Question

我有两个数字，N和L（比方说5和3）。

如何生成列表总和等于N（5）且每个list的长度为L（3）的每个可能list？

示例输出（在本例中）：

[0, 0, 5]
[0, 1, 4]
[0, 2, 3]
[0, 3, 2]
...
[0, 5, 0]
...
[1, 4, 0]
...
[5, 0, 0]

我已检出itertools及其combinations和permutations功能，但它们似乎不适合执行此任务。

Answer 1

您可以创建一个递归函数来生成具有给定条件的所有可能的排列，然后过滤以仅保留总和为所需值的列表：

def list_results(a, b):
   return [i for i in permutations(b) if sum(i) == a]

def permutations(d, current = []):
   if len(current) == d:
     yield current
   else:
     for i in range(10):
        yield from permutations(d, current+[i])

print(list_results(5, 3))

输出：

[[0, 0, 5], [0, 1, 4], [0, 2, 3], [0, 3, 2], [0, 4, 1], [0, 5, 0], [1, 0, 4], [1, 1, 3], [1, 2, 2], [1, 3, 1], [1, 4, 0], [2, 0, 3], [2, 1, 2], [2, 2, 1], [2, 3, 0], [3, 0, 2], [3, 1, 1], [3, 2, 0], [4, 0, 1], [4, 1, 0], [5, 0, 0]]

编辑：稍快一点，需要额外检查递归函数：

import time
def timeit(f):
   def wrapper(*args, **kwargs):
      c = time.time()
      results = list(f(*args, **kwargs))
      print("Result from function '{}' achieved in {}".format(f.__name__, abs(c-time.time())))
      return results
   return wrapper

@timeit
def outer_permutations():
   def permutations1(d, b, current = []):
     if len(current) == d:
       yield current
     else:
       for i in range(10):
         if len(current) < 2 or sum(current+[i]) == b:
           yield from permutations1(d, b, current+[i])
   yield from permutations1(3, 5)

@timeit
def list_results(a, b):
   return [i for i in permutations(b) if sum(i) == a]


v = outer_permutations()
v1 = list_results(3, 5)

输出：

Result from function 'outer_permutations' achieved in 0.0006079673767089844
Result from function 'list_results' achieved in 0.09148788452148438

请注意，这两个函数的输出是：

[[0, 0, 5], [0, 1, 4], [0, 2, 3], [0, 3, 2], [0, 4, 1], [0, 5, 0], [1, 0, 4], [1, 1, 3], [1, 2, 2], [1, 3, 1], [1, 4, 0], [2, 0, 3], [2, 1, 2], [2, 2, 1], [2, 3, 0], [3, 0, 2], [3, 1, 1], [3, 2, 0], [4, 0, 1], [4, 1, 0], [5, 0, 0]]

如何获得这个组合列表？

1 个答案: