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Sort and accumulate data in a matrix

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Elli
Elli on 27 Jan 2020
Commented: Stephan on 27 Jan 2020
Hi everyone,
I have written a code but it's to messy, I'm sure one could sum it up.
This is the data I have:
data= [0 -50
-10 -50
-10 -50
0 -40
-10 -40
10 -40
10 -30
-10 -30
0 -30
... and so on till +50.
The output I want should be seperated for the three values in the first column so that the output for 0, 10 and -10 should separatley look like this:
[ -50 0
-40 0
-30 0
-20 1
-10 1
0 2
10 6
20 6
30 6
40 6
50 6 ]
so thst the first column lists every value once and the second column the amount of occurance of ones for that specific value.
My code:
%%0 (delete others)
data(find(data(:,1)==10),:)=[]
data(find(data(:,1)==-10),:)=[]
%% 10 (delete others)
data(find(data(:,1)==0),:)=[]
data(find(data(:,1)==-10),:)=[]
%% -10 (delete others)
data(find(data(:,1)==10),:)=[]
data(find(data(:,1)==0),:)=[]
And then for every data (unfortenately) separately the following:
for i = size(data)
data1=sum(data(1:3,2))
data2=sum(data(4:6,2))
data3=sum(data(7:9,2))
data4=sum(data(10:12,2))
...
end
comp_data=[data1, data2, data3, data4]'
test_angle=unique(data(:,2));
dataAll=horzcat( test_angle,comp_data)
I am pretty sure one could sum it up by different for/if loops so that I don't have to run it for 0, 10 and -10 separately and independetly from fixed row values.
Thanks a lot!!!

  2 Comments

Walter Roberson
Walter Roberson on 27 Jan 2020
data(find(data(:,1)==10),:)=[]
data(find(data(:,1)==-10),:)=[]
%% 10 (delete others)
data(find(data(:,1)==0),:)=[]
data(find(data(:,1)==-10),:)=[]
?? The -10 are already gone -- you deleted them in the second statement.
Have you considered
data(ismember(data(:,1), [-10 0 10]), :) = [];
Elli
Elli on 27 Jan 2020
Your suggestion leads leads to an empty struct!
And you are right, I am sorry, it was a typo.
%% 10 (delete others)
data(find(data(:,1)==0),:)=[]
data(find(data(:,1)==-10),:)=[]

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Answers (1)

Stephan
Stephan on 27 Jan 2020
Edited: Stephan on 27 Jan 2020
Here is an example:
A = [-10 0 10 10 20 -20 30 -30 40 -50 50 -40 -10 0 0 10 20 -20 30 -30 40 -50 50 -40 40 40 40];
groups = findgroups(A);
occurences = splitapply(@numel,A,groups);
result = [unique(A); occurences].'

  3 Comments

Guillaume
Guillaume on 27 Jan 2020
Note that unique(A) is the 2nd output of findgroups.
Simpler code overall:
[groups, values] = findgroups(A(:)); %can also be done with [values, ~, groups] = unique(A);
result = [values, accumarray(groups, 1)]
or:
values = unique(A);
result = [values; histcounts(A, [values, Inf])].'
Elli
Elli on 27 Jan 2020
Hi Stephan, thank you!
This counts the occurance of the values in the first column (A), not the corresponding values in the second row!
Stephan
Stephan on 27 Jan 2020
@Elli This was just to give you a direction how to improveyour approach. Together with the commentfrom Guillaume you should be able to do this in a proper way.
@Guillaume Thank you for the hint, again a nice new fact i did not notice so far.

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