New variable based on a 'percentile' condition

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Maria
Maria 2014-8-26
回答： Kelly Kearney 2014-8-26
I have a double variable A with 6000 rows and 3 columns:
  • C1 gives the year;
  • C2 gives a code (no repetitions within each year);
  • C3 gives a number;
C1 C2 C3
A=[1983 11 54
1983 13 24
1983 16 32
1983 20 11
1983 25 14
1983 28 23
1983 29 19]
B, is a cell variable with 1 row and 31 columns (each column corresponds to a year) that gives the 90th percentile of A(:,3) per year:
B={53 49.3 51.7 49.2 48 40 41 44.6 …}
I am trying to obtain a new variable A2 that would be equal to A, but would only consider the cases in which A(:,3) > B(1,:).
For example, I have already done something similar but with different conditions, only in these cases my condition would not change from year to year:
numMean=A(A(:,3)>=42.8,[1 2 3]);
num25=A(A(:,3)>=25,[1 2 3]);
Thank you very much for your help.

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2014-8-26
编辑：Azzi Abdelmalek 2014-8-26
B={53 49.3 51.7 49.2 48 40 41}
A=[1983 11 54
1983 13 24
1983 16 32
1983 20 11
1983 25 14
1983 28 23
1983 29 19]
out=A(all(bsxfun(@gt, A(:,3),cell2mat(B)),2),:)

更多回答（1 个）

Kelly Kearney
Kelly Kearney 2014-8-26
Could do it all it one step with accumarray. Though this way doesn't preserve the initial order... not sure if that is necessary or not for your task.
% Fake data
n = 100;
A = [floor(rand(n,1)*5) + 1983 ones(n,1) round(rand(n,1)*100)];
% Find values >90% of their year
[unqyr, blah, ia] = unique(A(:,1));
nyr = length(unqyr);
atop = accumarray(ia, A(:,3), [nyr 1], @(x) {x(x > prctile(x,90))});
cell2mat(atop)

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