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

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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
motorspeed_detected(traj_cnt) = 0;
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:
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
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!

Risposta accettata

Timo Dietz
Timo Dietz il 1 Dic 2020
Modificato: Timo Dietz il 1 Dic 2020
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.
  2 Commenti
Rik il 1 Dic 2020
You can also have a look at this highlight post.
Nik Rocky
Nik Rocky il 1 Dic 2020
Thank you very much, I'm using [row,column]=size(Traj_interpl{1,209}) now!

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Più risposte (3)

Sriram Tadavarty
Sriram Tadavarty il 1 Dic 2020
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.

Image Analyst
Image Analyst il 1 Dic 2020
length() always gives you the LONGEST dimension. Since you have more columns (2) than rows (1), it will return 2.

Nik Rocky
Nik Rocky il 1 Dic 2020
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!
  2 Commenti
Rik il 1 Dic 2020
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.

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