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find a value & store in new variable (again)

조회 수: 22 (최근 30일)
Ubu
Ubu 2012년 3월 19일
Dear Matlab friends,
We recently posted a question about finding values and storing them in new variables:
Our question remains in the sense that we would like to know how to procede when we have a matrix rathen than a cell array.
Indeed, we have a 384 x 14 matrix A, and we would like to find all occurrences of number [2] in column 1, and save corresponded value in column 2 in a new variable "new_variable":
A =
3 5 6 ...
2 3 5
2 3 4
1 4 5
5 7 9
...
so the result would be:
new_variable =
2
2
The suggestion for cell arrays was
new_variable = A{2}(A{1} == 2);
We thank you very much for any suggestion!
Best,
Udiubu

답변 (3개)

Geoff
Geoff 2012년 3월 19일
Use logical indexing:
new_variable = A(A(:,1)==2, 2);
This indexes all those rows of A where column 1 is equal to 2, selects column 2 and assigns the result to new_variable.
  댓글 수: 13
Ubu
Ubu 2012년 3월 19일
Geoff that's so kind of you.. We'll be waiting!
Aldin
Aldin 2012년 3월 19일
@Geoff: How do you mean?

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Geoff
Geoff 2012년 3월 19일
Okay, since your matrix actually contains cells the equality operator doesn't work. I'll split the code up for extra clarity, since another reader insists ;-)
I'm still not sure if you want to replace column 2 with the value 2 or just copy the filtered column 2 into new_variable... The question was confusing. Anyway,
I would do this:
rowidx = cellfun( @(x) x==2, A(:,1) );
new_variable = cell2mat(A(rowidx,2));
cellfun maps a function over all elements of a cell-array or matrix. The '@(x) x==2' part is an anonymous (on-the-fly/adhoc) function. Here we use it on the first column. The output is a vector of booleans, which we then use as a logical index.
Edit:
The other way is to just extract the numeric columns that are of interest and use my other solution that works on matrices:
B = cell2mat(A(:,1:2));
new_variable = B( B(:,1)==2, 2);
  댓글 수: 6
Geoff
Geoff 2012년 3월 19일
That's odd. Did you use my edited version of the first code segment? Using A{rowidx,2} was incorrect.
I assume your matrix is generally numeric (ie the numbers themselves are not strings), but those NaNs get in the way. You could try converting them to numbers too:
A{cellfun(@(x) strcmpi(x,'nan'), A)} = NaN;
Ubu
Ubu 2012년 3월 19일
A{cellfun(@(x) strcmpi(x,'nan'), A)} = NaN; right hand side has too few values to satisfy the left hand side.
However, we managed to delete those NaN, transform to mat and execute our mean and SD calculations.
It was a real pleasure to have you. You made a couple of guys very happy tonight.
We thank you very much Geoff!

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Aldin
Aldin 2012년 3월 19일
Here, try this code:
A = [ 2 1 3 2 4; 4 5 3 6 2; 2 3 5 3 6; 1 2 5 3 6; 3 2 5 2 5]
for i = 1:5
if A(i,1) == 2
A(i,2) = 2;
end
end
  댓글 수: 4
Aldin
Aldin 2012년 3월 19일
oooo i know what you want :)
A = [ 2 1 3 2 4; 4 5 3 6 2; 2 3 5 3 6; 1 2 5 3 6; 3 2 5 2 5];
counter = 0;
for i = 1:5
if A(i,1) == 2
counter = counter + 1;
new_variable(count) = A(i,2);
end
end
Aldin
Aldin 2012년 3월 19일
Have I solved your problem?

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