2

看下面的代码，我用两种方法解决问题（简单递归和DP）。为什么DP方式较慢？

你有什么建议？

#!/usr/local/bin/python2.7
# encoding: utf-8

问题：有一个正整数数组。给定一个正整数 S，\ 找出其中数字之和为 S 的组合总数。

方法一： 

def find_sum_recursive(number_list, sum_to_find):
    count = 0

    for i in range(len(number_list)):        
        sub_sum = sum_to_find - number_list[i]
        if sub_sum < 0:
            continue
        elif sub_sum == 0:
            count += 1
            continue
        else:            
            sub_list = number_list[i + 1:]            
            count += find_sum_recursive(sub_list, sub_sum)       
    return count

方法二： 

def find_sum_DP(number_list, sum_to_find):
    count = 0

    if(0 == sum_to_find):
        count = 1
    elif([] != number_list and sum_to_find > 0):   
        count = find_sum_DP(number_list[:-1], sum_to_find) + find_sum_DP(number_list[:-1], sum_to_find - number_list[:].pop())     

    return count

运行它： 

def main(argv=None):  # IGNORE:C0111
    number_list = [5, 5, 10, 3, 2, 9, 8]
    sum_to_find = 15
    input_setup = ';number_list = [5, 5, 10, 3, 2, 9, 8, 7, 6, 4, 3, 2, 9, 5, 4, 7, 2, 8, 3];sum_to_find = 15'

    print 'Calculating...'
    print 'recursive starting'
    count = find_sum_recursive(number_list, sum_to_find)
    print timeit.timeit('count = find_sum_recursive(number_list, sum_to_find)', setup='from __main__ import find_sum_recursive' + input_setup, number=10)
    cProfile.run('find_sum_recursive(' + str(number_list) + ',' + str(sum_to_find) + ')')
    print 'recursive ended:', count    
    print 'DP starting'
    count_DP = find_sum_DP(number_list, sum_to_find)
    print timeit.timeit('count_DP = find_sum_DP(number_list, sum_to_find)', setup='from __main__ import find_sum_DP' + input_setup, number=10)
    cProfile.run('find_sum_DP(' + str(number_list) + ',' + str(sum_to_find) + ')')
    print 'DP ended:', count_DP        
    print 'Finished.'    

if __name__ == '__main__':
    sys.exit(main())

我重新编码了方法二，现在是：

def find_sum_DP(number_list, sum_to_find):
    count = [[0 for i in xrange(0, sum_to_find + 1)] for j in xrange(0, len(number_list) + 1)]    

    for i in range(len(number_list) + 1):
        for j in range(sum_to_find + 1):            
            if (0 == i and 0 == j):                
                count[i][j] = 1            
            elif (i > 0 and j > 0):                
                if (j > number_list[i - 1]):                    
                    count[i][j] = count[i - 1][j] + count[i - 1][j - number_list[i - 1]]
                elif(j < number_list[i - 1]):
                    count[i][j] = count[i - 1][j]
                else:
                    count[i][j] = count[i - 1][j] + 1                                
            else:                
                count[i][j] = 0

    return count[len(number_list)][sum_to_find]

比较方法一和二：

Calculating...
recursive starting
0.00998711585999
         92 function calls (63 primitive calls) in 0.000 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.000    0.000 <string>:1(<module>)
     30/1    0.000    0.000    0.000    0.000 FindSum.py:18(find_sum_recursive)
       30    0.000    0.000    0.000    0.000 {len}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
       30    0.000    0.000    0.000    0.000 {range}


recursive ended: 6
DP starting
0.00171685218811
         15 function calls in 0.000 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.000    0.000 <string>:1(<module>)
        1    0.000    0.000    0.000    0.000 FindSum.py:33(find_sum_DP)
        3    0.000    0.000    0.000    0.000 {len}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
        9    0.000    0.000    0.000    0.000 {range}


DP ended: 6
Finished.
4

2 回答 2

6

如果您使用的是 iPython，则 %prun 是您的朋友。

查看递归版本的输出：

         2444 function calls (1631 primitive calls) in 0.002 seconds

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    814/1    0.002    0.000    0.002    0.002 <ipython-input-1-7488a6455e38>:1(find_sum_recursive)
      814    0.000    0.000    0.000    0.000 {range}
      814    0.000    0.000    0.000    0.000 {len}
        1    0.000    0.000    0.002    0.002 <string>:1(<module>)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

现在，对于 DP 版本：

         10608 function calls (3538 primitive calls) in 0.007 seconds

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
   7071/1    0.007    0.000    0.007    0.007 <ipython-input-15-3535e3ab26eb>:1(find_sum_DP)
     3535    0.001    0.000    0.001    0.000 {method 'pop' of 'list' objects}
        1    0.000    0.000    0.007    0.007 <string>:1(<module>)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

7071比814高不少！

您的问题是您的动态编程方法不是动态编程！动态规划的要点是，当您遇到重叠子问题的问题时，就像您在这里所做的那样，您存储每个子问题的结果，然后当您再次需要结果时，您从该存储中获取它而不是重新计算。您的代码没有这样做：每次调用 find_sum_DP 时，您都在重新计算，即使已经完成了相同的计算。结果是您的 _DP 方法实际上不仅是递归的，而且是递归的，它比您的递归方法具有更多的函数调用。

（我目前正在写一个DP版本来演示）

编辑：

我需要补充一点，虽然我应该知道更多关于动态编程的知识，但我很尴尬地不知道。我也在深夜快速写这篇文章，有点作为自己的练习。然而，这里是函数的动态编程实现：

import numpy as np
def find_sum_realDP( number_list, sum_to_find ):
    memo = np.zeros( (len(number_list),sum_to_find+1) ,dtype=np.int)-1
    # This will store our results. memo[l][n] will give us the result
    # for number_list[0:l+1] and a sum_to_find of n. If it hasn't been
    # calculated yet, it will give us -1. This is not at all efficient
    # storage, but isn't terribly bad.

    # Now that we have that, we'll call the real function. Instead of modifying
    # the list and making copies or views, we'll keep the same list, and keep
    # track of the index we're on (nli).
    return find_sum_realDP_do( number_list, len(number_list)-1, sum_to_find, memo ),memo

def find_sum_realDP_do( number_list, nli, sum_to_find, memo ):
    # Our count is 0 by default.
    ret = 0

    # If we aren't at the sum to find yet, do we have any numbers left after this one?
    if ((sum_to_find > 0) and nli>0):
        # Each of these checks to see if we've already stored the result of the calculation.
        # If so, we use that, if not, we calculate it.
        if memo[nli-1,sum_to_find]>=0:
            ret += memo[nli-1,sum_to_find]
        else:
            ret += find_sum_realDP_do(number_list, nli-1, sum_to_find, memo)

        # This one is a bit tricky, and was a bug when I first wrote it. We don't want to
        # have a negative sum_to_find, because that will be very bad; we'll start using results
        # from other places in memo because it will wrap around.
        if (sum_to_find-number_list[nli]>=0) and memo[nli-1,sum_to_find-number_list[nli]]>=0:
            ret += memo[nli-1,sum_to_find-number_list[nli]]
        elif (sum_to_find-number_list[nli]>=0):
            ret += find_sum_realDP_do(number_list, nli-1, sum_to_find-number_list[nli], memo)

    # Do we not actually have any sum to find left?     
    elif (0 == sum_to_find):
        ret = 1

    # If we only have one number left, will it get us there?
    elif (nli == 0) and (sum_to_find-number_list[nli] == 0 ):
        ret = 1

    # Store our result.
    memo[nli,sum_to_find] = ret

    # Return our result.
    return ret

请注意，这使用numpy。您很可能没有安装它，但我不确定如何在没有它的情况下在 Python 中编写性能合理的动态编程算法；我认为 Python 列表的性能远不及 Numpy 数组。另请注意，这与您的代码处理零的方式不同，因此我不会调试它，而是说此代码用于数字列表中的非零正整数。现在，使用此算法，分析为我们提供：

         243 function calls (7 primitive calls) in 0.001 seconds

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    237/1    0.001    0.000    0.001    0.001 <ipython-input-155-4a624e5a99b7>:9(find_sum_realDP_do)
        1    0.000    0.000    0.001    0.001 <ipython-input-155-4a624e5a99b7>:1(find_sum_realDP)
        1    0.000    0.000    0.000    0.000 {numpy.core.multiarray.zeros}
        1    0.000    0.000    0.001    0.001 <string>:1(<module>)
        2    0.000    0.000    0.000    0.000 {len}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

243 甚至比递归版本好很多！但是您的示例数据足够小，它并没有真正展示动态编程算法有多好。

让我们试试nlist2 = [7, 6, 2, 3, 7, 7, 2, 7, 4, 2, 4, 5, 6, 1, 7, 4, 6, 3, 2, 1, 1, 1, 4, 2, 3, 5, 2, 4, 4, 2, 4, 5, 4, 2, 1, 7, 6, 6, 1, 5, 4, 5, 3, 2, 3, 7, 1, 7, 6, 6]，同样的sum_to_find=15。这有 50 个值，以及 900206 种方法来获得 15...

与find_sum_recursive：

         3335462 function calls (2223643 primitive calls) in 14.137 seconds

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
1111820/1   13.608    0.000   14.137   14.137 <ipython-input-46-7488a6455e38>:1(find_sum_recursive)
  1111820    0.422    0.000    0.422    0.000 {range}
  1111820    0.108    0.000    0.108    0.000 {len}
        1    0.000    0.000   14.137   14.137 <string>:1(<module>)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

现在有了find_sum_realDP：

         736 function calls (7 primitive calls) in 0.007 seconds

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    730/1    0.007    0.000    0.007    0.007 <ipython-input-155-4a624e5a99b7>:9(find_sum_realDP_do)
        1    0.000    0.000    0.007    0.007 <ipython-input-155-4a624e5a99b7>:1(find_sum_realDP)
        1    0.000    0.000    0.000    0.000 {numpy.core.multiarray.zeros}
        1    0.000    0.000    0.007    0.007 <string>:1(<module>)
        2    0.000    0.000    0.000    0.000 {len}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

所以我们的调用不到 1/1000，运行时间不到 1/2000。当然，您使用的列表越大，DP 算法的效果就越好。在我的电脑上，使用 sum_to_find 为 15 和从 1 到 8 的 600 个随机数的列表运行，realDP 只需要 0.09 秒，并且函数调用少于 10,000 个；大约在这一点上，我使用的 64 位整数开始溢出，我们还有各种其他问题。不用说，在计算机停止运行之前，递归算法永远无法处理接近该大小的列表，无论是由于其内部的材料分解还是宇宙的热寂。

于 2013-08-16T07:41:01.487 回答
1

一件事是你的代码做了很多列表复制。如果它只是传递一个或多个索引来定义一个“窗口视图”而不是复制所有列表会更快。对于第一种方法，您可以轻松添加参数starting_index并在 for 循环中使用它。在第二种方法中，您写入number_list[:].pop()和复制整个列表只是为了获取您可以简单地执行的最后一个元素 as number_list[-1]。您还可以添加一个参数ending_index并在您的测试中使用它（len(number_list) == ending_index而不是number_list != []，顺便说一句，即使只是简单number_list的也比针对空列表进行测试要好）。

于 2013-08-16T07:40:13.490 回答