How to sum all integer after running a loop

Question

I have written a script, but I don't know if it's possible to have the total. My script would give me total number per loop, but I want to have the lump sum of everything. For instance, if I input 6, I would expect the result to be 168. if I input 5, I would expect the result to be 105.

Thanks

My script is as follow:

def multiples(n, high):
    i = 0
    res = 0
    while i <= high:
        res = res + (n+i)
        i = i + 1
    print res
    return res

def p(high):
    i = 1
    while i <= high:
        multiples(i, i)
        i = i + 1

p(6)  # Expected Output: 168

Have you thought about using a cumulative sum function? docs.scipy.org/doc/numpy/reference/generated/numpy.cumsum.html — zglin
– zglin, Commented Aug 10, 2016 at 19:41

Adam Smith · Accepted Answer · 2016-08-10 19:46:52Z

1

A few things,

First off, this is a very old-school type of looping.

i = 0
while i <= some_number:
    do_something
    i = i + 1

The last language I used that required a pattern like that was Basic. Use for loops. A standard for loop (note: NOT Python's, but I'll get to that in a minute) looks like for i=0; i<=some_number; i++ { do_something }. That is:

Initialize i
Conditional for when to keep looping
What to do after each loop block

Python is even more clear. for loops over any iterable, so:

for element in [1,3,5,7,9]:
    ...

Gives you 1, then 3, then 5, then 7, then 9 as element in the loop body. Use this in conjunction with the range built-in to loop N times.

for i in range(high):
    # do something `high` number of times

A naive re-write looks like:

def multiples(k, n):
    res = 0
    for i in range(n):
        res += (k+i)
    print(res)
    return res

def p(n):
    for i in range(n):
        multiples(i, i)

However that doesn't really give you what you want either. What you WANT is to assign the value multiples(i, i) TO something.

def p(n):
    total = 0
    for i in range(n):
        total += multiples(i, i)
    return total

Now we're tracking your grand total inside p, and returning it afterwards.

result = p(6)
print(result)  # does what you want

Of course there's no good reason to break up these two functions. You could just as easily write:

def p(n):
    total = 0
    for i in range(n):
        for j in range(i):
            total += j+i
    return total

answered Aug 10, 2016 at 19:46

Adam Smith

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1 Comment

Adam Smith Over a year ago

Note that depending on what you're trying to do, I may have introduced an off-by-one error. Your original implementation loops from 1 -> N, while mine loops from 0 -> N-1. This isn't an issue in multiples` but might be in p. Fix this with for i in range(1, n+1) instead of for i in range(n)

Community · Accepted Answer · 2017-05-23 12:00:58Z

0

With the most minimal change, this would allow p(6) to return 168:

def multiples(n, high):
   i = 0
   res = 0
   while i <= high:
      res = res + (n+i)
      i = i + 1
   print res
   return res

def p(high):
   i = 1
   lumpSum = 0
   while i <= high:
      lumpSum += multiples(i, i)
      i = i + 1
   return lumpSum

print p(6)

Alternatively, you can shorten p:

def p(high):
    return sum([multiples(i, i) for i in range(1, high + 1)])

Edit: The entire code can indeed be reduced to this: Adam Smith pointed out in his comment.

p = lambda n: sum(i+j for i in range(n) for j in range(i))

edited May 23, 2017 at 12:00

CommunityBot

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answered Aug 10, 2016 at 19:45

user823447

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1 Comment

Adam Smith Over a year ago

or indeed p = lambda n: sum(i+j for i in range(n) for j in range(i))

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