running for loop gives error " Row index exceeds table dimentions "

i want to extraxt specific range of data from a table names as "Raw", and finally find mean of selected data assigned to variable answer. the below codes works frine for single range, but i want to mention an array of min & max range and so that when a for loop runs it should select new range on every iteration but it gives error " Row index exceeds table dimentions .

Raw = table([1 2 3 4 5 6 7 8 9]',[3.6 3.4 5.3 6.8 8.4 3.6 4.8 8.8 7.3]') % the simple code works fine

minrange = 1; maxrange = 3;

data = table2array(varfun(@(x)((x>=minrange)&(x<=maxrange)), Raw(:,1)));

data = Raw(data,:)

answer = mean(table2array(data(:,2)))

##### #######

Raw = table([1 2 3 4 5 6 7 8 9]',[3.6 3.4 5.3 6.8 8.4 3.6 4.8 8.8 7.3]') % this code gives error when minrange changed

answer = zeros(3,1);

minrange = [1 4]; maxrange = [3 6];

for i = 1:2

data = table2array(varfun(@(x)((x>=minrange)&(x<=maxrange)), Raw(:,1)));

data = Raw(data,:)

answer(i,1) = mean(table2array(data(:,2)))

end

Answer 1

[Abolfazl Chaman Motlagh]

your minrange and maxrange now are vecotors with two value.if you want to put it in for loop, in data you should mention wich component you want.

 Raw = table([1 2 3 4 5 6 7 8 9]',[3.6 3.4 5.3 6.8 8.4 3.6 4.8 8.8 7.3]') ;  % this code gives error when minrange changed
answer = zeros(2,1);
minrange = [1 4];    maxrange = [3 6];
for i = 1:2
    data = table2array(varfun(@(x)((x>=minrange(i))&(x<=maxrange(i))), Raw(:,1)));
    data = Raw(data,:);
    answer(i,1) = mean(table2array(data(:,2)));
end
answer

answer = 2×1

4.1000 6.2667

this would solve your problem.

or you could :

Raw = table([1 2 3 4 5 6 7 8 9]',[3.6 3.4 5.3 6.8 8.4 3.6 4.8 8.8 7.3]') ;  % this code gives error when minrange changed
answer = zeros(2,1);
minrange = [1 4];    maxrange = [3 6];
data = table2array(varfun(@(x)((x>=minrange)&(x<=maxrange)), Raw(:,1)));
for i = 1:2
    data_ = Raw(data(:,i),:);
    answer(i,1) = mean(table2array(data_(:,2)));
end
answer

answer = 2×1

4.1000 6.2667

but for an easier way to do this, i suggest: (don't use table)

Raw = [3.6 3.4 5.3 6.8 8.4 3.6 4.8 8.8 7.3]';
minrange = [1 4];    maxrange = [3 6];
for i=1:2
answer(i) = mean(Raw(minrange(i):maxrange(i)));
end
answer

answer = 2×1

4.1000 6.2667

[taimour sadiq]

Thanks Abolfazl Chaman Motlagh, ur Answer is exactly according to my Requirements, especially u guide with multipe method that is very helpful for me and hope for others also.

Answer 2

[Peter Perkins]

Not sure what purpose the first vector serves, but if it always defines non-overlapping groups, then there's no need for a loop:

>> Raw = table([1 2 3 4 5 6 7 8 9]',[3.6 3.4 5.3 6.8 8.4 3.6 4.8 8.8 7.3]')

>> Raw.Group = [1;1;1;2;2;2;3;3;3]
>> varfun(@mean,Raw,'GroupingVariables','Group','InputVariables','Var2')
ans =
3×3 table
Group    GroupCount       mean_Var2    
_____    __________    ________________
1          3                      4.1
2          3         6.26666666666667
3          3         6.96666666666667