问如何加速Numpy数组过滤/选择？
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Stack Overflow用户

提问于 2018-11-22 15:56:53

回答 2查看 554关注 0票数 1

我有大约40k行，我想在这些行上测试各种选择组合。我所说的选择是指布尔掩码。掩模/滤镜的数量约为250 is。

当前简化的代码如下：

np_arr = np.random.randint(1, 40000, 40000)
results = np.empty(250000000)
filters = np.random.randint(1, size=(250000000, 40000))
for i in range(250000000):
    row_selection = np_arr[filters[i].astype(np.bool_)] # Select rows based on next filter
    # Performing simple calculations such as sum, prod, count on selected rows and saving to result
    results[i] = row_selection.sum() # Save simple calculation result to results array

我尝试了Numba和Multiprocessing，但由于大多数处理都是在过滤器选择中，而不是计算中，这没有太大帮助。

解决这个问题最有效的方法是什么？有没有办法将其并行化？就我所见，我需要遍历每个过滤器，然后分别计算sum，prod，count等，因为我不能并行应用过滤器(即使应用过滤器后的计算非常简单)。

感谢您在性能改进/加速方面的任何建议。

python

performance

numpy

parallel-processing

multiprocessing

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回答 2

Stack Overflow用户

发布于 2018-11-22 17:09:26

要在Numba中获得良好的性能，只需避免掩蔽，从而避免非常昂贵的阵列拷贝。您必须自己实现筛选器，但您提到的筛选器应该没有任何问题。

并行化也很容易做到。

示例

import numpy as np
import numba as nb

max_num = 250000 #250000000
max_num2 = 4000#40000
np_arr = np.random.randint(1, max_num2, max_num2)
filters = np.random.randint(low=0,high=2, size=(max_num, max_num2)).astype(np.bool_)

#Implement your functions like this, avoid masking
#Sum Filter
@nb.njit(fastmath=True)
def sum_filter(filter,arr):
  sum=0.
  for i in range(filter.shape[0]):
    if filter[i]==True:
      sum+=arr[i]
  return sum

#Implement your functions like this, avoid masking
#Prod Filter
@nb.njit(fastmath=True)
def prod_filter(filter,arr):
  prod=1.
  for i in range(filter.shape[0]):
    if filter[i]==True:
      prod*=arr[i]
  return sum

@nb.njit(parallel=True)
def main_func(np_arr,filters):
  results = np.empty(filters.shape[0])
  for i in nb.prange(max_num):
    results[i]=sum_filter(filters[i],np_arr)
    #results[i]=prod_filter(filters[i],np_arr)
  return results

票数 3

Stack Overflow用户

发布于 2018-11-22 16:40:53

改进的一种方法是将as_type移到循环之外。在我的测试中，它将执行时间减少了一半以上。为了进行比较，请检查以下两个代码：

import numpy as np
import time

max_num = 250000 #250000000
max_num2 = 4000#40000
np_arr = np.random.randint(1, max_num2, max_num2)
results = np.empty(max_num)
filters = np.random.randint(1, size=(max_num, max_num2))
start = time.time()
for i in range(max_num):
    row_selection = np_arr[filters[i].astype(np.bool_)] # Select rows based on next filter
    # Performing simple calculations such as sum, prod, count on selected rows and saving to result
    results[i] = row_selection.sum() # Save simple calculation result to results array

end = time.time()
print(end - start)

采用2.12

而

import numpy as np
import time

max_num = 250000 #250000000
max_num2 = 4000#40000
np_arr = np.random.randint(1, max_num2, max_num2)
results = np.empty(max_num)
filters = np.random.randint(1, size=(max_num, max_num2)).astype(np.bool_)
start = time.time()
for i in range(max_num):
    row_selection = np_arr[filters[i]] # Select rows based on next filter
    # Performing simple calculations such as sum, prod, count on selected rows and saving to result
    results[i] = row_selection.sum() # Save simple calculation result to results array

end = time.time()
print(end - start)