Assignment has more non-singleton rhs dimensions than non-singleton subscripts help

Hi, I am trying to make something like a table so for every iteration (1:200) a new column is added in 'list_val' but with a variable number of elements in each column. I get the error on the 'list_val(:,it_3)=val_set(o_ue);' line. The following code is embedded in a for loop wiith iteration (it_3=) 1:200. 'list_val', 'val_set' and 'o_ue' are pre-defined and in the first iteration where the error appears the rhs has 3 values.

if length(o_ue)>0
list_val(:,it_3)=val_set(o_ue);
end

[James Tursa]

What do you mean by "... variable number of elements in each column ..."? What are the dimensions of list_val to begin with? What is the size of the val_set(o_ue) result? Are you trying to stuff a variable number of elements in a brand new column with 0 padding on the end?

[ajk1]

Hi there, the number of results of val_set(o_ue) is different after each iteration. After the first iteration the size is 3x1 and in the second iteration it will be a different size (nx1 result where n is a integer). I want these values to be stored in list_val, each succeeding iteration being stored in a new column. I haven't assigned a dimension to list_val because I am trying to make the dimensions to be dynamic. I'm trying to store a variable number of elements in a new column with 0 padding. Thanks.

Answer 1

[Stephen23]

The trick to including different-length columns into a matrix is to define the subscript indexing, and not just using the colon operator to allocate the whole column:

dat = {[1,2,3],4,[5,6,7,8,9],[]};
out = [];
for k = numel(dat):-1:1
    vec = dat{k};
    out(1:numel(vec),k) = vec;
end

produces this matrix:

>> out
out =
   1     4     5     0
   2     0     6     0
   3     0     7     0
   0     0     8     0
   0     0     9     0

[ajk1]

Thanks for your help.