the loop does not display results

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Lev Mihailov
Lev Mihailov 2020-1-27
评论： Lev Mihailov 2020-1-27
Hello! There is a matrix and I need to find the first value less than 90
clear all
close all
X=[ 80 90 100 110 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20]' ;
[~,loc]=max(X); % search j
i=1:length(X(:,1)); % search i
j=loc;
while X(i,j)>90
jx(i)=j(i)+1 % find out how many j are running in iteration 1
end
jx is not shown at all, although the condition is true
while X(i,j)<90
jy(i)=j(i)-1
end
first cycle find out the quantity j while condition X> 90 at each iteration
second cycle find out the quantity j while condition X <90 at each iteration
and the problem I do not see jx and jy
  2 个评论
Adam
Adam 2020-1-27
i and j are vectors so
while X(i,j)
will not do what you want. You can either loop around all the values testing the condition on each or you can use a vectorised approach.
X(i,j) < 90
returns a logical 8x9 array with a 1 (true) in the places that fulfil the condition and 0 otherwise. I'm not really sure what you want jy to output though since as it currently is it doesn't make sense as j is a row vector and i an 8x9 array.
Lev Mihailov
Lev Mihailov 2020-1-27
I do not understand,
% for the first iteration
while X(i,j)>90 % X(1,4) [110 100 95 90 20]
jx(i)=j(i)+1 % here I do not know what to do
% % jx(1)=j(1)+1 % but I need to do while the whole condition is satisfied
% [110 100 95 90 20] %
% jx(1)=j(1)+1
% jx(1)=j(1)+1 this condition for the first iteration must be met twice
end

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回答（1 个）

Philippe Lebel
Philippe Lebel 2020-1-27
编辑：Philippe Lebel 2020-1-27
You are not looping over the values of "i" nor "j"
i=1:length(X(:,1))
i =
1 2 3 4 5 6 7 8
j =
4 4 4 4 4 4 4 4 4
If you try to evaluate X(i,j), it is equivalent to say that you try to evaluate X([1 2 3 4 5 6 7 8], [4 4 4 4 4 4 4 4 4])
If you want to do what i think you want to do, here is the code:
clear all
close all
X=[ 80 90 100 110 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20
80 90 100 150 100 95 90 20]' ;
[~,loc]=max(X); % search j
j=loc;
for i = 1:length(X(:,1))
for k = 1:length(j)
jx(i)=j(i)+1 % find out how many j are running in iteration 1
if X(i,j(k))>90
break
end
end
if X(i,j(k))>90
break
end
end
this is not optimal but it is easier to understand (i think)
  3 个评论
显示 1更早的评论
Philippe Lebel
Philippe Lebel 2020-1-27
this is not relevant to the question asked in this post.
I don't know what is the intent behind your code so i don't know how to respond to this.
Lev Mihailov
Lev Mihailov 2020-1-27
code goal get jx per column.
And the next question is, can a matlab work with its iterations?

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