So I am currently using 'accumarray' to find the averages of a range of numbers wich correspond to matching ID's. Ex Input:
ID----Value
1 215
1 336
1 123
2 111
2 246
2 851
My current code finds the unweighted average of the above values, using the ID as the 'seperator' so that I don't get the average for all of the values together as one number, but rather seperate results for just values which have corresponding ID's. EX Output:
ID----Value
1 224.66
2 402.66
To achieve this I am using this code:
[ID, ~, Groups] = unique(StarData2(:,1),'stable');
app = accumarray(Groups, StarData2(:,2), [], @mean);
With StarData2 being the input of the function. This is working perfectly for my purposes until now, I need to know if accumarray can be made to give me a weighted average, such that each point in app (before the average is found) can be assigned a weight or that the @mean can be replaced with a function that can achieve this. The new input will look like this:
ID----Value----Weight
1 215 12
1 336 17
1 123 11
2 111 6
2 246 20
2 851 18
The new code must do the sum(val(i)*weight(i))/sum(weight) instead of just the standard mean. Thanks for any assistance.
You can use the row index as the "vals" (second input to accumarray
) and define your own function that does the weighted mean on group of the data:
Weights = data(:,3); Vals = data(:,2); % pick your columns here
WeightedMeanFcn = @(ii) sum(Vals(ii).*Weights(ii))/sum(Weights(ii));
wmeans = accumarray(Groups, 1:numel(Groups), [], WeightedMeanFcn)
Demonstration
Starting with data
(the new input with your weights) and your unique
command:
data = [1,215,12; 1,336,17; 1,123,11; 2,111,6; 2,246,20; 2,851,18];
[ID, ~, Groups] = unique(data(:,1),'stable');
The accumarray
usage is as follows (redefine WeightedMeanFcn
every time you change data
!):
>> Weights = data(:,3); Vals = data(:,2); % pick your columns here
>> WeightedMeanFcn = @(ii) sum(Vals(ii).*Weights(ii))/sum(Weights(ii));
>> app = accumarray(Groups, 1:numel(Groups), [], WeightedMeanFcn)
app =
241.1250
475.0909
Checking manually, with the first group:
ig = 1;
sum(data(Groups==ig,2).*data(Groups==ig,3))/sum(data(Groups==ig,3))
ans =
241.1250
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