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Find mean of rows containing decimal numbers in between integers in a column

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Aleya Marzuki
Aleya Marzuki on 17 Aug 2019
Edited: Aleya Marzuki on 18 Aug 2019
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
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)

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Aleya Marzuki
Aleya Marzuki on 17 Aug 2019
Works great! There just needs to be an integer at the end of the array and all the output is there. Thank you :)
madhan ravi
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
Aleya Marzuki
Aleya Marzuki on 17 Aug 2019
Cool, thanks again!
*EDIT*
I accepted this answer in the end because it accounts for there occasionally being no spaces between integers, which is what my actual data has (again should have specified this in my question, apologies).

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More Answers (3)

Stephen Cobeldick
Stephen Cobeldick on 17 Aug 2019
Edited: Stephen Cobeldick 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.

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Andrei Bobrov
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
Bruno Luong on 17 Aug 2019
Compact little gem, though I'm not fan of "~~x", IMO "x~=0" is clearer and perhaps faster.

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darova
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

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KALYAN ACHARJYA
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.

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madhan ravi
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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