for loop store values?

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Phil Whitfield
Phil Whitfield 2018-8-1
回答： Ashraf Rayed 2020-5-12
so I am convinced there is a better way for me to approach the following code:
ftse = xlsread('ftse123.xlsx');
L = length(ftse);
s = floor(L/10);
f25 = ftse(:,3);
k=1;
h=1;
for i = 1:10
j=s*i;
h=s*(i-1);
if h== 0
h=1;
else
h=s*(i-1);
end
y{i} = f25(h:j,:);
end
x1=cell2mat(y(1));
x2=cell2mat(y(2));
x3=cell2mat(y(3));
x4=cell2mat(y(4));
x5=cell2mat(y(5));
x6=cell2mat(y(6));
x7=cell2mat(y(7));
x8=cell2mat(y(8));
x9=cell2mat(y(9));
x10=cell2mat(y(10));
I used copy and paste after struggling to save the values x1 etc, within the for loop.
Also noticed that x1 is has a length of 522 and not 523 like the rest?
The code achieves what I want/need but must be long and drawn out.
any advice would be great thanks.
  2 个评论
Stephen23
Stephen23 2018-8-1
编辑：Stephen23 2018-8-1
"I used copy and paste after struggling to save the values x1 etc, within the for loop."
Copy-and-pasting code is a sign that you are doing something wrong. Using numbered variables is a sign that you are doing something wrong. The solution to both of these: use indexing.
Please upload ftse123.xlsx so that we have something to work with.
Phil Whitfield
Phil Whitfield 2018-8-1
that's when I posted after I started copy and pasting because I knew it was wrong.

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采纳的回答

Stephen23
Stephen23 2018-8-1
编辑：Stephen23 2018-8-1
I would recommend that you use readtable or importdata and use a table to store, group, and process your data. In lieu of that, here is some simple code to group the third column into ten groups:
>> N = size(ftse,1);
>> V = linspace(1,11,N+1);
>> V = fix(V(1:end-1));
>> C = accumarray(V(:),ftse(:,3),[],@(v){v});
It is much simpler and more efficient to access the cells of C using indexing. Do NOT create numbered variables.
"Also noticed that x1 is has a length of 522 and not 523 like the rest?"
Because ftse has 5225 rows. There is no way to split this into ten equally sized groups:
>> size(ftse)
ans =
5225 4
>> size(ftse,1)/10
ans =
522.5
Half the groups will have 252, half have 253 elements. My code spreads these evenly through C:
>> cellfun('size',C,1)
ans =
523
522
523
522
523
522
523
522
523
522
  2 个评论
Phil Whitfield
Phil Whitfield 2018-8-1
编辑：Phil Whitfield 2018-8-1
Thanks, is there a way to create a 523x7 and 523x3 matrix of C from the index?
normally I would do :
xx=[x1,x2,x3,x4,x5,x6,x7];
xxx=[x8,x9,x10];
also I appreciate the 522 thing, should've seen that thanks.
Stephen23
Stephen23 2018-8-1
编辑：Stephen23 2018-8-1
"Thanks, is there a way to create a 523x7 and 523x3 matrix of C from the index?"
Not easily, because that does not match how many of each size data group that you actually have: five of each. So you could easily get 522x5 and 523x5 matrices:
M523 = [C{1:2:end}]
M522 = [C{2:2:end}]

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更多回答（1 个）

Ashraf Rayed
Ashraf Rayed 2020-5-12
i have this code, i have to check the leaf area of some pictures in a folder and then i have save the area of those leaves by sequence in an excel sheet. Can you please help to sort out the problem?
D = 'F:\MATLAB\R2018a\bin\rayed mat lab\experimental';
S = dir(fullfile(D,'*.jpg'));
for k = 1:numel(S)
F = fullfile(D,S(k).name);
I = imread(F);
ih=rgb2hsv(I);
ih1=ih(:,:,1);
binary=ih1>0.6;
leafArea = sum(binary(:))
subplot(2, 2, 2);
imshow(binary, []);
message = sprintf('The leaf area = %d pixels.\n pixels = %.1f%%',...
leafArea, defectArea/leafArea*100);
uiwait(msgbox(message))
end

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