Problems with a if statement in a for-loop

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Tatjana Mü
Tatjana Mü 2022-4-21
评论： KSSV 2022-4-21
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 个评论
Chunru
Chunru 2022-4-21
What are sizes of SpalteSr87 and SpalteSr86?
Tatjana Mü
Tatjana Mü 2022-4-21
Both are 260x1 double
Anzahl_runs is 260 because the file I read in is 260 rows to 303 columns.

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采纳的回答

KSSV
KSSV 2022-4-21
编辑：KSSV 2022-4-21
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.
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Tatjana Mü
Tatjana Mü 2022-4-21
But then I would say, select this row if there is a value over 1, isn't it? But I want to say: If there are two or more 1 in a row, do this calculations.
KSSV
KSSV 2022-4-21
nnz((intens_RL(k,:) > 1))
The above gives how many values are greater than 1.

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更多回答（1 个）

Dyuman Joshi
Dyuman Joshi 2022-4-21
编辑：Dyuman Joshi 2022-4-21
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
  1 个评论
Tatjana Mü
Tatjana Mü 2022-4-21
Thank you for this explanation. You are right, I didnt think about this point. Thank you a lot!

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