Question

以下是我划分两个矩阵的代码：

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time.time（），似乎方法2-“divideM2”比方法1更快 - “divideM1”，但为什么呢？是否有更好的划分方法？

EDIT1：有趣的是，当我使用time.time（）时：

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"

即使我自己比较“divide1”，我总是得到“方法2更快”。有人可以解释一下吗？

Answer 1

我认为方法2更快，因为你只使用2个循环而不是你在第一个方法中使用的所有for循环。

关于以其他方式执行此操作，您可以尝试numpy.split。这是doc。

在8个小矩阵中划分两个矩阵的更快方法

1 个答案: