# How do I find the maximum/last index value from a list of indices

3 Ansichten (letzte 30 Tage)
Alex am 24 Dez. 2016
Beantwortet: Alex am 18 Mär. 2017
I have imported some data to form an array called 'newData1.data' and then find the row/col indices for each value 1-49 in a vector called 'numbers'
numbers=1:1:49;
for i=1:length(numbers)
[row,col] = find(newData1.data' == i);
end
How do I find the greatest/maximum/last column number?
test(i) = max(col)
does not work within the loop. There are plenty of questions and answers on how to get the indices of the maximum value in the array, but that is not what I am looking for.
Best regards,
Alex
##### 0 Kommentare-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden

Melden Sie sich an, um zu kommentieren.

### Akzeptierte Antwort

Image Analyst am 24 Dez. 2016
Try this:
numbers = 1 : 1 : 49;
for k = 1 : length(numbers)
[rows, columns] = find(newData1.data' == numbers(k));
maxColumns(k) = max(columns)
end
##### 2 KommentareKeine anzeigenKeine ausblenden
Alex am 24 Dez. 2016
This is the same thing I was doing. It returns the error, "Index exceeds matrix dimensions."
Image Analyst am 24 Dez. 2016
It's not exactly the same thing you were doing since my code is more robust in that it compares "numbers" instead of i. Your code defines numbers but then never uses it. Anyway there is nothing wrong so the only way we can continue is if you attach your data via a .mat file.

Melden Sie sich an, um zu kommentieren.

### Weitere Antworten (1)

Alex am 18 Mär. 2017
Solution: I found that I created a variable called min/max: max=max(data) min=min(data) By changing these lines to: dmax=max(data) dmin=min(data) the max(idx), min(idx) function as expected.
##### 0 Kommentare-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden

Melden Sie sich an, um zu kommentieren.

### Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by