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matlab和python哪个运行快_matlab vs python: 跑循环的速度对比

發布時間:2025/3/20 python 29 豆豆
生活随笔 收集整理的這篇文章主要介紹了 matlab和python哪个运行快_matlab vs python: 跑循环的速度对比 小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

測試1

matlab代碼

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);

end

plot(iters, time)

xlabel('iter')

ylabel('time(/s)')

python代碼

N = range(20,26)

iters = [2**n for n in N]

ts = []

a, b = 0.111, 0.222

for 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)')

結果對比

將兩者數據畫到一起,方便對比。

結論:隨著循環增多,兩者消耗時間都線性增大。對于這個測試案例(兩個乘法和一個加法)。python約比matlab慢60倍

測試2

matlab代碼

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);

end

figure;

plot(iters, time)

xlabel('iter')

ylabel('time(/s)')

python代碼

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, 0.222

for 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)')

結果對比

將兩者數據畫到一起,方便對比。

結論:

隨著循環增多,兩者消耗時間都線性增大。python約比matlab慢110倍

將此測試結果與測試1對比, 可猜想:僅僅是在2*2矩陣中索引一個數,python也要比matlab很多倍,猜想慢110-60=50倍。再通過一個測試3來驗證下猜想。

測試3

matlab代碼

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);

end

figure;

plot(iters, time)

xlabel('iter')

ylabel('time(/s)')

python代碼

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, 0.222

for 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)')

結果對比

猜想正確,僅僅是2*2矩陣索引一個數,python也比matlab慢50倍。

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