Removing rows from array using indices

A = randi(9, 7, 3)
A = 7×3
     8     6     8
     3     6     3
     8     6     9
     7     6     2
     2     7     4
     1     5     8
     8     9     3
Rowidx = [2; 5];
B = A;
B(Rowidx,:) = []
B = 5×3
     8     6     8
     8     6     9
     7     6     2
     1     5     8
     8     9     3

A different option, using ‘logical indexing’ would be —

Lidx = true(size(A,1),1);
Lidx(Rowidx) = false
Lidx = 7×1 logical array
   1
   0
   1
   1
   0
   1
   1
B = A(Lidx,:)
B = 5×3
     8     6     8
     8     6     9
     7     6     2
     1     5     8
     8     9     3

I am not certain what ‘different dimensions’ means here.

If all the arrays have the same row sizes with different numbers of columns, the advantage to the ‘logical indexing’ approach is that ‘Lidx’ could be used with every subsequent array without first having to copy the output arrays.

However if the row sizes are different, the first approach would be best.

.

Star Strider 2021-11-1

在 MATLAB Online 中打开

‘Doesn't this mean that in row 1, take columns 13,14,15 and in row 2, take columns 25,26,27 and so on?’

Yes.

‘Then why the output doesn't have the same dimensions (8916x3)?’

It should. It turns out that the sort of vectorised indexing is not appropriate here.

A simple for loop solves that problem —

A = randi(9, 5, 45)
A = 5×45
     7     3     5     9     9     6     2     8     4     3     5     6     3     3     8     9     1     8     9     5     8     7     9     9     8     2     1     8     9     9
     2     8     3     2     9     6     8     5     4     1     1     3     1     7     6     5     8     4     6     9     7     8     5     5     5     9     6     2     7     2
     3     5     7     3     2     2     9     7     1     3     6     1     3     7     4     4     6     8     6     6     3     2     7     9     6     9     6     6     4     1
     8     1     6     7     8     8     9     9     2     2     8     3     3     7     4     4     9     7     3     5     7     8     1     4     9     4     4     4     5     1
     5     8     1     7     4     3     9     8     9     4     4     9     8     3     6     5     2     6     2     2     1     5     4     8     2     5     2     5     1     3
Rowidx = randi(15, 5,1)
Rowidx = 5×1
     7
    15
     4
    10
     2
Select = (Rowidx-1)*3+(1:3)
Select = 5×3
    19    20    21
    43    44    45
    10    11    12
    28    29    30
     4     5     6
for k = 1:size(A,1)
    B(k,:) = A(k,Select(k,:));
end
B
B = 5×3
     9     5     8
     9     3     1
     3     6     1
     4     5     1
     7     4     3

‘Where does 26748 come from?’

The vectorised indexing approach will not produce the correct result. It is necessary to use a for loop to get it to work as I intend it to work.

.

Pelajar UM 2021-11-1

Wonderful! Thanks a lot for your time. I really appreciate it.

Star Strider 2021-11-1

Thank you!

As always, my pleasure!

.

请先登录，再进行评论。

Removing rows from array using indices

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

10 个评论
显示 8更早的评论隐藏 8更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

Removing rows from array using indices

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

10 个评论 显示 8更早的评论隐藏 8更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

10 个评论
显示 8更早的评论隐藏 8更早的评论