Add sum of numbers except certain digit / python

Question

New to using python. Was wondering how can I create a function to get the sum of numbers before a certain number in a list using while loop. For example, I want to get the sum of all numbers before the number 3 pops up in the list [1,2,5,2,3,2] should result in 10. Using my faulty code, my function doesn't consider the 3 and just adds all the numbers.

def nothree(nt):
    while nt != 3: 
        list_sum = sum(nt)
        return list_sum

Answer 1

def nothree(nt):
  i = 0
  sum = 0
  while nt[i]: 
    if nt[i] is 3:
       break
    sum += nt[i]
    i += 1
  return sum

this is so you can keep the while loop for whatever reason. But also in python you could something like:

def nothree(nt):
  for i in nt[:-2]: 
     sum += [i]
  return list_sum

Answer 2

You can use itertools.takewhile for this pattern:

from itertools import takewhile
def nothree(nt):
    return sum(takewhile(lambda x: x != 3, nt))

>>> nothree([1, 2, 5, 2, 3, 1])
10

While Fejs has you covered with an instructive loop-based solution, I might add this one-line hack that works without any libraries:

return sum(x if x != 3 else next(iter([])) for x in nt)

where the StopIteration raised by next(iter([])) will stop the generator on the first 3 .

Answer 3

Something like this:

def nothree(nt):
    sum = 0
    for n in nt:
        if n == 3:
            break
        sum += n
    return sum

If there is 3 in list, it will break the loop and return sum of numbers before 3 was reached. If there is no 3 in list, the sum will be sum of the list. And finally, if there is no numbers in list, sum will be 0.

Answer 4

You can do it one line as well

nt = [1,2,5,2,3,2]
x = sum(nt[:nt.index(3)])

nt[:nt.index(3)] will give you the list before 3 appears for the first time in the list. ie. [1, 2, 5, 2]