How to emulate sum() using a list comprehension?

Question

Is it possible to emulate something like sum() using list comprehension ?

For example - I need to calculate the product of all elements in a list :

l         


        

Answer 1

No; a list comprehension produces a list that is just as long as its input. You will need one of Python's other functional tools (specifically reduce() in this case) to fold the sequence into a single value.

Answer 2

List comprehension always creates another list, so it's not useful in combining them (e.g. to give a single number). Also, there's no way to make an assignment in list comprehension, unless you're super sneaky.

The only time I'd ever see using list comprehensions as being useful for a sum method is if you only want to include specific values in the list, or you don't have a list of numbers:

list = [1,2,3,4,5]
product = [i for i in list if i % 2 ==0] # only sum even numbers in the list
print sum(product)

or another example":

# list of the cost of fruits in pence
list = [("apple", 55), ("orange", 60), ("pineapple", 140), ("lemon", 80)]
product = [price for fruit, price in list]
print sum(product)

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Super sneaky way to make an assignment in a list comprehension

dict = {"val":0}
list = [1, 2, 3]
product = [dict.update({"val" : dict["val"]*i}) for i in list]
print dict["val"] # it'll give you 6!

...but that's horrible :)

Answer 3

>>> from operator import mul
>>> nums = [1, 2, 3]
>>> reduce(mul, nums)
6

_{On Python 3 you will need to add this import: from functools import reduce}

Implementation Artifact

In Python 2.5 / 2.6 You could use vars()['_[1]'] to refer to the list comprehension currently under construction. This is horrible and should never be used but it's the closest thing to what you mentioned in the question (using a list comp to emulate a product).

>>> nums = [1, 2, 3]
>>> [n * (vars()['_[1]'] or [1])[-1] for n in nums][-1]
6

Answer 4

Found the magic on http://code.activestate.com/recipes/436482/.

>>> L=[2, 3, 4]
>>> [j for j in [1] for i in L for j in [j*i]][-1]
24

It should be the logic like the following code.

L=[2, 3, 4]
P=[]
for j in [1]:
    for i in L:
        for j in [j*i]:
            P.append(j)
print(P[-1])

Answer 5

Something like this should suffice:

list = [1, 2, 3]
product = product_of([i for i in list])  

Answer 6

I complement the answer of Ignacio Vazquez-Abrams with some code that uses the reduce operator of Python.

list_of_numbers = [1, 5, 10, 100]
reduce(lambda x, y: x + y, list_of_numbers)

which can also be written as

list_of_numbers = [1, 5, 10, 100]

def sum(x, y):
    return x + y

reduce(sum, list_of_numbers)

Bonus: Python provides this functionality in the built-in sum function. This is the most readable expression imo.

list_of_numbers = [1, 5, 10, 100]
sum(list_of_numbers)