Is it possible to detect and replace abnormal/wrong numbers in an array?
2 ビュー (過去 30 日間)
古いコメントを表示
I have a dataset which is an array containing values, and which are plotted in the figure below. A section of the array for the figure is: A = [.... 25.9, 25.9, 26.2, 27, 28, 29, 29.3, 29.6, 3, 30.4, 30.5, 30.4, 30.3, 30.3, ....]; Here is number 3 an abnormality in the array. Is it possible to detect these abnormal numbers and replace them with the average value of the number before and after? So for the number 3 in A I want it to be (29.6+30.4)/2.

0 件のコメント
採用された回答
Kevin Phung
2019 年 3 月 20 日
A = [25.9, 25.9, 26.2, 27, 28, 29, 29.3, 29.6, 3, 30.4, 30.5, 30.4, 30.3, 30.3]
abnorm =mean(A) - 3*std(A); % you can set however many standard deviations to constitute 'abnormal'
n = find(A<abnorm); %locations of abnormals
for i = 1:numel(n)
if or(n(i) == 1, n(i)==numel(A)) %if first or last index, replace with NaN;
A(n(i)) = nan;
else
A(n(i)) = (A(n(i) - 1) + A(n(i) + 1)) / 2;
end
end
A(isnan(A)) = []; %remove the endpoints that are NaN
let me know if this works for you
7 件のコメント
Kevin Phung
2019 年 3 月 20 日
編集済み: Kevin Phung
2019 年 3 月 20 日
happy to help!
edit: if number of lines of code is important to you at all, you can actually condense the first 3 lines to one
n = find(isoutlier(A,'movmedian',5)); %locations of abnormals
for i = 1:numel(n)
if or(n(i) == 1, n(i)==numel(A)) %if first or last index, replace with NaN;
A(n(i)) = nan;
else
A(n(i)) = (A(n(i) - 1) + A(n(i) + 1)) / 2;
end
end
A(isnan(A)) = []; %remove the endpoints that are NaN
その他の回答 (1 件)
Walter Roberson
2019 年 3 月 20 日
2 件のコメント
Chris Turnes
2019 年 3 月 21 日
You may also be interested in the filloutliers function, which lets you not only identify outliers but replace them as well.
参考
カテゴリ
Help Center および File Exchange で Data Preprocessing についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!