Question

我有一个R x C矩阵填充到第k行并且在该行下面是空的。我需要做的是填补剩余的行。为了做到这一点，我有一个函数，它将2个整行作为参数，处理这些行并输出2个新行（这些输出将填充矩阵的空行，批量为2）。我有一个固定矩阵，包含要处理的所有'对'行，但我的for循环没有帮助性能：

# the processRows function:
processRows = function(r1, r2)
{
  # just change a little bit the two rows and return it in a compact way    
  nr1 = r1 * 0.1
  nr2 = -r2 * 0.1

  matrix (c(nr1, nr2), ncol = 2)
 }

# M is the matrix
# nrow(M) and k are even, so nLeft is even

M = matrix(1:48, ncol = 3)
# half to fill (can be more or less, but k is always even)
k = nrow(M)/2

# simulate empty rows to be filled
M[-(1:k), ] = 0

cat('before fill')
print(M)

# number of empty rows to fill
nLeft  = nrow(M) - k
nextRow = k + 1

# each row in idxList represents a 'pair' of rows to be processed 
# any pairwise combination of non-empty rows could happen
# make it reproducible

set.seed(1)
idxList = matrix (sample(1:k, k), ncol = 2, byrow = TRUE)

for ( i in 1 : (nLeft / 2))
{
   row1 = M[idxList[i, 1],]
   row2 = M[idxList[i, 2],]

   # the two columns in 'results' will become 2 rows in M
   results = processRows(row1, row2)

   # fill the matrix
   M[nextRow, ] = results[, 1]
   nextRow = nextRow + 1
   M[nextRow, ] = results[, 2]
   nextRow = nextRow + 1
 }

 cat('after fill')
 print(M)

Answer 1

好的，这是您的代码。我们运行这个，以便我们有一个“真实”矩阵的副本，我们希望更快地重现。

#### Original Code (aka Gold Standard) ####
M = matrix(1:48, ncol = 3)
k = nrow(M)/2
M[-(1:k), ] = 0
nLeft  = nrow(M) - k
nextRow = k + 1
idxList = matrix(1:k, ncol = 2)
for ( i in 1 : (nLeft / 2))
{
    row1 = M[idxList[i, 1],]
    row2 = M[idxList[i, 2],]
    results = matrix(c(2*row1, 3*row2), ncol = 2)
    M[nextRow, ] = results[, 1]
    nextRow = nextRow + 1
    M[nextRow, ] = results[, 2]
    nextRow = nextRow + 1
}

现在这里是矢量化代码。基本的想法是，如果您正在处理4行。而不是一次一个地将它们作为向量传递，而是立即执行。那就是：

(1:3) * 2
(1:3) * 2
(1:3) * 2
(1:3) * 2

与（

）相同（但速度较慢）

c(1:3, 1:3, 1:3, 1:3) * 2

首先，我们将使用相同的设置代码，然后将要处理的行创建为两个长向量（其中所有4个原始行仅在上面的简单示例中串联在一起）。然后，我们获取这些结果，并将它们转换为具有适当尺寸的矩阵。最后一招是将结果分成两步。您可以一次分配矩阵的多行，因此我们使用seq()来获取奇数和偶数，因此分别将结果的第一列和第二列分配给。

#### Vectorized Code (testing) ####
M2 = matrix(1:48, ncol = 3)
k2 = nrow(M2)/2
M2[-(1:k2), ] = 0
nLeft2  = nrow(M2) - k2
nextRow2 = k2 + 1
idxList2 = matrix(1:k2, ncol = 2)

## create two long vectors of all rows to be processed
row12 <- as.vector(t(M2[idxList2[, 1],]))
row22 <- as.vector(t(M2[idxList2[, 2],]))

## get all results
results2 = matrix(c(2*row12, 3*row22), ncol = 2)

## add results back
M2[seq(nextRow2, nextRow2 + nLeft2-1, by = 2), ] <- matrix(results2[,1], nLeft2/2, byrow=TRUE)
M2[seq(nextRow2+1, nextRow2 + nLeft2, by = 2), ] <- matrix(results2[,2], nLeft2/2, byrow=TRUE)

## check that vectorized code matches your examples
all.equal(M, M2)

在我的机器上给出了：

> all.equal(M, M2)
[1] TRUE

在R中矢量化矩阵运算

1 个答案: