Replace nested for loop with list comprehension in Python

Question

Given a list of integers c, I want to perform the following:

Count how many elements of c are less than index i, where i is from 0 to len(c). For example: c=[1, 0, 2, 1], then the output to the above will be b=[1, 3, 4, 4] because only the 1 element of c is less than or equal to i=0, then there are 3 elements of c less than or equal to i=1, then there are 4 elements of c less than or equal to i=2, and lastly there are 4 elements of c less than or equal to i=3.
Find a=[b[0] b[1]-1 b[2]-2 ... ]
Find math.prod(a)

I manage to do it using nested for loops, but I am learning the list comprehensions and want to transform them into nested for loops inside the list comprehensions.

#this code works
def solve(c):
    a=1
    for i in range(len(c)):
        a=a*(sum(element<=i for element in c)-i)
    return(a)
#input c=[1, 0, 2, 1]
#output 4

But this one fails:

$this code does not work
def solve(c):
    a=1
    return([(a=a*(sum(element<=i for element in c)-i)) for i in range(len(c))])

Having an assignment inside the list comprehension doesn't make sense, and you don't seem to be using math.prod at all. Start smaller, try making b with a list comprehension or converting [1, 2, 2, 1] to 4 with math.prod, then put the pieces together. — jonrsharpe
– jonrsharpe, Commented Aug 24, 2022 at 15:52
@jonrsharpe b=[sum(element<=i for element in c) for i in range(len(c))] — Lee
– Lee, Commented Aug 24, 2022 at 15:59
@jonrsharpe thanks, I got the answer now. a=[(b[i]-i) for i in range(len(c))] and then return(math.prod(a)). — Lee
– Lee, Commented Aug 24, 2022 at 16:10

Sami Tahri · Accepted Answer · 2022-08-24 16:03:39Z

1

[[item for item in c if item <= ind] for ind in range(len(c))] give you the list of item that are < for the index.

[len([item for item in c if item <= ind]) - ind for ind in range(len(c))] will give you your second point.

answered Aug 24, 2022 at 16:03

Sami Tahri

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