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How to extract array from the mean directly using indexing?

2 vues (au cours des 30 derniers jours)
William
William le 6 Juil 2023
A = [2.29441228002842 17.3351873969651 2.79928860040389 7.17554281085515;
3.16200415659481 16.9975006219209 3.18395042061777 7.45747601536461;
4.55387378846938 13.4948344868957 3.22594708715312 7.49001605579868]
meanA = mean(A,1)
output: [3.3368 15.9425 3.0697 7.3743]
how can i get second element of mean in one line?
ex:
mean(A,1)(2)
  • (2) is index
But, it is not working. Thanks.

Réponse acceptée

Nikhil Baishkiyar
Nikhil Baishkiyar le 6 Juil 2023
Modifié(e) : Nikhil Baishkiyar le 6 Juil 2023
A = [2.29441228002842 17.3351873969651 2.79928860040389 7.17554281085515;
3.16200415659481 16.9975006219209 3.18395042061777 7.45747601536461;
4.55387378846938 13.4948344868957 3.22594708715312 7.49001605579868]
A = 3×4
2.2944 17.3352 2.7993 7.1755 3.1620 16.9975 3.1840 7.4575 4.5539 13.4948 3.2259 7.4900
mean(A(:,2))
ans = 15.9425
You mean something like this?
subindex = @(matrix, r) matrix(r); % An anonymous function for indexing
value = subindex(mean(A,1), 2) % Use the function to index the vector
value = 15.9425
  3 commentaires
William
William le 6 Juil 2023
Modifié(e) : William le 6 Juil 2023
Yeah.. But we only got second column. It seems we will use mean four times. if there is thousands?
Actually, all of data is needed. and indexing is what can solve the problem and can ellimitate the time complexity. or we can use two lines to make the mean variable.
Nikhil Baishkiyar
Nikhil Baishkiyar le 6 Juil 2023
Modifié(e) : Nikhil Baishkiyar le 6 Juil 2023
I read the post and updated my answer based on that. If you want every mean value you will have to store it in a temporary variable I am afraid

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