python判断矩阵中每行非零个数的方法-创新互联

如下所示：

10年积累的成都网站制作、网站建设经验，可以快速应对客户对网站的新想法和需求。提供各种问题对应的解决方案。让选择我们的客户得到更好、更有力的网络服务。我虽然不认识你，你也不认识我。但先网站设计制作后付款的网站建设流程，更有邵原免费网站建设让你可以放心的选择与我们合作。

# -*- coding: utf-8 -*-
# @Time  : 2018/5/17 15:05
# @Author : Sizer
# @Site  : 
# @File  : test.py
# @Software: PyCharm
import time
import numpy as np

# data = np.array([
# [5.0, 3.0, 4.0, 4.0, 0.0],
# [3.0, 1.0, 2.0, 3.0, 3.0],
# [4.0, 3.0, 4.0, 3.0, 5.0],
# [3.0, 3.0, 1.0, 5.0, 4.0],
# [1.0, 5.0, 5.0, 2.0, 1.0]
# ])
data = np.random.random((1000, 1000))
print(data.shape)
start_time = time.time()
# avg = [float(np.mean(data[i, :])) for i in range(data.shape[0])]
# print(avg)


start_time = time.time()
avg = []
for i in range(data.shape[0]):
  sum = 0
  cnt = 0
  for rx in data[i, :]:
   if rx > 0:
     sum += rx
     cnt += 1
  if cnt > 0:
   avg.append(sum/cnt)
  else:
   avg.append(0)
end_time = time.time()
print("op 1:", end_time - start_time)

start_time = time.time()
avg = []
isexist = (data > 0) * 1
for i in range(data.shape[0]):
  sum = np.dot(data[i, :], isexist[i, :])
  cnt = np.sum(isexist[i, :])
  if cnt > 0:
   avg.append(sum / cnt)
  else:
   avg.append(0)
end_time = time.time()
print("op 2:", end_time - start_time)
#
# print(avg)
factor = np.mat(np.ones(data.shape[1])).T
# print("facotr :")
# print(factor)
exist = np.mat((data > 0) * 1.0)
# print("exist :")
# print(exist)
# print("res  :")
res = np.array(exist * factor)
end_time = time.time()
print("op 3:", end_time-start_time)

start_time = time.time()
exist = (data > 0) * 1.0
factor = np.ones(data.shape[1])
res = np.dot(exist, factor)
end_time = time.time()
print("op 4:", end_time - start_time)

分享标题：python判断矩阵中每行非零个数的方法-创新互联
本文链接：http://cdiso.cn/article/edhoe.html