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matlab vs python: 跑循环的速度对比
阅读量:157 次
发布时间:2019-02-28

本文共 2144 字,大约阅读时间需要 7 分钟。

测试1

matlab代码

N = 20:25;iters = 2.^N;time = zeros(1,length(N));a = 0.111;b = 0.222;for k = 1:length(N)    r = 0;    t1 = clock;        for i = 1:2^N(k)        r = 0.5*a + 0.6*b;    end    t2 = clock;    time(k) = etime(t2,t1);    endplot(iters, time)xlabel('iter')ylabel('time(/s)')

python代码

N = range(20,26)iters = [2**n for n in N]ts = []a, b = 0.111, 0.222for n in N:        t1 = time.time()    for i in range(2**n):        r = 0.5*a + 0.6*b    t2 = time.time()    ts.append(t2-t1)_, ax = plt.subplots()ax.plot(iters, ts)ax.set_xlabel('iter')ax.set_ylabel('time(/s)')

结果对比

将两者数据画到一起,方便对比。
结论:随着循环增多,两者消耗时间都线性增大。对于这个测试案例(两个乘法和一个加法)。python约比matlab慢60倍
在这里插入图片描述

测试2

matlab代码

N = 20:25;iters = 2.^N;time = zeros(1,length(N));a = 0.111;b = 0.222;M = [0.111,0.222;0.111,0.222];for k = 1:length(N)    r = 0;    t1 = clock;        for i = 1:2^N(k)        r = M(1,1)*a + M(1,2)*b;    end    t2 = clock;    time(k) = etime(t2,t1);    endfigure;plot(iters, time)xlabel('iter')ylabel('time(/s)')

python代码

N = range(20,26)iters = [2**n for n in N]ts = []M = np.array([[0.111, 0.222],[0.111, 0.222]])a, b = 0.111, 0.222for n in N:        t1 = time.time()    for i in range(2**n):        r = M[0,0]*a + M[0,1]*b    t2 = time.time()    ts.append(t2-t1)_, ax = plt.subplots()ax.plot(iters, ts)ax.set_xlabel('iter')ax.set_ylabel('time(/s)')

结果对比

将两者数据画到一起,方便对比。
结论:

  • 随着循环增多,两者消耗时间都线性增大。python约比matlab慢110倍
  • 将此测试结果与测试1对比, 可猜想:仅仅是在2*2矩阵中索引一个数,python也要比matlab很多倍,猜想慢110-60=50倍。再通过一个测试3来验证下猜想。
    在这里插入图片描述

测试3

matlab代码

N = 20:25;iters = 2.^N;time = zeros(1,length(N));a = 0.111;b = 0.222;M = [0.111,0.222;0.111,0.222];for k = 1:length(N)    r = 0;    t1 = clock;        for i = 1:2^N(k)        r = M(1,1);    end    t2 = clock;    time(k) = etime(t2,t1);    endfigure;plot(iters, time)xlabel('iter')ylabel('time(/s)')

python代码

N = range(20,26)iters = [2**n for n in N]ts = []M = np.array([[0.111, 0.222],[0.111, 0.222]])a, b = 0.111, 0.222for n in N:        t1 = time.time()    for i in range(2**n):        r = M[0,0]    t2 = time.time()    ts.append(t2-t1)_, ax = plt.subplots()ax.plot(iters, ts)ax.set_xlabel('iter')ax.set_ylabel('time(/s)')

结果对比

猜想正确,仅仅是2*2矩阵索引一个数,python也比matlab慢50倍。
在这里插入图片描述

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