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

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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

采纳的回答

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
  1 个评论
taimour sadiq
taimour sadiq 2021-9-7
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.

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更多回答(1 个)

Peter Perkins
Peter Perkins 2022-3-2
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

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