Selecting Numbers from a String in a Matrix
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Katy Weihrich
il 26 Feb 2018
Commentato: Katy Weihrich
il 28 Feb 2018
I used the
B = regexp(A,'\d*','Match');
comment to identify the year, month, day, hour, min, sec and msec of a weird Timestamp (written as: 2018-02-13T14:07:14.000Z).
It worked well when I used
Timestamp = regexp('2018-02-13T14:07:14.000Z','\d*','Match')
But as soon as I tried to apply it to a matrix by using
Date = Matrix_Text(:,2)
Timestamp = regexp(Date,'\d*','Match')
I only got {1×7 cell} as the answer of each cell.
When I tried to identify the problem by only including one value at first:
Date = Matrix_Text(2,2)
Timestamp = regexp(Date,'\d*','Match')
The output was {1×7 cell} again.
Does someone know what the error could be?
1 Commento
Jan
il 26 Feb 2018
Why do you assume, that there is an error? If Matrix_Text is a cell array, this is the expected behavior. This would be immediately clear, if you post, what this array is. Of course we cannot guess this.
Risposta accettata
Stephen23
il 26 Feb 2018
Modificato: Stephen23
il 26 Feb 2018
"The output was {1×7 cell} again."
"Does someone know what the error could be?"
There is no error, because that is the expected behavior. The regexp help states that "If either str or expression is a cell array of character vectors or a string array, and the other is a character vector or a string scalar, the output is a cell array of row vectors. The output cell array has the same dimensions as the input array".
So when you provide the input as a cell array of strings/char vectors, then the output is a cell array containing the outputs that you would expect if the input was just one char vector. It is basically equivalent to doing this:
out{1} = regexp(inp{1},...)
out{2} = regexp(inp{2},...)
out{3} = regexp(inp{3},...)
...
so if you want to access the output data then you will need to dig one cell deeper using cell array indexing:
out{1}
out{2}
...
out = regexp(Matrix_Text(:,2),...);
out = vertcat(out{:});
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