# Problems with a if statement in a for-loop

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Tatjana Mü on 21 Apr 2022
Commented: KSSV on 21 Apr 2022
Hi,
My plan for the code is following:
• read all rows from 1 to anzahl_runs (=number of rows)
• If there is a 1 in the row -> Divide two double vectors through each other, if there is no 1, delete the row
• Save this ratios in a new vector.
I tried it this way:
for k=1:anzahl_runs
if any (intens_RL(k,:)==1)
Ratio(k,1)=SpalteSr87(k,:)/SpalteSr86(k,:);
else
intens_RL(k,:)=[];
end
end
Ratio should be a new created vector.
But at the moment the line Ratio.... gives me following error:
Index in position 1 exceeds array bounds (must not exceed 181).
Error in TOF_Skript_SingleParticlesSrNdOs (line 96)
if any (intens_RL(k,:)==1)
I appreciate any help :-)
##### 2 CommentsShowHide 1 older comment
Tatjana Mü on 21 Apr 2022
Both are 260x1 double
Anzahl_runs is 260 because the file I read in is 260 rows to 303 columns.

KSSV on 21 Apr 2022
Edited: KSSV on 21 Apr 2022
id = true(anzahl_runs,1) ;
for k=1:anzahl_runs
if any (intens_RL(k,:)==1)
Ratio(k,1)=SpalteSr87(k,:)/SpalteSr86(k,:);
id(k) = 0 ;
end
end
intens_RL(id,:)=[];
I suspect, as you are removing the rows in else condition, the number of rows in intens_RL will reduce, if the condition is not met. So this error popped out. Get the indices where the condition is not met and after the loop, you remove these rows from the array. You may try the above code.
KSSV on 21 Apr 2022
nnz((intens_RL(k,:) > 1))
The above gives how many values are greater than 1.

Dyuman Joshi on 21 Apr 2022
Edited: Dyuman Joshi on 21 Apr 2022
So, Anzahl_runs is 260 meaning the for loop will run from 1 to 260.
Suppose your condition did not find a 1 in any row, so it will delete that row from the matrix. Now your matrix doesn't have 260 rows. So when the loop variable reaches a value greater than the number of rows your matrix has, this error shows up.
So, If x number of rows do not have 1 at all, your matrix will be reduced to 260-x rows. As soon as k reaches 260-x+1 (181 in your case), this error will be displayed and your code will stop running.
You can use a while loop or get the index of the rows from for loop and delete them after the loop
Tatjana Mü on 21 Apr 2022
Thank you for this explanation. You are right, I didnt think about this point. Thank you a lot!