Hi everyone,
i have a 1x4 cell array like this:
{[0.1]} {1×3 double} {[0.02]} {1×3 double}
and i want to extract only the 1x3 double cells (of which I do not know the exact location into cell array).
Thank's a lot!
Riccardo Rossi

 Akzeptierte Antwort

Stephen23
Stephen23 am 15 Jan. 2019
Bearbeitet: Stephen23 am 15 Jan. 2019

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Where C is your cell array:
X = cellfun('length',C)==3;
out = C(X)

7 Kommentare

Why Not
X = cellfun(@length,C)==3;
Riccardo Rossi
Riccardo Rossi am 15 Jan. 2019
Thank you very much!
Stephen23
Stephen23 am 15 Jan. 2019
Bearbeitet: Stephen23 am 15 Jan. 2019
@TADA: that would also work perfectly, although if using a function handle I would recommend to use numel rather than length. The reason to use 'length' is purely one of speed: the inbuilt "Backward Compatibility" syntaxes are faster than calling cellfun with a function handle.
TADA
TADA am 15 Jan. 2019
So It should Be Faster Because There's No Overloading Involved?
Riccardo Rossi
Riccardo Rossi am 15 Jan. 2019
I'm sorry, just another question.
If i have a cell array like this:
{1×3 double} {1×3 double} {2×3 double} {1×3 double} {4×3 double}
how can i extract only the 1x3 double cells (of which I do not know the exact location into cell array)?
Thank's a lot!
Stephen23
Stephen23 am 15 Jan. 2019
F = @(v)isnumeric(v)&&isrow(v)&&numel(v)==3;
X = cellfun(F,C);
out = C(X)
Guillaume
Guillaume am 15 Jan. 2019
So It should Be Faster Because There's No Overloading Involved?
It is faster because cellfun use a different algorithm when using the backward compatible syntax. The built-in code does not call any other function when iterating over the elements, it's got its own implementation of length.
One side effect of that is that if you've overloaded length for a type in the cell array, that overload won't get called. For that reason, I personally wouldn't recommend using the old syntax unless you're 100% certain that your cell array will never ever contain objects. There's a reason the old syntax has been deprecated.

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