下面是我的两个矩阵除法代码:
def divideM1(X,Y):
n=len(X)
a=[[col for col in row[:len(row)/2]] for row in X[:n/2]]
b=[[col for col in row[len(row)/2:]] for row in X[:n/2]]
c=[[col for col in row[:len(row)/2]] for row in X[n/2:]]
d=[[col for col in row[len(row)/2:]] for row in X[n/2:]]
e=[[col for col in row[:len(row)/2]] for row in Y[:n/2]]
f=[[col for col in row[len(row)/2:]] for row in Y[:n/2]]
g=[[col for col in row[:len(row)/2]] for row in Y[n/2:]]
h=[[col for col in row[len(row)/2:]] for row in Y[n/2:]]
return a,b,c,d,e,f,g,h
def divideM2(X,Y):
n=len(X)
a=[[0 for i in range(n/2)] for j in range(n/2)]
b=[[0 for i in range(n/2)] for j in range(n/2)]
c=[[0 for i in range(n/2)] for j in range(n/2)]
d=[[0 for i in range(n/2)] for j in range(n/2)]
f=[[0 for i in range(n/2)] for j in range(n/2)]
e=[[0 for i in range(n/2)] for j in range(n/2)]
g=[[0 for i in range(n/2)] for j in range(n/2)]
h=[[0 for i in range(n/2)] for j in range(n/2)]
for i in range(n/2):
for j in range(n/2):
a[i][j]=X[i][j]
b[i][j]=X[i][j+n/2]
c[i][j]=X[i+n/2][j]
d[i][j]=X[i+n/2][j+n/2]
e[i][j]=Y[i][j]
f[i][j]=Y[i][j+n/2]
g[i][j]=Y[i+n/2][j]
h[i][j]=Y[i+n/2][j+n/2]
return a,b,c,d,e,f,g,h
如果我使用时间。时间()看起来方法2“divideM2”比方法1“divideM1”快,但为什么呢? 有没有更好的划分方法?你知道吗
编辑1: 有趣的是当我使用时间。时间():
start = time.time()
print("method1")
for i in range(10000):
1>2
divideM2(a1,a1)
end = time.time()
t1=end-start
print t1 ,"m1"
start = time.time()
print("method2")
for j in range(10000):
1>2
divideM1(a2,a2)
end = time.time()
t2= end-start
print t2, "m2"
if t1>t2:
print "method 2 is faster"
else:
print "method 1 is faster"
我总是得到“方法2更快”,即使我自己比较“divide1”。有人能解释一下吗?你知道吗
我认为方法2更快,因为您只使用2个循环,而不是第一个方法中使用的所有
for
循环。你知道吗关于用另一种方式做,你可以试试
numpy.split
。这是doc。你知道吗相关问题 更多 >
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