How to create a loop with unknown number of iterations?
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I would like a help about a script. I have a file with three columns, each of them are numbers. I would like to satisfy a condition via a loop.I want, based on a specific equation/relationship:
mask = z_input<fin & z_input>=(fin-1);
in order to get only the elements from the file that satisfy this equation and not get an "emplty" file.
If not, then the relationship should be converted to
mask= z_input<(fin-1) & z_input>=(fin-2);
One again not then the relationship should be converted to
mask= z_input<(fin-2) & z_input>=(fin-3);
One again not then the relationship should be converted to
mask= z_input<(fin-3) & z_input>=(fin-4);
etc
I have tried the following commands:
clc
clear
filename1= 'mydata.txt';
[d1,tex]= importdata(filename1);
y_input=d1.data(:,2);
x_input=d1.data(:,1);
z_input=d1.data(:,4);
mask = z_input<fin & z_input>=(fin-1);
selected= d1.data(mask,:);
y_B=selected(:,2);
x_B=selected(:,1);
z_input_B=selected(:,4);
fin=max(z_input);
if isempty(z_input_B)
mask= z_input<(fin-1) & z_input>=(fin-2);
selected = d1.data(mask,:);
end
if isempty(z_input_B)
mask= z_input<(fin-2) & z_input>=(fin-3);
selected = d1.data(mask,:);
end
if isempty(z_input_B)
mask= z_input<(fin-3) & z_input>=(fin-4);
selected = d1.data(mask,:);
end
I need something in (i) that takes values from 0 to whatever is needed to satisfy the relation/condition and not get an "empty" file.
I mean something like
mask= z_input<(fin-i) & z_input>=(fin-(i+1));
Could you please help me?
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채택된 답변
Jan
2022년 11월 3일
편집: Jan
2022년 11월 3일
ready = false;
k = 1;
while ~ready
mask = z_input < fin & z_input >= (fin - k);
if any(mask)
ready = true;
else
k = k + 1;
end
end
But remember, that the value of min(z_input - fin) should help to find k directly without using a loop.
Another option is using maxk(z_input, 2) to get the 2 largest outputs, which let you determine k also.
댓글 수: 4
Jan
2022년 11월 5일
편집: Jan
2022년 11월 5일
@Ivan Mich: You ignore my answer and my comments. I have mentioned, that you modify mask , but check for isempty(z_input_B).
Checking a binary mask by isempty will fail also, because this counts the number of elements without considering their values. Look in the code I've posted: any(x) does something different from isempty(mask).
Posting an example file is not useful, because I cannot guess, what I should do with this file.
My code does already, what you have asked for and you explain repeatedly, that another code does not work. Even if I mention the problem of the other code, you do not react. So I repeat another time, that a loop is a waste of time only, because maxk(z_input, 2) reveals the value of k directly: It replies the 2 largest values and you have to subtract them only.
I've posted some code, which is working as far as I can see, and mentioned a much more efficient solution. I do not see, how I could help you further. Why do you hestitate to use these information?
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