jake,
I do the same thing but I like to have the correct length and capacity.
func NewMatrix(r, c int) [][]int {
a := make([]int, r*c)
m := make([][]int, r)
lo, hi := 0, c
for i := range m {
m[i] = a[lo:hi:hi]
lo, hi = hi, hi+c
}
return m
}
https://play.golang.org/p/ifWtRaiVSQ3
Peter
On Wednesday, August 5, 2020 at 9:40:36 PM UTC-4, jake...@gmail.com wrote:
>
> Not a direct answer, but one simple improvement you could do is to
> allocate all the rows together, then break them up by slicing. So instead
> of your:
>
> T := make([][]byte, n)
> for i := 0; i < n; i++ {
> T[i] = make([]byte, m)
> }
>
> Do something like:
>
> T := make([][]byte, n)
> Q := make([]byte, m*n)
> for i := 0; i < n; i++ {
> T[i] = Q[i*m : (i+1)*m]
> }
>
> In my benchmarks, with your matrix size, this gives a 20-30 percent
> speedup. In part because it avoids many allocations, but also (I suspect)
> because the data is contiguous.
>
> On Wednesday, August 5, 2020 at 2:30:28 PM UTC-4 shenw...@gmail.com wrote:
>
>> Hi all,
>>
>> Matrix transpose in pure golang is slow in HPC cases, and using package
>> gonum needs structure transformation which costs extra time. So a
>> assembly version may be a better solution.
>>
>> Sizes of the matrix vary ([][]byte) or can be fixed for example (
>> [64][512]byte), and the element type may be int32 or int64 for general
>> scenarios.
>>
>> Below is a golang version:
>>
>> m := 64
>> n := 512
>>
>> // orignial matrix
>> M := make([][]byte, m)
>> for i := 0; i < m; i++ {
>> M[i] = make([]byte, n)
>> }
>>
>> func transpose(M [][]byte) [][]byte {
>> m := len(M)
>> n := len(M[0])
>>
>> // transposed matrix
>> T := make([][]byte, n)
>> for i := 0; i < n; i++ {
>> T[i] = make([]byte, m)
>> }
>>
>> var row []byte // a row in T
>> for i := 0; i < n; i++ {
>> row = T[i]
>> for j = 0; j < m; j++ {
>> row[j] = M[j][i]
>> }
>> }
>>
>> return T
>> }
>>
>>
>>
>
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