Why do I get NaN from corrcoef12?

조회 수: 7(최근 30일)
Peter Mills
Peter Mills 2017년 4월 6일
답변: Walter Roberson 2017년 4월 7일
Why does this give me NaN as the result?
load('Varablesuv.mat') %see attached
rho = corrcoef12(invnormcdf(u),invnormcdf(v));
Here is the code for corrcoef12:
function xy = corrcoef(x,y)
%CORRCOEF Correlation coefficients.
% CORRCOEF(X) is a matrix of correlation coefficients formed
% from array X whose each row is an observation, and each
% column is a variable.
% CORRCOEF(X,Y), where X and Y are column vectors is the same as
% If C is the covariance matrix, C = COV(X), then CORRCOEF(X) is
% the matrix whose (i,j)'th element is
% C(i,j)/SQRT(C(i,i)*C(j,j)).
% See also COV, STD.
% J. Little 5-5-86
% Revised 6-9-88 LS 2-13-95 BJ
% Copyright (c) 1984-98 by The MathWorks, Inc.
% $Revision: 5.5 $ $Date: 1997/11/21 23:23:34 $
switch nargin
case 1
c = cov(x);
case 2
c = cov(x,y);
error('Not enough input arguments.');
d = diag(c);
xy = c./sqrt(d*d');
xy = xy(1,2);
From debugging this I can see that the problem is because: c = cov(x,y); gives NaN's however I'm not clear why cov is giving NaN's?
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Peter Mills
Peter Mills 2017년 4월 7일
Yes thanks that's the the code I'm using for invnormcdf

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Walter Roberson
Walter Roberson 2017년 4월 7일
Your data contains 1.0 values. The invnormcdf formula is x = erfinv(2*p-1)*sqrt(2); which for 1.0 would be erfinv(2*1-1) which is erfinv(1) which is inf. Therefore you would be taking cov() of vectors with infinite values, and that is leading to NaN.

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