wrong number of rows of cell array using length()-function

Hello dear community,

I get some error by using length in one Cell-Array on Traj_interpl{1,209}:

for traj_cnt = 1:length(Traj_interpl) %for every Traj-Cell
    for t_row_cnt = 1:length(Traj_interpl{traj_cnt}) %for any row of Traj 
        if t_interpl(t_ref_cnt) == Traj_interpl{traj_cnt}(t_row_cnt,1) %if right row of Traj is chosen - HERE IS ERROR!!!
            motorspeed_detected(traj_cnt) = Traj_interpl{traj_cnt}(t_row_cnt,2); %write detected value to array
            break;
        else
            motorspeed_detected(traj_cnt) = 0;
        end
    end
end

Error: Index in position 1 exceeds array bounds (must not exceed 1).

My cell array is: Traj_interpl = 1×209 cell array

I found that:

Traj_interpl{1,59}
=
   18.6300  185.2204
   18.6400  185.5478
   18.6500  185.8752
   18.6600  186.0040
   18.6700  185.6136
   18.6800  185.2232
   18.6900  184.8328
   
length(Traj_interpl{1,59}) =   7 % number of rows; works great

but

Traj_interpl{1,209}
 =
   65.9000  118.0449

length(Traj_interpl{1,209}) =   2 % number of rows + 1/ number of elements; Error

Why in the 59 cell the length give me a right amount of rows and in a 209 cell a wrong one? How can I avoid this? Thank you!

Answer 1

[Timo Dietz]

You should use the size command to distinguish between row and col.

To be more precise:

length(X) returns the length of vector X. It is equivalent

to MAX(SIZE(X)) for non-empty arrays and 0 for empty ones.

[Rik]

You can also have a look at this highlight post.

[Nik Rocky]

Thank you very much, I'm using [row,column]=size(Traj_interpl{1,209}) now!

Answer 2

[Sriram Tadavarty]

Hi Nik,

length returns the length of the largest array dimension in X. For vectors, the length is simply the number of elements. For arrays with more dimensions, the length is max(size(X)). The length of an empty array is zero.

In your case, {1,59} cell is a matrix with more number of rows, so length provided that value, whereas cell {1,209} has more number of columns, thus, length provided value 2.

To get the number of rows of a matrix, use can use size(X,1).

Hope this helps.

Regards,

Sriram

[Nik Rocky]

Hi Sriram, thank you, now I understand a principle! I'm happy now!

[Nik Rocky]

Best regards, Nik

Answer 3

[Image Analyst]

length() always gives you the LONGEST dimension. Since you have more columns (2) than rows (1), it will return 2.

[Nik Rocky]

I understand now, thanks!

Answer 4

[Nik Rocky]

If i want to build it in my for-loop, what is the nice way to do it?

Old part of code

for traj_cnt = 1:length(Traj_interpl) 
    for t_row_cnt = 1:length(Traj_interpl{traj_cnt})
        if t_interpl(t_ref_cnt) == Traj_interpl{traj_cnt}(t_row_cnt,1) 

new code:

for traj_cnt = 1:length(Traj_interpl) % here is allways 1 x multiple columns array, so length is ok
    for t_row_cnt = 1:size(Traj_interpl{traj_cnt},1) %-new part
        if t_interpl(t_ref_cnt) == Traj_interpl{traj_cnt}(t_row_cnt,1) 
                     

okay? Thank you!

[Rik]

I would discourage the use of length. Either use size, or height/width, or numel. That last one is very flexible, as it will simply let you loop over all elements without having to worry about the shape.

[Nik Rocky]

Okay, I will change it! Thank you Rik!