Indexing an array with find(contains()) while retaining duplicate elements.

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Joes Edders
Joes Edders 2022-4-1
评论： Voss 2022-4-1
Hello,
I have two cell arrays containing strings. 'A' contains a list of unique identifying numbers (as strings) and 'B' contains a list of those IDs with the potential for duplicates.
I am attempting to produce an array, the size of 'B', which contains the indices of the relative elements of 'B' within 'A'.
I have previously been using find(contains(X,y)) to extract singular array indices but I am unsure how to expand this to take two arrays while also retaining the duplicate indices, or even if it's appropriate to attempt to use these functions to do so.
I have had success in getting what I want with the code below but would welcome any suggestions on how to do this in-line.
Thanks
A = {'000000699';'000002373';'000002374';'000002378';'000002385';'000002387';'000002389';'000000895';'000001037';'000001038';'000001041';'000001043';'000001045';'000001052';'000001371';'000001746';'000000670';'000000537';'000001193';'000002284';'000002285';'000002286';'000002298';'000002299';'000002300';'000002301';'000002302';'000002315';'000002317';'000002330';'000002468';'000002513';'000000479'};
B = {'000000479';'000000537';'000000670';'000000699';'000000895';'000001037';'000001038';'000001041';'000001043';'000001043';'000001045';'000001052';'000001193';'000001371';'000001371';'000001746';'000002284';'000002285';'000002286';'000002298';'000002299';'000002300';'000002300';'000002301';'000002302';'000002302';'000002315';'000002317';'000002317';'000002330';'000002373';'000002374';'000002374';'000002374';'000002378';'000002385';'000002387';'000002389';'000002468';'000002513';'000002513'};
for i = 1:length(B)
C(i) = find(contains(A,B(i)));
end

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采纳的回答

Voss
Voss 2022-4-1
Ran in:
You can use ismember:
A = {'000000699';'000002373';'000002374';'000002378';'000002385';'000002387';'000002389';'000000895';'000001037';'000001038';'000001041';'000001043';'000001045';'000001052';'000001371';'000001746';'000000670';'000000537';'000001193';'000002284';'000002285';'000002286';'000002298';'000002299';'000002300';'000002301';'000002302';'000002315';'000002317';'000002330';'000002468';'000002513';'000000479'};
B = {'000000479';'000000537';'000000670';'000000699';'000000895';'000001037';'000001038';'000001041';'000001043';'000001043';'000001045';'000001052';'000001193';'000001371';'000001371';'000001746';'000002284';'000002285';'000002286';'000002298';'000002299';'000002300';'000002300';'000002301';'000002302';'000002302';'000002315';'000002317';'000002317';'000002330';'000002373';'000002374';'000002374';'000002374';'000002378';'000002385';'000002387';'000002389';'000002468';'000002513';'000002513'};
% using contains()
for i = 1:length(B)
C(i) = find(contains(A,B(i)));
end
C
C = 1×41
33 18 17 1 8 9 10 11 12 12 13 14 19 15 15 16 20 21 22 23 24 25 25 26 27 27 28 29 29 30
% using ismember()
[~,C_new] = ismember(B,A)
C_new = 41×1
33 18 17 1 8 9 10 11 12 12
isequal(C_new.',C)
ans = logical
1

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