sum of products of couples in a list

Question

I want to find out the sum of products of couples in a list. For example a list is given [1, 2, 3, 4] . What I want to get is answer = 1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4 .

I do it using brute-force, it gives me the time-out error for very large lists. I want an efficient way to do this. Kindly tell me, how can I do this?

Here is my code, this is working but i need more efficient one:

def proSum(list):
    count  = 0
    for i in range(len(list)- 1):
        for j in range(i + 1, len(list)):
            count +=  list[i] * list[j]
    return count

Answer 1

Here it is:

In [1]: def prodsum(xs):
   ...:     return (sum(xs)**2 - sum(x*x for x in xs)) / 2
   ...: 

In [2]: prodsum([1, 2, 3, 4]) == 1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4
Out[2]: True

Let xs = a1, a2, .., an , then

(a1+a2+...+an)^2 = 2(a1a2+a1a3+...+an-1an) + (a1^2+...+an^2)

So we have

a1a2+...+an-1an = {(a1+a2+...+an)^2 - (a1^2+...+an^2)}/2

---

Compare the performance of @georg's method and mine

The result and the test codes as following(The less time used is better):

In [1]: import timeit

In [2]: import matplotlib.pyplot as plt

In [3]: def eastsunMethod(xs):
   ...:     return (sum(xs)**2 - sum(x*x for x in xs)) / 2
   ...: 

In [4]: def georgMethod(given):
   ...:     sum = 0
   ...:     res = 0
   ...:     cur = len(given) - 1
   ...: 
   ...:     while cur >= 0:
   ...:         res += given[cur] * sum
   ...:         sum += given[cur]
   ...:         cur -= 1
   ...:     return res
   ...: 

In [5]: sizes = range(24)

In [6]: esTimes, ggTimes = [], []

In [7]: for s in sizes:
   ...:     t1 = timeit.Timer('eastsunMethod(xs)', 'from __main__ import eastsunMethod;xs=range(2**%d)' % s)
   ...:     t2 = timeit.Timer('georgMethod(xs)', 'from __main__ import georgMethod;xs=range(2**%d)' % s)
   ...:     esTimes.append(t1.timeit(8))
   ...:     ggTimes.append(t2.timeit(8))

In [8]: fig, ax = plt.subplots(figsize=(18, 6));lines = ax.plot(sizes, esTimes, 'r', sizes, ggTimes);ax.legend(lines, ['Eastsun', 'georg'], loc='center');ax.set_xlabel('size');ax.set_ylabel('time');ax.set_xlim([0, 23])

Answer 2

Use itertools.combinations to generate unique pairs:

# gives [(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
unique_pairs = list(itertools.combinations([1, 2, 3, 4], 2))

Then use a list comprehension to get the product of each pair:

products = [x*y for x, y in unique_pairs] # => [2, 3, 4, 6, 8, 12]

Then use sum to add up your products:

answer = sum(products) # => 35

This can be all wrapped up in a one-liner like so:

answer = sum(x*y for x,y in itertools.combinations([1, 2, 3, 4], 2))

In making it a one-liner the result of combinations is used without casting to a list . Also, the brackets around the list comprehension are discarded, transforming it generator expression .

Note : Eastsun's answer and georg's answer use much better algorithms and will easily outpreform my answer for large lists.

Answer 3

Note : actually @Eastsun's answer is better.

Here's another, more "algorithmical" way to deal with that. Observe that given

a ₀ , a ₁ , ..., a _n

the desired sum is (due to the distributive law)

a ₀ (a ₁ + a ₂ + ... + a _n ) + a ₁ (a ₂ + a ₃ + ... + a _n ) + ... + a _n-2 (a _n-1 + a _n ) + a _n-1 a _n

which leads to the following algorithm:

• let sum be 0 and current be the last element
• on each step
  • multiply sum and current and add to the result
  • add current to sum
  • let current be the previous of current

In python:

sum = 0
res = 0
cur = len(given) - 1

while cur >= 0:
    res += given[cur] * sum
    sum += given[cur]
    cur -= 1

print res

Answer 4

With no external library, you can use map and lambda to calculate * pairwise, and then sum everything up

l=[1, 2, 3, 4]
sum(map(lambda x,y:x*y, l, l[1:]+[l[0]]))

But since you are dealing with big data, I suggest you use numpy.

import numpy as np

l = np.array([1, 2, 3, 4])

print sum(l*np.roll(l, 1))
# 24

EDIT: to keep up with the updated question of OP

import numpy as np

l = [1, 2, 3, 4]
sums = 0
while l:
    sums+=sum(l.pop(0)*np.array(l))

print sums
#35

So what it does is taking out the first element of list and * with the rest. Repeating until there is nothing to take out from the list.

Answer 5

def sumOfProductsOfCouples（l）：返回sum（l [i-1] * l [i]，代表inumerate（l）中的i，n）

Answer 6

from itertools import combinations
l=[1, 2, 3, 4]
cnt=0
for x in combinations(l,2):
    cnt+=x[0]*x[1]
print (cnt)

Output;

>>> 
35
>>> 

combinations() will give your pairs like you want. Then do your calculates.

Debug it like ;

l=[1, 2, 3, 4]
for x in combinations(l,2):
    print (x)

>>> 
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
>>> 

See that your pairs are here. Actually you will find the sum of this combinations pairs.

Answer 7

Use the permutations method from the itertools module:

from itertools import *

p = permutations([1, 2, 3, 4], 2) # generate permutations of two
p = [frozenset(sorted(i)) for i in p] # sort items and cast 
p = [list(i) for i in set(p)] # remove duplicates, back to lists

total = sum([i[0] * i[1] for i in p]) # 35 your final answer

Answer 8

you can use map, sum functions.

>>> a = [1, 2, 3, 4]
>>> sum(map(sum, [map(lambda e: e*k, l) for k, l in zip(a, (a[start:] for start, _ in enumerate(a, start=1) if start < len(a)))]))
35

Dividing above expression in parts,

>>> a = [1, 2, 3, 4]
>>> c = (a[start:] for start, _ in enumerate(a, start=1) if start < len(a))
>>> sum(map(sum, [map(lambda e: e*k, l) for k, l in zip(a, c)]))
35