How can I return the minimum value that is not zero inside a for loop?

2 Ansichten (letzte 30 Tage)
Thomas
Thomas am 12 Feb. 2016
Bearbeitet: Fangjun Jiang am 12 Feb. 2016
Hi,
I have a for loop that find the min, max and mean of a set of data for each month that is entered and I want the minimum function to return the minimum value that is not zero. I know if i have matrix A i can use,
min(A(A>0);
But in this loop it does not allow me to do this, here is the code I am using,
for i = unique(months)' %allows function to work with any number of months
maxs(i,:) = max(hourdata(months==i,1:end));
mins(i,:) = min(hourdata(months==i,1:end));
means(i,:) = mean(hourdata(months==i,1:end));
end
I tried using the following for the minimum value but it does not like it,
mins(i,:) = min(hourdata(hourdata>0(months==i,1:end)));
I get the error 'Unbalanced or unexpected parenthesis or bracket'.
Any help would be greatly appreciated.

Antworten (1)

Fangjun Jiang
Fangjun Jiang am 12 Feb. 2016
Bearbeitet: Fangjun Jiang am 12 Feb. 2016
How about using a temp variable?
temp=hourdata(months==i,1:end);
temp(temp<=0)=inf;
mins(i,:) = min(temp,[],1);
clear temp
  2 Kommentare
Thomas
Thomas am 12 Feb. 2016
The result I am expecting is a 12x24 double (12 months of the year and over 24 hours) and using the temp variable gives a 12x1 double. This might be because that is where the only values greater than zero are. However, I tried with,
if true
temp = hourdata(months==i,1:end);
mins(i,:) = min(temp(temp>=0));
clear temp
This rendered the same result of a 12x1 double.
Fangjun Jiang
Fangjun Jiang am 12 Feb. 2016
Bearbeitet: Fangjun Jiang am 12 Feb. 2016
I see. To keep the matrix formation, you can temporarily replace those zero and negative values with a very large number and then do the min() operation.
I also see a potential problem when a particular month has only one row of data. In that case, the statement inside the for-loop will return a 1x1 scalar, rather than a 1x24 vector which is desired. So please use the optional input argument of min() function.
See if the update code works.

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