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

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?

Answer 1

[madhan ravi]

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)

[KALYAN ACHARJYA]

+1 Great! @Madhan still 3 value is not appearing??

[madhan ravi]

See the edited answer , while copy pasting missed out the last value in Y. That's the reason you were not able to see the third one.

[KALYAN ACHARJYA]

yes

[madhan ravi]

Aleya my code contains bugs, Stephen answer is the one you need to follow.

[Aleya Marzuki]

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]

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]

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).

Answer 2

[Stephen23]

>> 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.

[Aleya Marzuki]

Sorry I tried this on a long array of over 9000 rows and when I ran the cumsum function it showed this error:

'Error using horzcat

Dimensions of matrices being concatenated are not consistent.'

[madhan ravi]

X = cumsum(~~[1;diff(~mod(Y(:),1))]);
Z = accumarray(X(:),Y(:),[],@mean)

[Stephen23]

"Sorry I tried this on a long array of over 9000 rows and when I ran the cumsum function it showed this error: 'Error using horzcat Dimensions of matrices being concatenated are not consistent.'"

This error occurs because your example data in your question is a row vector, i.e. has size 1xN, whereas apparently the data you are actually using is a column vector, i.e. has size Nx1. These are not the same.

But of course it is easy to adapt my code to work with vectors of either orientation:

>> X = cumsum(~~[1;diff(~mod(Y(:),1))]);

[Aleya Marzuki]

Gotcha! Sorry for not specifying I was using a column vector, thank you :)

[Andrei Bobrov]

+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]

Compact little gem, though I'm not fan of "~~x", IMO "x~=0" is clearer and perhaps faster.

Answer 3

[darova]

Use ismembertol() or ismember() (round values to some tolerance if needed) to find indices of integer values

Use loop to find mean

Answer 4

[KALYAN ACHARJYA]

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.

[madhan ravi]

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];

[KALYAN ACHARJYA]

#Edited