Reputation: 4683
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
Reputation: 61643
To sum a list of numbers, use sum
:
xs = [1, 2, 3, 4, 5]
print(sum(xs))
This outputs:
15
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
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
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
Reputation: 1088
This question has been answered here
a = [1,2,3,4]
sum(a)
sum(a) returns 10
Upvotes: 17
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
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
Reputation: 1841
Let us make it easy for Beginner:-
global
keyword will allow the global variable message to be assigned within the main function without producing a new local variablemessage = "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
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
Reputation: 101
Thanks to Karl Knechtel i was able to understand your question. My interpretation:
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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