layo
layo

Reputation: 4683

Sum a list of numbers in Python

Given a list of numbers such as:

[1, 2, 3, 4, 5, ...]

How do I calculate their total sum:

1 + 2 + 3 + 4 + 5 + ...

How do I calculate their pairwise averages:

[(1+2)/2, (2+3)/2, (3+4)/2, (4+5)/2, ...]

Upvotes: 461

Views: 2253044

Answers (26)

Karl Knechtel
Karl Knechtel

Reputation: 61643

Question 1:

To sum a list of numbers, use sum:

xs = [1, 2, 3, 4, 5]
print(sum(xs))

This outputs:

15

Question 2:

So you want (element 0 + element 1) / 2, (element 1 + element 2) / 2, ... etc.

We make two lists: one of every element except the first, and one of every element except the last. Then the averages we want are the averages of each pair taken from the two lists. We use zip to take pairs from two lists.

I assume you want to see decimals in the result, even though your input values are integers. By default, Python does integer division: it discards the remainder. To divide things through all the way, we need to use floating-point numbers. Fortunately, dividing an int by a float will produce a float, so we just use 2.0 for our divisor instead of 2.

Thus:

averages = [(x + y) / 2.0 for (x, y) in zip(my_list[:-1], my_list[1:])]

Upvotes: 381

Rafe Kettler
Rafe Kettler

Reputation: 76975

To sum a list of numbers:

sum(list_of_nums)

Generate a new list with adjacent elements averaged in xs using a list comprehension:

[(x + y) / 2 for x, y in zip(xs, xs[1:])]

Sum all those adjacent elements into a single value:

sum((x + y) / 2 for x, y in zip(xs, xs[1:]))

Upvotes: 145

Zak
Zak

Reputation: 3283

So many solutions, but my favourite is still missing:

>>> import numpy as np
>>> arr = np.array([1,2,3,4,5])

a numpy array is not too different from a list (in this use case), except that you can treat arrays like numbers:

>>> ( arr[:-1] + arr[1:] ) / 2.0
[ 1.5  2.5  3.5  4.5]

Done!

explanation

The fancy indices mean this: [1:] includes all elements from 1 to the end (thus omitting element 0), and [:-1] are all elements except the last one:

>>> arr[:-1]
array([1, 2, 3, 4])
>>> arr[1:]
array([2, 3, 4, 5])

So adding those two gives you an array consisting of elements (1+2), (2+3) and so on. Note that I'm dividing by 2.0, not 2 because otherwise Python believes that you're only using integers and produces rounded integer results.

advantage of using numpy

Numpy can be much faster than loops around lists of numbers. Depending on how big your list is, several orders of magnitude faster. Also, it's a lot less code, and at least to me, it's easier to read. I'm trying to make a habit out of using numpy for all groups of numbers, and it is a huge improvement to all the loops and loops-within-loops I would otherwise have had to write.

Upvotes: 6

Vijay
Vijay

Reputation: 1088

This question has been answered here

a = [1,2,3,4]
sum(a) 

sum(a) returns 10

Upvotes: 17

vijayraj34
vijayraj34

Reputation: 2415

You can also do the same using recursion:

Python Snippet:

def sumOfArray(arr, startIndex):
    size = len(arr)
    if size == startIndex:  # To Check empty list
        return 0
    elif startIndex == (size - 1): # To Check Last Value
        return arr[startIndex]
    else:
        return arr[startIndex] + sumOfArray(arr, startIndex + 1)


print(sumOfArray([1,2,3,4,5], 0))

Upvotes: 2

chepner
chepner

Reputation: 532418

In Python 3.8, the new assignment operator can be used

>>> my_list = [1, 2, 3, 4, 5]
>>> itr = iter(my_list)
>>> a = next(itr)
>>> [(a + (a:=x))/2 for x in itr]
[1.5, 2.5, 3.5, 4.5]

a is a running reference to the previous value in the list, hence it is initialized to the first element of the list and the iteration occurs over the rest of the list, updating a after it is used in each iteration.

An explicit iterator is used to avoid needing to create a copy of the list using my_list[1:].

Upvotes: 3

Aamir M Meman
Aamir M Meman

Reputation: 1841

Let us make it easy for Beginner:-

  1. The global keyword will allow the global variable message to be assigned within the main function without producing a new local variable
    message = "This is a global!"


def main():
    global message
    message = "This is a local"
    print(message)


main()
# outputs "This is a local" - From the Function call
print(message)
# outputs "This is a local" - From the Outer scope

This concept is called Shadowing

  1. Sum a list of numbers in Python
nums = [1, 2, 3, 4, 5]

var = 0


def sums():
    for num in nums:
        global var
        var = var + num
    print(var)


if __name__ == '__main__':
    sums()

Outputs = 15

Upvotes: 3

StickySli
StickySli

Reputation: 101

Thanks to Karl Knechtel i was able to understand your question. My interpretation:

  1. You want a new list with the average of the element i and i+1.
  2. You want to sum each element in the list.

First question using anonymous function (aka. Lambda function):

s = lambda l: [(l[0]+l[1])/2.] + s(l[1:]) if len(l)>1 else []  #assuming you want result as float
s = lambda l: [(l[0]+l[1])//2] + s(l[1:]) if len(l)>1 else []  #assuming you want floor result

Second question also using anonymous function (aka. Lambda function):

p = lambda l: l[0] + p(l[1:]) if l!=[] else 0

Both questions combined in a single line of code :

s = lambda l: (l[0]+l[1])/2. + s(l[1:]) if len(l)>1 else 0  #assuming you want result as float
s = lambda l: (l[0]+l[1])/2. + s(l[1:]) if len(l)>1 else 0  #assuming you want floor result

use the one that fits best your needs

Upvotes: 1

Sai G
Sai G

Reputation: 33

Try the following -

mylist = [1, 2, 3, 4]   

def add(mylist):
    total = 0
    for i in mylist:
        total += i
    return total

result = add(mylist)
print("sum = ", result)

Upvotes: -4

Jesse
Jesse

Reputation: 213

A simple way is to use the iter_tools permutation

# If you are given a list

numList = [1,2,3,4,5,6,7]

# and you are asked to find the number of three sums that add to a particular number

target = 10
# How you could come up with the answer?

from itertools import permutations

good_permutations = []

for p in permutations(numList, 3):
    if sum(p) == target:
        good_permutations.append(p)

print(good_permutations)

The result is:

[(1, 2, 7), (1, 3, 6), (1, 4, 5), (1, 5, 4), (1, 6, 3), (1, 7, 2), (2, 1, 7), (2, 3, 
5), (2, 5, 3), (2, 7, 1), (3, 1, 6), (3, 2, 5), (3, 5, 2), (3, 6, 1), (4, 1, 5), (4, 
5, 1), (5, 1, 4), (5, 2, 3), (5, 3, 2), (5, 4, 1), (6, 1, 3), (6, 3, 1), (7, 1, 2), 
(7, 2, 1)]

Note that order matters - meaning 1, 2, 7 is also shown as 2, 1, 7 and 7, 1, 2. You can reduce this by using a set.

Upvotes: 0

Md. Rezwanul Haque
Md. Rezwanul Haque

Reputation: 2950

You can try this way:

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
sm = sum(a[0:len(a)]) # Sum of 'a' from 0 index to 9 index. sum(a) == sum(a[0:len(a)]
print(sm) # Python 3
print sm  # Python 2

Upvotes: 36

Samueltommzy
Samueltommzy

Reputation: 19

Loop through elements in the list and update the total like this:

def sum(a):
    total = 0
    index = 0
    while index < len(a):
        total = total + a[index]
        index = index + 1
    return total

Upvotes: 1

Aashutosh jha
Aashutosh jha

Reputation: 628

I use a while loop to get the result:

i = 0
while i < len(a)-1:
   result = (a[i]+a[i+1])/2
   print result
   i +=1

Upvotes: 1

Ely
Ely

Reputation: 11174

All answers did show a programmatic and general approach. I suggest a mathematical approach specific for your case. It can be faster in particular for long lists. It works because your list is a list of natural numbers up to n:

Let's assume we have the natural numbers 1, 2, 3, ..., 10:

>>> nat_seq = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

You can use the sum function on a list:

>>> print sum(nat_seq)
55

You can also use the formula n*(n+1)/2 where n is the value of the last element in the list (here: nat_seq[-1]), so you avoid iterating over elements:

>>> print (nat_seq[-1]*(nat_seq[-1]+1))/2
55

To generate the sequence (1+2)/2, (2+3)/2, ..., (9+10)/2 you can use a generator and the formula (2*k-1)/2. (note the dot to make the values floating points). You have to skip the first element when generating the new list:

>>> new_seq = [(2*k-1)/2. for k in nat_seq[1:]]
>>> print new_seq
[1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5]

Here too, you can use the sum function on that list:

>>> print sum(new_seq)
49.5

But you can also use the formula (((n*2+1)/2)**2-1)/2, so you can avoid iterating over elements:

>>> print (((new_seq[-1]*2+1)/2)**2-1)/2
49.5

Upvotes: 14

vasanth kumar
vasanth kumar

Reputation: 1

n = int(input("Enter the length of array: "))
list1 = []
for i in range(n):
    list1.append(int(input("Enter numbers: ")))
print("User inputs are", list1)

list2 = []
for j in range(0, n-1):
    list2.append((list1[j]+list1[j+1])/2)
print("result = ", list2)

Upvotes: 0

user4183543
user4183543

Reputation: 329

>>> a = range(10)
>>> sum(a)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'int' object is not callable
>>> del sum
>>> sum(a)
45

It seems that sum has been defined in the code somewhere and overwrites the default function. So I deleted it and the problem was solved.

Upvotes: 28

Jochen Ritzel
Jochen Ritzel

Reputation: 107786

Generators are an easy way to write this:

from __future__ import division
# ^- so that 3/2 is 1.5 not 1

def averages( lst ):
    it = iter(lst) # Get a iterator over the list
    first = next(it)
    for item in it:
        yield (first+item)/2
        first = item

print list(averages(range(1,11)))
# [1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5]

Upvotes: 3

Shapi
Shapi

Reputation: 5613

Using a simple list-comprehension and the sum:

>> sum(i for i in range(x))/2. #if x = 10 the result will be 22.5

Upvotes: 16

tags
tags

Reputation: 4060

import numpy as np    
x = [1,2,3,4,5]
[(np.mean((x[i],x[i+1]))) for i in range(len(x)-1)]
# [1.5, 2.5, 3.5, 4.5]

Upvotes: 3

Evans Murithi
Evans Murithi

Reputation: 3267

Question 2: To sum a list of integers:

a = [2, 3, 5, 8]
sum(a)
# 18
# or you can do:
sum(i for i in a)
# 18

If the list contains integers as strings:

a = ['5', '6']
# import Decimal: from decimal import Decimal
sum(Decimal(i) for i in a)

Upvotes: 79

shashankS
shashankS

Reputation: 1073

The simplest way to solve this problem:

l =[1,2,3,4,5]
sum=0
for element in l:
    sum+=element
print sum

Upvotes: 6

Orane
Orane

Reputation: 2261

I'd just use a lambda with map()

a = [1,2,3,4,5,6,7,8,9,10]
b = map(lambda x, y: (x+y)/2.0, fib[:-1], fib[1:])
print b

Upvotes: 0

Michael J. Barber
Michael J. Barber

Reputation: 25052

Short and simple:

def ave(x,y):
  return (x + y) / 2.0

map(ave, a[:-1], a[1:])

And here's how it looks:

>>> a = range(10)
>>> map(ave, a[:-1], a[1:])
[0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5]

Due to some stupidity in how Python handles a map over two lists, you do have to truncate the list, a[:-1]. It works more as you'd expect if you use itertools.imap:

>>> import itertools
>>> itertools.imap(ave, a, a[1:])
<itertools.imap object at 0x1005c3990>
>>> list(_)
[0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5]

Upvotes: 2

tzot
tzot

Reputation: 96071

Using the pairwise itertools recipe:

import itertools
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = itertools.tee(iterable)
    next(b, None)
    return itertools.izip(a, b)

def pair_averages(seq):
    return ( (a+b)/2 for a, b in pairwise(seq) )

Upvotes: 4

kevpie
kevpie

Reputation: 26108

In the spirit of itertools. Inspiration from the pairwise recipe.

from itertools import tee, izip

def average(iterable):
    "s -> (s0,s1)/2.0, (s1,s2)/2.0, ..."
    a, b = tee(iterable)
    next(b, None)
    return ((x+y)/2.0 for x, y in izip(a, b))

Examples:

>>>list(average([1,2,3,4,5]))
[1.5, 2.5, 3.5, 4.5]
>>>list(average([1,20,31,45,56,0,0]))
[10.5, 25.5, 38.0, 50.5, 28.0, 0.0]
>>>list(average(average([1,2,3,4,5])))
[2.0, 3.0, 4.0]

Upvotes: 1

TartanLlama
TartanLlama

Reputation: 65770

Try using a list comprehension. Something like:

new_list = [(old_list[i] + old_list[i+1])/2 for i in range(len(old_list-1))]

Upvotes: 0

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