Hello,
I have this for loop. K is a table containing times in its first column and BC is a vector with zeros and ones.
I want a new Vector B1 to be created, every time that BC is one. In the vector, I want the corresponding value of the first column in K. I know something in my table must be missing, but I dont get an error message, my loop just gets ignored. Whats going on?
Thank you for your help!

4 Kommentare

Dyuman Joshi
Dyuman Joshi am 4 Nov. 2021
There is a syntax error in your for loop definition. Why are you writing it as 24(K) or 1(K)?
Rik
Rik am 4 Nov. 2021
This looks like you would benefit from doing a basic Matlab tutorial, like the Onramp tutorial.
Noush
Noush am 4 Nov. 2021
@Dyuman Joshi because I would like to refer to my table, K. I have another loop that works like that:
Is there a better way to do that?
Rik
Rik am 4 Nov. 2021
I don't know why this even works, but this
for m=2:4(K)
is equivalent to this:
for m=2:4
(K)
So this is probably not doing what you think it's doing.
You are also ignoring all the mlint warnings. You should make sure there aren't any in your final code.
You are also using size without a second input argument, meaning it will return a vector. What do you think this line does?
for l=1:[5 3]
Are you sure without trying which values your variable l will have?

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Antworten (2)

Voss
Voss am 17 Dez. 2021

0 Stimmen

Maybe this?
B1 = [];
for l = 1:size(K,1)
if BC(l) == 1
B1(end+1) = K{l,1};
end
end

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am 4 Nov. 2021

Beantwortet:

am 17 Dez. 2021

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