python - 从列表列表中获取所有唯一组合，直到第 n 个组合

Question

我有一个列表列表，其中每个内部列表中的变量是图像的路径。通常每个内部列表的长度约为 35，并且在一个列表中将有 9 个这样的列表。例如

 [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35]]

无

我想从这个列表列表中生成 9 的唯一组合。如果我使用 itertools.product 它可以工作，但花费的时间太长 - 它会使我的计算机崩溃。我需要的是能够一直上升到第 n 个组合的东西，其中 n 可能约为 200。我试过这个...

  list(itertools.product(*z))[:200]

其中 z 是我的列表列表，但它不起作用，因为它在执行切片之前先生成所有组合(太慢了)。

还有其他有效的方法来运行它吗？

编辑:我应该补充一点，我需要将其转换为列表列表...

基准测试:

布拉德:

 def combos():
     my_iter = itertools.product(*z)
     print([next(my_iter) for i in range(1, 10000)]

cProfile.run('print(combos())')
     10006 function calls in 0.021 seconds

Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    1    0.000    0.000    0.021    0.021 <string>:1(<module>)
    1    0.000    0.000    0.021    0.021 gcm.py:17(combos)
    1    0.002    0.002    0.004    0.004 gcm.py:19(<listcomp>)
    1    0.000    0.000    0.021    0.021 {built-in method builtins.exec}
 9999    0.003    0.000    0.003    0.000 {built-in method builtins.next}
    2    0.016    0.008    0.016    0.008 {built-in method builtins.print}
    1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

DeepSpaces 更新的答案:

 cProfile.run('print(list(my_gen(z, 10000)))')

 10005 function calls in 0.019 seconds

 Ordered by: standard name

 ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    1    0.001    0.001    0.019    0.019 <string>:1(<module>)
10001    0.004    0.000    0.004    0.000 gcm.py:10(my_gen)
    1    0.000    0.000    0.019    0.019 {built-in method builtins.exec}
    1    0.014    0.014    0.014    0.014 {built-in method builtins.print}
    1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

为了迈出这一步，我做了这个......

   def my_gen(z, limit):
      count = 0
      for i in itertools.product(*z):
         if count < limit:
            if count % 1000 == 0:
               yield i
               count += 1
            else:
               count += 1
               continue
      else:
        raise StopIteration

Answer 1

试试这个:

my_iter = itertools.product(*z)
[next(my_iter) for i in range(200)]