【下载文档:  python判断矩阵中每行非零个数的方法.txt 】

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python 判断矩阵中每行非零个数的方法
如下所示：
# -*- 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)
经过多次验证, 第四种实现方式的事件效率最高！
以上这篇python 判断矩阵中每行非零个数的方法就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持中文源码网。