I have code of the following kind in MATLAB:
indices = find([1 2 2 3 3 3 4 5 6 7 7] == 3)
This returns 4,5,6 - the indices of elements in t
This is not an answer - I am just comparing the running time of the three solutions suggested by chappjc and Daniel R.
N = 5e7; % length of vector
p = 0.99; % probability
KK = 100; % number of instances
rntm1 = zeros(KK, 1); % runtime with ismember
rntm2 = zeros(KK, 1); % runtime with ismembc
rntm3 = zeros(KK, 1); % runtime with Daniel's function
for kk = 1:KK
x = cumsum(rand(N, 1) > p);
searchfor = x(ceil(4*N/5));
tic
[tf,loc]=ismember(x, searchfor);
inds1 = find(tf);
rntm1(kk) = toc;
tic
tf = ismembc(x, searchfor);
inds2 = find(tf);
rntm2(kk) = toc;
tic
a=1;
b=numel(x);
c=1;
d=numel(x);
while (a+1
Daniel's binary search is very fast.
% Mean of running time
mean([rntm1 rntm2 rntm3])
% 0.631132275892504 0.295233981447746 0.000400786666188
% Percentiles of running time
prctile([rntm1 rntm2 rntm3], [0 25 50 75 100])
% 0.410663611685559 0.175298784336465 0.000012828868032
% 0.429120717937665 0.185935198821797 0.000014539383770
% 0.582281366154709 0.268931132925888 0.000019243302048
% 0.775917520641649 0.385297304740352 0.000026940622867
% 1.063753914942895 0.592429428396956 0.037773746662356