drop same row from one matrix based on another matrix

6 vues (au cours des 30 derniers jours)
zhengyang shang
zhengyang shang le 25 Mar 2021
Commenté : Walter Roberson le 26 Mar 2021
i hava a matrix a(253*2) , the first column is the id number and i want drop the same id in matrix b(30*1) , keep the rest data in matrix c(223*2) how should i do?

Réponses (1)

Walter Roberson
Walter Roberson le 25 Mar 2021
c = a;
c(ismember(c(:,1), b),:) = [];
  2 commentaires
zhengyang shang
zhengyang shang le 25 Mar 2021
i am sorry it seems not ture. because it just drop the last 30 rows of a but not contain b
Walter Roberson
Walter Roberson le 26 Mar 2021
Your findings do not agree with my tests.
%test data.
%define a and b so a(:,1) includes all of b exactly once,
%and that the other things in a(:,1) are not in b.
%the order of entries is scrambled so that we can be sure
%that the entries are being removed by value and not by position
b = randperm(99,30).';
not_b = setdiff(1:99,b).';
a1 = [b; not_b(randi(length(not_b), 253-length(b),1))];
a1 = a1(randperm(length(a1)));
a2 = randi([0 9], size(a1));
a = [a1,a2];
%let us get an idea of what is in a
a(1:10,:)
ans = 10×2
65 4 39 8 45 2 69 1 66 4 55 5 59 6 90 7 4 1 51 6
%let us get an idea of what is in b
b(1:10,:)
ans = 10×1
45 1 27 85 96 42 37 43 22 34
%now do the task we were asked to do, remove all the entries in a that
%have a column 1 that occurs somewhere in b.
c = a;
c(ismember(c(:,1), b),:) = [];
%check sizes to be sure we got the size wanted
size(a), size(b), size(c)
ans = 1×2
253 2
ans = 1×2
30 1
ans = 1×2
223 2
%check to be sure that some entries in a were removed and that the order of
%the remaining entries remained the same
c(1:10,:)
ans = 10×2
65 4 39 8 69 1 66 4 55 5 59 6 90 7 4 1 51 6 28 0
%check to be sure there are no column 1 entries in c that are the same as
%any entry in b
nnz(ismember(c(:,1),b))
ans = 0

Connectez-vous pour commenter.

Catégories

En savoir plus sur Marine and Underwater Vehicles dans Help Center et File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by