如何在执行 itertools.combinations 时限制列表中的元素数量？

Question

Hi im writting a small project. One of the elements of my code is creating combinations of wages. What I try to do is to get all possible combinations of 4 numbers (from 0.0 to 1.0) that will give me a sum of 1.0. I loop with step = 5 to get it fast.

for i in range(0,101,5):
  wage = i/100
  l_wages.append(wage)
numbers = l_wages
result = [list(seq) for i in range(len(numbers), 0, -1) for seq in itertools.combinations(numbers, i) if sum(seq) == 1.0]
result

I want to have wages that sum up to 1. I already achived that. The only thing i need to do now is to have a situation when in every list there are 4 elements. There is often an output like "0.9,0.1". And i would like it to be "0.9,0.1,.0.0,0.0".

Hope that somebody will help me solve that problem.

Answer 1

if you want to be in this format: [0.9, 0.1, 0.0, 0.0] .

it means that the values are going to repeat .

the meaning of combinations is that they are sorted and they dont repeat . that is the first thing you learn at backtracking at highschool.

there is a quick solution for you:

this solution doesnt implement combinations

l_wages = []
for i in range(0,101,5):
    wage = i/100
    l_wages.append(wage)

solutions = []
for x in l_wages:
    for y in l_wages:
        for z in l_wages:
            for w in l_wages:
                sol = [x, y, z, w]
                if sum(sol) == 1.0:
                    solutions.append(sol)

for s in solutions:
    print(s)
print(len(solutions))

output (there are a lot of solutions )

computation takes around 2-3 seconds (if you print them)

[0.0, 0.0, 0.0, 1.0]
[0.0, 0.0, 0.05, 0.95]
[0.0, 0.0, 0.1, 0.9]
[0.0, 0.0, 0.15, 0.85]
[0.0, 0.0, 0.2, 0.8]
[0.0, 0.0, 0.25, 0.75]
[0.0, 0.0, 0.3, 0.7]
...
...
...
[0.9, 0.0, 0.05, 0.05]
[0.9, 0.0, 0.1, 0.0]
[0.9, 0.05, 0.0, 0.05]
[0.9, 0.05, 0.05, 0.0]
[0.9, 0.1, 0.0, 0.0]
[0.95, 0.0, 0.0, 0.05]
[0.95, 0.0, 0.05, 0.0]
[0.95, 0.05, 0.0, 0.0]
[1.0, 0.0, 0.0, 0.0]
1680