Incremental indexing to create an array

Hi,
I have a 1097x1097 array (called list_of_distances) and would like to index into that array repeatedly; I only need certain values from the array. By doing this repeated indexing, I'd like to create new array that is 1x1097 of values. I included a screen shot of the values I'm looking to get into a new array.
Any help would be appreciated
Michael
Michael
R = 2
S = 1
for i = length(list_of_distances)
final_list(i) = list_of_distances(R,S)
R = R+1
S = S+1
end

 Respuesta aceptada

Peng Li
Peng Li el 15 de Abr. de 2020

1 voto

using diag(list_of_distances, -1) to get the diagonal you want.

5 comentarios

Peng Li
Peng Li el 15 de Abr. de 2020
and btw, this is off main diagonal so you won't be able to get a 1*1097 vector, 1096 instead.
Michael King
Michael King el 15 de Abr. de 2020
Thanks so much for the quick answer! looking forward to testing this out!
Michael King
Michael King el 15 de Abr. de 2020
It worked really well. A much more concise solution! Would you (or anyone) be able to take a minute to suggest what I was doing wrong with my for loop?
the loop doesn't run as your i is fixed at length(list_of_distances), and you prob want to do for i = 1:length(list_of_distances), but this way you are up to an indexing error as I commented above, you aren't able to have a vector of 1097 elements. Do for i = 1:size(list_of_distances, 1)-1 instead.
And you don't necessarily need to do a loop, although you are not familiar with some of these builtin functions. For this case, you can do list_of_distances(2:size(list_of_distances, 1)+1:end), which will give you the same results as diag(list_of_distances, -1).
list_of_distances = magic(10)
list_of_distances =
92 99 1 8 15 67 74 51 58 40
98 80 7 14 16 73 55 57 64 41
4 81 88 20 22 54 56 63 70 47
85 87 19 21 3 60 62 69 71 28
86 93 25 2 9 61 68 75 52 34
17 24 76 83 90 42 49 26 33 65
23 5 82 89 91 48 30 32 39 66
79 6 13 95 97 29 31 38 45 72
10 12 94 96 78 35 37 44 46 53
11 18 100 77 84 36 43 50 27 59
>> b = diag(list_of_distances, -1)
b =
98
81
19
2
90
48
31
44
27
>> c = list_of_distances(2:size(list_of_distances)+1:end)
c =
98 81 19 2 90 48 31 44 27
Michael King
Michael King el 15 de Abr. de 2020
thanks, I've learned a lot here! much appreciated

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