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?

采纳的回答

Jan
Jan 2022-11-3
编辑:Jan 2022-11-3
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.
  4 个评论
Ivan Mich
Ivan Mich 2022-11-5
fin is the maximum value of the z_input variable. The point is that I would like to increase the k in the while loop with 1, until the z_input_B is not empty anymore. (according to this equation
mask= z_input_B<(fin-k) & z_input>=(fin-(k+1));, with k=0,1,.....etc)
I am uploading an example of input file in order to undersatnd what I mean
Jan
Jan 2022-11-5
编辑:Jan 2022-11-5
@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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更多回答(1 个)

Jim Riggs
Jim Riggs 2022-11-3
编辑:Jim Riggs 2022-11-3
What you are describing sounds like a "while" loop.
You specify a logical condition which will terminate the loop
i=0;
condition=false;
while condition == false
...
...
i=i+1;
condition = % set your termination condition
end
  5 个评论
Jim Riggs
Jim Riggs 2022-11-3
Based on what you are asking for, it looks like @Jan's answer is good, except you said that you wanted the index (i) to start at zero, but i'm not sure that that is really what you want. Is ther somthing about @Jan's answer that you don't like?
Ivan Mich
Ivan Mich 2022-11-5
Thank you, but I actually not works...
I have tried the following commands:
ready = false;
k = 0;
while ~ready
mask= z_input_B<(fin-k) & z_input>=(fin-(k+1));
if isempty(z_input_B)
ready = true;
else
k = k + 1;
end
end
but matlab crashed (I mean that do not stop executing). What I type wrong? I am absolutely sure that I have a mistake in syntax...
Could you please help me??

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