Find mean of rows containing decimal numbers in between integers in a column
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I have a column with mostly decimal numbers and some integers
Example (the integers are in bold). - Y = [1 0.098 0.00076 0.01 2 0.099 0.007 2 0.003 0.04 0.1 4]. The integers range from 1 - 4.
I want to find the average of the numbers in between the integers, essentially [1 mean(0.098 0.00076 0.01) 2 mean (0.099 0.007) 2 mean(0.003 0.04 0.1) 4].
What would be the best way to do this? Is it recommended I turn the integers into a grouping variable?
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Accepted Answer
madhan ravi
on 17 Aug 2019
Edited: madhan ravi
on 17 Aug 2019
Y = [1 0.098 0.00076 0.01 2 0.099 0.007 2 0.003 0.04 0.1 4];
Y=Y(:);
ix=diff(find(~mod(Y,1)))-1;
assert(nnz(~mod(Y,1))>2,'atleast one gap between integers is required')
y=Y(mod(Y,1)~=0);
V=mat2cell(y(1:sum(ix)),ix);
Wanted=cellfun(@mean,V)
7 Comments
madhan ravi
on 17 Aug 2019
Edited: madhan ravi
on 17 Aug 2019
Try the below, there was a minor error before:
Y=Y(:);
ii=~mod(Y,1);
ix=diff(find(ii))-1;
y=Y(mod(Y,1)~=0);
assert(~isempty(nonzeros(ix)),'Atleast one gap between integers is required')
V=mat2cell(y(1:sum(ix)),ix);
W=cellfun(@mean,V);
Wanted = zeros(nnz(ii)+numel(W),1);
Wanted(1:2:end) = Y(ii);
Wanted(2:2:end) = W
More Answers (3)
Stephen23
on 17 Aug 2019
Edited: Stephen23
on 18 Aug 2019
>> Y = [1,0.098,0.00076,0.01,2,0.099,0.007,2,0.003,0.04,0.1,4];
>> X = cumsum([1;diff(~mod(Y(:),1))]~=0);
>> Z = accumarray(X(:),Y(:),[],@mean)
Z =
1
0.03625333333333333
2
0.05300000000000001
2
0.04766666666666667
4
EDIT: minor changes based on comments below.
6 Comments
Andrei Bobrov
on 17 Aug 2019
+1
Y = [1,0.098,0.00076,0.01,2,0.099,0.007,2,0.003,0.04,0.1,4];
out = accumarray(cumsum([1;diff(mod(Y(:),1) == 0)~=0]),Y(:),[],@mean)
Bruno Luong
on 17 Aug 2019
Edited: Bruno Luong
on 17 Aug 2019
Compact little gem, though I'm not fan of "~~x", IMO "x~=0" is clearer and perhaps faster.
darova
on 17 Aug 2019
Use ismembertol() or ismember() (round values to some tolerance if needed) to find indices of integer values
Use loop to find mean
0 Comments
KALYAN ACHARJYA
on 17 Aug 2019
Edited: KALYAN ACHARJYA
on 17 Aug 2019
Its just the Jugaar non-efficient code
# Recomended not to use, I tried few minutes, hence I posted here
Y=[1 0.098 0.00076 0.01 2 0.099 0.007 2 0.003 0.04 0.1];
data1=find(Y>=1);
mean_data=zeros(1,length(data1));
for i=1:length(data1)
if i==length(data1)
mean_data(i)=mean(Y(data1(i)+1:end));
else
mean_data(i)=mean(Y(data1(i)+1:data1(i+1)-1));
end
end
mean_data
Result:
mean_data =
0.0363 0.0530 0.0477
*Elapsed time is 0.013581 seconds.
2 Comments
madhan ravi
on 17 Aug 2019
Edited: madhan ravi
on 17 Aug 2019
Kalyan the result is completely wrong. How did you assume it was 3??
mean_data=zeros(1,3);
How would you preallocate mean_data for the below example???
Y=[1 0.098 0.00076 0.01 2 0.099 0.007 2 0.003 0.04 0.1 4 .3 .1 .4 3 0 3];
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