speed up for long list matching

2012 1 5
3 个回答
更新时间：2021 8 20
1 次查看（30 天）

0 个投票

Is there a better way to do this as the length of idtsall is more than 30k? Thanks in advance.
for n = 1:length(idtsall)
ind = find(ismember(idts, idtsall{n}));
if(ind > 0)
ranktsall(n) = rankts(ind);
else
ranktsall(n) = 0;
end
end

回答（3 个）

Walter Roberson
Walter Roberson 2012-1-5

0 个投票

The two-output ismember will return the index in the second output, allowing you to skip the find() step. The first output of insmember() will indicate whether it was found or not.
[foundit, ind] = ismember(idts, idtsall{n});
if foundit
ranksall(n) = rankts(ind);
else
ranksall(n) = 0;
end
Now, to check: idtsall is a cell array of vectors? All the same size or different sizes?

3 个评论
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Eric
Eric 2012-1-5
Thanks, Walter. The idtsall is a vector of cell array. The string is of the same size.
Jan
Jan 2012-1-5
Which string is of the same size as what other string?
Eric
Eric 2012-1-5
The strings in the cell array of both idtsall and idts are of the same size.
Jan
Jan 2012-1-5

0 个投票

What type and dimension is idts?
ranktsall = zeros(1, length(idtsall); % Pre-allocate!
for n = 1:length(idtsall)
ind = find();
if all(ismember(idts, idtsall{n})) % Explicite ALL
ranktsall(n) = rankts(ind);
end
end
With a pre-allocation the processing is much faster and the else branch is not required anymore.
There can be much faster methods depending of the size and dimensions of idts and the contents of idtsall.
Eric
Eric 2012-1-5

0 个投票

I'm thinking.. could we have something like this..
ranktsall = zeros(length(idtsall), 1);
[idx, val] = matching(idts, idtsall); % I have no idea about this part
ranktsall(idx) = val;
This would be perfect! idts is a subset of idtsall.
CAN SOMEONE HELP???

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Eric
Eric
2012-1-5

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