How to remove elements from a matrix?

I would like to delete random elements from this matrix: U = [ 't' 'h' 'e' 's' 't' 'r' 'i' 'n' 'g']
How can i do it?

2 Comments

Note that [] is a concatenation operator, so this
[ 't' 'h' 'e' 's' 't' 'r' 'i' 'n' 'g']
is simply identical to this:
'thestring'

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 Accepted Answer

the cyclist
the cyclist on 6 Jun 2017
Edited: the cyclist on 6 Jun 2017
If you have the Statistics and Machine Learning Toolbox, you can do
numberToKeep = 4;
U = U(sort(randsample(length(U),numberToKeep)));
If you do not have that toolbox, you can accomplish the same thing using a different random function, like randi, but you'll need to guard against repeated elements. Let us know if you get stuck.

4 Comments

Thanks u very much for the answer, how can i do the exact same thing without 'numberToKeep'? For instance, delete random elements without limit in numberToKeep.
U = [];
would be the quickest way!
John, are you saying that you don't even want to specify the number of elements to be removed at random ahead of time? You want that to be random, also?
If that is the case, then you can just define
numberToKeep = randi(length(U))
Thank u very much, u helped me a lot!

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More Answers (1)

Stephen23
Stephen23 on 6 Jun 2017
Edited: Stephen23 on 6 Jun 2017
Use randperm (it has no repeats):
nkeep = 4;
str = 'thestring';
str(randperm(numel(str),nkeep))

3 Comments

Sorry for my silly questions but im new at Matlab. I tried this code but i cant find the solution of my error.
str = 'thestring';
nkeep = randperm(length(str));
str = str(randperm(numel(str),nkeep));
disp(str)
I want nkeep to have random number from the length of the string. For example if the length of str is 13 then nkeep will be randomly 8.
If you want to keep the same order, and have random number of characters removed, then you could do this:
>> str = 'thestring';
>> N = numel(str);
>> str(randperm(N,randperm(N,1))) = []
str = teting
or use randi instead:
>> str = 'thestring';
>> N = numel(str);
>> str(randperm(N,randi(N,1))) = []
str = ttrin
Thank you very much for the answers. You helped me a lot! :)

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on 6 Jun 2017

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on 6 Jun 2017

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