Borrar filtros
Borrar filtros

Delete m consecutive rows every n rows

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Enrico Gambini
Enrico Gambini el 25 de En. de 2022
Comentada: Enrico Gambini el 25 de En. de 2022
Hello! I'm looking to find a way to delete certain rows in a big table that i'm working with. I would like to find a smart way to delete m consecutive rows every n rows. In my case I have m=6 and n=24--> i want rows 25:30 to be erased while rows 31:54 preserved, up till the end. Hope that the question is clear. Thank you!
  2 comentarios
Enrico Gambini
Enrico Gambini el 25 de En. de 2022
Hi matieu.
That answer seems not useful fot this topic

Iniciar sesión para comentar.

Respuesta aceptada

Voss
Voss el 25 de En. de 2022
Editada: Voss el 25 de En. de 2022
Here's one way to do it, with a matrix. The logic would be the same for a table.
m = 6;
n = 24;
N = 87;
data = (1:N).'+(0:10:20); % some data
disp(data);
1 11 21 2 12 22 3 13 23 4 14 24 5 15 25 6 16 26 7 17 27 8 18 28 9 19 29 10 20 30 11 21 31 12 22 32 13 23 33 14 24 34 15 25 35 16 26 36 17 27 37 18 28 38 19 29 39 20 30 40 21 31 41 22 32 42 23 33 43 24 34 44 25 35 45 26 36 46 27 37 47 28 38 48 29 39 49 30 40 50 31 41 51 32 42 52 33 43 53 34 44 54 35 45 55 36 46 56 37 47 57 38 48 58 39 49 59 40 50 60 41 51 61 42 52 62 43 53 63 44 54 64 45 55 65 46 56 66 47 57 67 48 58 68 49 59 69 50 60 70 51 61 71 52 62 72 53 63 73 54 64 74 55 65 75 56 66 76 57 67 77 58 68 78 59 69 79 60 70 80 61 71 81 62 72 82 63 73 83 64 74 84 65 75 85 66 76 86 67 77 87 68 78 88 69 79 89 70 80 90 71 81 91 72 82 92 73 83 93 74 84 94 75 85 95 76 86 96 77 87 97 78 88 98 79 89 99 80 90 100 81 91 101 82 92 102 83 93 103 84 94 104 85 95 105 86 96 106 87 97 107
idx = (n+1:n+m).'+(n+m)*(0:ceil(N/(n+m))-1); % index of rows to delete
disp(idx);
25 55 85 26 56 86 27 57 87 28 58 88 29 59 89 30 60 90
idx(idx > N) = []; % don't go off the end
disp(idx);
25 26 27 28 29 30 55 56 57 58 59 60 85 86 87
data(idx,:) = []; % delete the dang rows
disp(data);
1 11 21 2 12 22 3 13 23 4 14 24 5 15 25 6 16 26 7 17 27 8 18 28 9 19 29 10 20 30 11 21 31 12 22 32 13 23 33 14 24 34 15 25 35 16 26 36 17 27 37 18 28 38 19 29 39 20 30 40 21 31 41 22 32 42 23 33 43 24 34 44 31 41 51 32 42 52 33 43 53 34 44 54 35 45 55 36 46 56 37 47 57 38 48 58 39 49 59 40 50 60 41 51 61 42 52 62 43 53 63 44 54 64 45 55 65 46 56 66 47 57 67 48 58 68 49 59 69 50 60 70 51 61 71 52 62 72 53 63 73 54 64 74 61 71 81 62 72 82 63 73 83 64 74 84 65 75 85 66 76 86 67 77 87 68 78 88 69 79 89 70 80 90 71 81 91 72 82 92 73 83 93 74 84 94 75 85 95 76 86 96 77 87 97 78 88 98 79 89 99 80 90 100 81 91 101 82 92 102 83 93 103 84 94 104

Más respuestas (1)

Matt J
Matt J el 25 de En. de 2022
Editada: Matt J el 25 de En. de 2022
m = 6;
n = 24;
N = 87;
data = (1:N).'+(0:10:20); % some data
discard=repelem( [false;true] , [n;m], ceil(N/(m+n)) );
data(discard(1:N),:)=[]
data = 72×3
1 11 21 2 12 22 3 13 23 4 14 24 5 15 25 6 16 26 7 17 27 8 18 28 9 19 29 10 20 30

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