对Python中的数字列表求和

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.

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.

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)

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)

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

>>> 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.

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

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

The simplest way to solve this problem:

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

This question has been answered here

a = [1,2,3,4] sum(a) returns 10

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]

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]

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

2. 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

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) )

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]

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 elemens (1+2), (2+3) and so on. Not 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 lops-within-loops I would otherwise have had to make.

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

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

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

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

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))

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))]

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]

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)

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.

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:].

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)