Sum a list of numbers in Python

Question

I have a list of numbers such as  1 2 3 4 5      and I want to calculate  1 2  2 and for the second   2 3  2 and the third   3 4  2  and so on  How can I do that   I would like to sum the first number with the second and divide it by 2  then sum the second with the third and divide by 2  and so on   Also  how can I sum a list of numbers   a    1  2  3  4  5         Is it   b   sum a  print b   to get one number   This doesn t work for me

User · Answer

Loop through elements in the list and update the total like this   def sum a       total   0     index   0     while index  lt  len a           total   total   a index          index   index   1     return total

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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    gt  gt  gt  nat seq    1  2  3  4  5  6  7  8  9  10    You can use the sum function on a list    gt  gt  gt  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    gt  gt  gt  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    gt  gt  gt  new seq     2 k-1  2  for k in nat seq 1     gt  gt  gt  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    gt  gt  gt  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    gt  gt  gt  print    new seq -1  2 1  2   2-1  2 49 5

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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

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So many solutions  but my favourite is still missing   gt  gt  gt  import numpy as np  gt  gt  gt  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   gt  gt  gt    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   gt  gt  gt  arr  -1  array  1  2  3  4    gt  gt  gt  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

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The simplest way to solve this problem   l   1 2 3 4 5  sum 0 for element in l      sum  element print sum

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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

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This question has been answered here a    1 2 3 4  sum a    sum a  returns 10

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Using the pairwise itertools recipe   import itertools def pairwise iterable        s - gt   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

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Short and simple   def ave x y     return  x   y    2 0  map ave  a  -1   a 1      And here s how it looks    gt  gt  gt  a   range 10   gt  gt  gt  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    gt  gt  gt  import itertools  gt  gt  gt  itertools imap ave  a  a 1     lt itertools imap object at 0x1005c3990 gt   gt  gt  gt  list     0 5  1 5  2 5  3 5  4 5  5 5  6 5  7 5  8 5

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Let us make it easy for Beginner -    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   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

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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

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I use a while loop to get the result   i   0 while i  lt  len a -1     result    a i  a i 1   2    print result    i   1

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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

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Sum list of numbers   sum list of nums    Calculating half of n and n - 1  if I have the pattern correct   using a list comprehension     x    x - 1     2 for x in list of nums    Sum adjacent elements  e g    1   2    2      2   3    2        using reduce and lambdas  reduce lambda x  y   x   y    2  list of nums

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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

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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

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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

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gt  gt  gt  a   range 10   gt  gt  gt  sum a  Traceback  most recent call last     File   lt stdin gt    line 1  in  lt module gt  TypeError   int  object is not callable  gt  gt  gt  del sum  gt  gt  gt  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

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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

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Question 1  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       Question 2   That use of sum should work fine  The following works   a   range 10     0 1 2 3 4 5 6 7 8 9  b   sum a  print b   Prints 45   Also  you don t need to assign everything to a variable at every step along the way  print sum a  works just fine   You will have to be more specific about exactly what you wrote and how it isn t working

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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

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In the spirit of itertools  Inspiration from the pairwise recipe   from itertools import tee  izip  def average iterable        s - gt   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    gt  gt  gt list average  1 2 3 4 5     1 5  2 5  3 5  4 5   gt  gt  gt list average  1 20 31 45 56 0 0     10 5  25 5  38 0  50 5  28 0  0 0   gt  gt  gt list average average  1 2 3 4 5      2 0  3 0  4 0

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In Python 3 8  the new assignment operator can be used   gt  gt  gt  my list    1  2  3  4  5   gt  gt  gt  itr   iter my list   gt  gt  gt  a   next itr   gt  gt  gt    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

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Using a simple list-comprehension and the sum    gt  gt  sum i for i in range x   2   if x   10 the result will be 22 5

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Thanks to Karl Knechtel i was able to understand your question  My interpretation    You want a new list with the average of the element i and i 1  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  gt 1 else      assuming you want result as float s   lambda l    l 0  l 1    2    s l 1    if len l  gt 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  gt 1 else 0   assuming you want result as float s   lambda l   l 0  l 1   2    s l 1    if len l  gt 1 else 0   assuming you want floor result   use the one that fits best your needs

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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

[python] Sum a list of numbers in Python

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