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Selecting Numbers from a String in a Matrix

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Katy Weihrich
Katy Weihrich 2018-2-26
评论： Katy Weihrich 2018-2-28
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 个评论
Jan
Jan 2018-2-26
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.

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Stephen23
Stephen23 2018-2-26
编辑：Stephen23 2018-2-26
"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}
...
Tip: vertcat probably gives you what you want:
out = regexp(Matrix_Text(:,2),...);
out = vertcat(out{:});

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