How do I store data that meet conditions of an if statement

44 Ansichten (letzte 30 Tage)
NA
NA am 1 Feb. 2021
Kommentiert: NA am 3 Feb. 2021
Hi All,
I'm slightly struggling trying to get my code to work, and I hope someone can point me in the right direction.
Very briedly, I have two vectors, 'xx' and 'yy'. I created a matrix using these two vectors, 'xy'. If xy(:,1) is ==1, I would like to store the corresponding cell in xy(:,2) in a seperate variable, 'data'. Otherwise, if the logical expression = 0, I'd like that cell to be a NaN
xx=[-3; -3; -3; -3; -3; -3; 1; -2; -2; -3; -3; -2; 1; 1; 1; -2; 1; 1; ];
yy=[26.28; 36.96; 90.00; 90.00; 90.00; 90.00; 27.85; 29.97; 47.03; 90.00; 67.22; 78.87; 19.60; 9.00; 2.00; 3.41; 1.88; 2.50];
xy=[xx yy];
for i=1:size(xy(:,1),1)
if xy(:,1)==1
data=xy(:,2);
else
data=NaN;
end
end
Thanks in advance.
Cheers

Akzeptierte Antwort

Jan
Jan am 1 Feb. 2021
Bearbeitet: Jan am 1 Feb. 2021
xx = [-3; -3; -3; -3; -3; -3; 1; -2; -2; -3; -3; -2; 1; 1; 1; -2; 1; 1; ];
yy = [26.28; 36.96; 90.00; 90.00; 90.00; 90.00; 27.85; 29.97; ...
47.03; 90.00; 67.22; 78.87; 19.60; 9.00; 2.00; 3.41; 1.88; 2.50];
xy = [xx yy];
data = nan(size(xx));
match = (xx == 1);
data(match) = xy(match, 2);
This "vectorized" method is faster and nicer than a loop. But with a loop:
data = nan(size(xy, 1), 1);
for i = 1:size(xy, 1) % better than: size(xy(:,1),1)
if xy(i, 1) == 1
data = xy(i, 2);
end
end
  5 Kommentare
dpb
dpb am 2 Feb. 2021
I posted a demonstration of what happens with dynamic allocation just a day or so ago at<Comment_1298378>
NA
NA am 3 Feb. 2021
Thanks dpb, thats very useful information. Cheers

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

dpb
dpb am 1 Feb. 2021
Don't need any loops nor the explicity xy array that is duplicate of existing data.
data=nan(size(x,1),2); % initialize to NaN
ix=(xx==1); % logical addressing vector of wanted locations
data(ix,:)=[xx(ix) yy(ix)]; % set data values
Alternatively,
data=[xx yy]; % initialize to data
data(data(:,1),:)=nan; % set unwanted values to NaN
  1 Kommentar
NA
NA am 1 Feb. 2021
I wrote it usuing a for loop because it's a subset of a larger code (and needs to be incorporated within a loop). But thank you for your feedback :) Cheers

Melden Sie sich an, um zu kommentieren.


J Chen
J Chen am 1 Feb. 2021
You over wrote the variable data. One soluation is to change data to data(i).
  1 Kommentar
dpb
dpb am 1 Feb. 2021
The if condition will never be satisfied. if expression is true IFF all elements of expression are true.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by