1)])
36

Hope it helped :)

Note: Answer edited to meet OP's changes.

Upvotes: 1

user2390182
user2390182

Reputation: 73460

Simplest:

lst = [1,2,3]  # do not shadow `list`
sum(i*v for i, v in enumerate(lst, 1))
# 14

Your reduce approach fails as it returns an int which is not the same type that it expects as its two inputs (tuple). You could do it with:

reduce(lambda a, b: (a[0]*a[1] + b[0]*b[1], 1), enumerate(lst, 1))[0]
# 14

Note the (..., 1) structure of the lambda expression where the 1 serves as an auxiliary factor for the next step.

Update: As you actually want the homogenous product of all the elements of your nested iterable, the following is simpler:

from itertools import chain
from operator import mul

reduce(mul, chain(*enumerate(lst, 1)))
# 36

Upvotes: 4

DimK
DimK

Reputation: 331

ans = sum(i * x for i, x in enumerate(list, 1))

Not exactly what you are asking for, but it gives the desired result.

Upvotes: 0

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