Python 循环或任何迭代以找到满足条件的所有组合

Question

I want to find all possible combinations of n numbers such that the sum is = 100 in Python

A sample of 2 numbers:

x=[]
for i, j in itertools.product(range(0,101), range(0,101)):
    if i+j==100:
       x.append([i,j])

Any alternative and clever way to do this with a variable number of iteration variables and get the outcome in the form of this: n=5:

[[10,10,10,30,40], [100,0,0,0,0], [1,1,2,3,97] .......]

Answer 1

itertools.product takes a repeat argument, which you can use to repeat the range(1, 101) iterator repeat number of times. This way you don't need to specify the iterator multiple times or generate the desired number of arguments. For example, for 5 times:

[i for i in itertools.product(range(1, 101), repeat=5) if sum(i) == 100]

Answer 2

A Pure Python Solution (ie without itertools.product )

The main difficulty here is executing a variable number of for-loops inside a function. The way we can get around this easily is using recursion which involves a function calling itself.

If we use recursion, then inside any instance of the function, only one for-loop is actually iterated through. So to apply this to the problem at hand, we want our function to take two parameters: the target for what number we are trying to sum to, and n - the number of positive integers we have available to use.

Each function will then return (given a target and n numbers), all the combinations that will make that target - in the form of a two-dimensional list .

The only special case that we must consider is the "leaf nodes" of our recursive tree (the cases where we have a certain target, but n == 1 , so we only have one number to make the target with). This is easy to handle, we just need to remember that we should always return all combinations that make the target so in this case, there is only one "combination" which is the target .

Then (if n > 1 ) the rest is self explanatory, we are simply looping through every number less than target and adding to a list of combinations ( cs ) with the results of calling the function again.

However, before we concatenate these combos onto our list , we need to use a comprehension to add i (the next number) to the start of every combination.

And that's it! Hopefully you can see how the above translates into the following code:

def combos(target, n):
    if n == 1:
        return [[target]]
    cs = []
    for i in range(0, target+1):
        cs += [[i]+c for c in combos(target-i, n-1)]
    return cs

---

and a test (with target as 10 and n as 3 to make it clearer) shows it works:

>>> combos(10, 3)
[[0, 0, 10], [0, 1, 9], [0, 2, 8], [0, 3, 7], [0, 4, 6], [0, 5, 5], [0, 6, 4], [0, 7, 3], [0, 8, 2], [0, 9, 1], [0, 10, 0], [1, 0, 9], [1, 1, 8], [1, 2, 7], [1, 3, 6], [1, 4, 5], [1, 5, 4], [1, 6, 3], [1, 7, 2], [1, 8, 1], [1, 9, 0], [2, 0, 8], [2, 1, 7], [2, 2, 6], [2, 3, 5], [2, 4, 4], [2, 5, 3], [2, 6, 2], [2, 7, 1], [2, 8, 0], [3, 0, 7], [3, 1, 6], [3, 2, 5], [3, 3, 4], [3, 4, 3], [3, 5, 2], [3, 6, 1], [3, 7, 0], [4, 0, 6], [4, 1, 5], [4, 2, 4], [4, 3, 3], [4, 4, 2], [4, 5, 1], [4, 6, 0], [5, 0, 5], [5, 1, 4], [5, 2, 3], [5, 3, 2], [5, 4, 1], [5, 5, 0], [6, 0, 4], [6, 1, 3], [6, 2, 2], [6, 3, 1], [6, 4, 0], [7, 0, 3], [7, 1, 2], [7, 2, 1], [7, 3, 0], [8, 0, 2], [8, 1, 1], [8, 2, 0], [9, 0, 1], [9, 1, 0], [10, 0, 0]]

---

Improving performance

If we consider the case where we are trying to make 10 with 4 numbers. At one point, the function will be called with a target of 6 after say 1 and 3 . The algorithm will as we have already explained and return the combinations using 2 numbers to make 6 . However, if we now consider another case further down the line when the function is asked to give the combinations that make 6 (same as before) having been called with say 2 and 2 . Notice how even though we will get the right answer (through our recursion and the for-loop ), we will return the same combinations as before - when we were called with 1 and 3 . Furthermore, this scenario will happen extremely often: the function will be called from different situations but be asked to give the same combinations that have already been previously calculated at a different time.

This gives way to a great optimisation technique called memoization which essentially just means storing the results of our function as a key: value pair in a dictionary ( mem ) before returning.

We then just check at the start of every function call if we have ever been called before with the same parameters (by seeing if the key is in the dictionary) and if it is, then we can just return the result we got last time.

This speeds up the algorithm dramatically.

mem = {}
def combos(target, n):
    k = (target, n)
    if k in mem:
        return mem[k]
    if n == 1:
        return [[target]]
    cs = []
    for i in range(0, target+1):
        cs += [[i]+c for c in combos(target-i, n-1)]
    mem[k] = cs
    return cs

Answer 3

This is a trivial generalisation / partial optimisation of your algorithm.

Edit : @heemayl's alternative which uses repeat is preferable to this solution.

import itertools

n = 3
x = []

x = [list(i) for i in itertools.product(*(range(0,101) \
             for _ in range(n))) if sum(i) == 100]