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如何改进效率低下的解决方案?[复制]

python-3.x

alex108 · 技术社区 · 5 年前

solution 它使用动态规划方法。根据他的定性描述,我用python翻译了他的解决方案。我正试图优化这个更大的列表,这会占用我很多内存。有人能推荐一些优化或其他技术来解决这个问题吗?下面是我在python中的尝试:

import random
from time import time
from itertools import product

time0 = time()

# create a zero matrix of size a (row), b(col)
def create_zero_matrix(a,b):
    return [[0]*b for x in xrange(a)]

# generate a list of size num with random integers with an upper and lower bound
def random_ints(num, lower=-1000, upper=1000):
    return [random.randrange(lower,upper+1) for i in range(num)]

# split a list up into N and P where N be the sum of the negative values and P the sum of the positive values.
# 0 does not count because of additive identity
def split_sum(A):
    N_list = []
    P_list = []
    for x in A:
        if x < 0:
            N_list.append(x)
        elif x > 0:
            P_list.append(x)
    return [sum(N_list), sum(P_list)]

# since the column indexes are in the range from 0 to P - N
# we would like to retrieve them based on the index in the range N to P
# n := row, m := col
def get_element(table, n, m, N):
    if n < 0:
        return 0
    try:
        return table[n][m - N]
    except:
        return 0

# same definition as above
def set_element(table, n, m, N, value):
    table[n][m - N] = value

# input array
#A = [1, -3, 2, 4]
A = random_ints(200)

[N, P] = split_sum(A)

# create a zero matrix of size m (row) by n (col)
#
# m := the number of elements in A
# n := P - N + 1 (by definition N <= s <= P)
#
# each element in the matrix will be a value of either 0 (false) or 1 (true)
m = len(A)
n = P - N + 1;
table = create_zero_matrix(m, n)

# set first element in index (0, A[0]) to be true
# Definition: Q(1,s) := (x1 == s). Note that index starts at 0 instead of 1.
set_element(table, 0, A[0], N, 1)

# iterate through each table element
#for i in xrange(1, m): #row
#    for s in xrange(N, P + 1): #col
for i, s in product(xrange(1, m), xrange(N, P + 1)):
    if get_element(table, i - 1, s, N) or A[i] == s or get_element(table, i - 1, s - A[i], N):
        #set_element(table, i, s, N, 1)
        table[i][s - N] = 1

# find zero-sum subset solution
s = 0
solution = []
for i in reversed(xrange(0, m)):
    if get_element(table, i - 1, s, N) == 0 and get_element(table, i, s, N) == 1:
        s = s - A[i]
        solution.append(A[i])

print "Solution: ",solution

time1 = time()

print "Time execution: ", time1 - time0

0 回复 | 直到 14 年前

Martin Carames Abente 14 年前

我不太确定你的解是精确解还是PTA(poly-time approximation)。

也就是说,每个已知的(精确的)算法对输入的大小都有指数时间行为。

也就是说,如果你能在0.01纳秒内处理1个操作,那么对于一个包含59个元素的列表,它需要:

2^59 ops -->     2^59     seconds -->     2^26      years -->      1 year
            --------------           ---------------
            10.000.000.000           3600 x 24 x 365

另一方面,如果你用集合中数的值的界来限制问题(另一个问题),那么问题的复杂性就降低到多项式时间。但即使这样,所消耗的内存空间也将是一个非常高阶的多项式。
甚至比你硬盘上的几兆字节还要大。

(对于集合中元素值的边界的小值)

也许这就是你的动态规划算法的情况。

在我看来,您在构建初始化矩阵时使用了1000的边界。

祝你好运!

skorks 14 年前

Hacker News上有人提出了以下解决问题的方法,我很喜欢。它正好在python中:):

def subset_summing_to_zero (activities):
  subsets = {0: []}
  for (activity, cost) in activities.iteritems():
      old_subsets = subsets
      subsets = {}
      for (prev_sum, subset) in old_subsets.iteritems():
          subsets[prev_sum] = subset
          new_sum = prev_sum + cost
          new_subset = subset + [activity]
          if 0 == new_sum:
              new_subset.sort()
              return new_subset
          else:
              subsets[new_sum] = new_subset
  return []

我花了几分钟用它,效果很好。

mitchus 14 年前

有一篇关于优化python代码的有趣文章 here . 基本上,主要的结果是你应该内联你的频繁循环,所以在你的例子中,这意味着不要调用 get_element 每次循环两次,将该函数的实际代码放入循环中,以避免函数调用开销。

希望有帮助!干杯

Luka Rahne 14 年前

,第一次吸引眼球

def split_sum(A):
  N_list = 0
  P_list = 0
  for x in A:
    if x < 0:
        N_list+=x
    elif x > 0:
        P_list+=x
  return [N_list, P_list]

一些建议:

尝试使用1D list和bitarray来减少内存占用(http://pypi.python.org/pypi/bitarray)所以只需更改get/set函数。这样至少可以减少64个内存占用(列表中的整数是指向整型的指针,所以可以是因子3*32)
避免使用try-catch,但在开始时找出合适的范围,您可能会发现您将获得巨大的速度。

devDeejay 9 年前

from itertools import chain, combinations
def powerset(iterable):
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

nums = input("Enter the Elements").strip().split()
inputSum = int(input("Enter the Sum You want"))

for i, combo in enumerate(powerset(nums), 1):
    sum = 0
    for num in combo:
        sum += int(num)
    if sum == inputSum:
        print(combo)

Enter the Elements 1 2 3 4
Enter the Sum You want 5
('1', '4')
('2', '3')

Eric Aya 8 年前

def subsetsum(cs,k,r,x,w,d):
    x[k]=1
    if(cs+w[k]==d):
        for i in range(0,k+1):

            if x[i]==1:
                print (w[i],end=" ")
        print()

    elif cs+w[k]+w[k+1]<=d :
        subsetsum(cs+w[k],k+1,r-w[k],x,w,d)

    if((cs +r-w[k]>=d) and (cs+w[k]<=d)) :
        x[k]=0
        subsetsum(cs,k+1,r-w[k],x,w,d)
#driver for the above code
w=[2,3,4,5,0]
x=[0,0,0,0,0]

subsetsum(0,0,sum(w),x,w,7)