How to convert cell to cell array of character vectors.

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Hi,
So I have a basic table that looks like this but has over 60 000 rows:
A = [48.44, 48.16].';
B = ['A', 'B'].';
C = table(A, B);
The problem is that when I type unique(C.B) Matlab complains : Cell array input must be a cell array of character vectors.cell array of character vectors.
The commands class(btl_data.POLYGON) returns 'cell', and iscellstr(btl_data.POLYGON) returns 0. The commands char(C.B) and cellstr(C.B) fail (Element 1 is not a string scalar or character array. All elements of cell input must be string scalars or character arrays.)
So I dont know. I know this is really basic but nothing works. Column B really should be only characters.
Thoughts ?
  3 Comments
Guillaume
Guillaume on 13 Aug 2019
Note that with your example, B is not a cell array. It's a 2x1 char vector, and unique would work fine with that.
Proper demo data:
A = [48.44, 48.16].';
B = {'A', 'BB'}.'; %with vectors of varying size you would not have been able to create a char matrix
C = table(A, B);
clearly, if iscellstr returns false, then as Adam explains, there's a non char vector in your cell array (maybe [] instead ''?). As the error message tells you, it's at least the first element that is a problem.
Stephen23
Stephen23 on 14 Aug 2019
Note that this line
B = ['A', 'B'].';
is unlikely to be useful. Square brackets are a concatenation operator, so this
['A', 'B']
is exactly equivalent to this:
'AB'

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Accepted Answer

Adam Danz
Adam Danz on 13 Aug 2019
Edited: Adam Danz on 13 Aug 2019
At least one of the elements in btl_data.POLYGON is not a character vector. This demo below recreates the error.
A = [48.44, 48.16, 50].';
B = ['A', 'B', {5:10}].';
C = table(A, B);
unique(C.B)
% Error using cell/unique (line 85)
% Cell array input must be a cell array of character vectors.
To find the row numbers of table C that do not contain a char vector in column "B",
notChar_rowIndex = find(~cellfun(@ischar,C.B))
% ans =
% 3
Now you can look at those rows,
C(notChar_rowIndex,:)
  1 Comment
Guillaume
Guillaume on 13 Aug 2019
The error message says it's the first element that's problematic. There may be more so there's indeed value in getting these row indices.

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More Answers (1)

Blue
Blue on 14 Aug 2019
Thank you all for your input.

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