Find mean of rows containing decimal numbers in between integers in a column
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Aleya Marzuki
il 17 Ago 2019
Modificato: Aleya Marzuki
il 18 Ago 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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madhan ravi
il 17 Ago 2019
Modificato: madhan ravi
il 17 Ago 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 Commenti
madhan ravi
il 17 Ago 2019
Modificato: madhan ravi
il 17 Ago 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
Più risposte (3)
Stephen23
il 17 Ago 2019
Modificato: Stephen23
il 18 Ago 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 Commenti
Andrei Bobrov
il 17 Ago 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
il 17 Ago 2019
Modificato: Bruno Luong
il 17 Ago 2019
Compact little gem, though I'm not fan of "~~x", IMO "x~=0" is clearer and perhaps faster.
darova
il 17 Ago 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
il 17 Ago 2019
Modificato: KALYAN ACHARJYA
il 17 Ago 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 Commenti
madhan ravi
il 17 Ago 2019
Modificato: madhan ravi
il 17 Ago 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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