Hii Marco Ciapparelli,
I understand you want to get a solution to the issue regarding Spearman and Pearson.
The difference in output that you are observing when using 'partialcorr' with the 'Spearman' type between single-precision and double-precision formats could be due to the algorithm used for computing the partial rank correlation coefficient.
The 'Spearman' type in 'partialcorr' computes the partial rank correlation coefficient using the Spearman's rank correlation formula. The algorithm for computing rank correlations involves sorting the data, assigning ranks, and then calculating the correlation based on the ranks. So,when you use single-precision data, there can be differences in the sorting and ranking process due to the limited precision of single-precision numbers.
The 'Pearson' type in 'partialcorr' calculates the partial correlation coefficient using Pearson's correlation formula. Pearson's correlation is based on the covariance and standard deviations of the data, which are not affected by the precision of the numbers.Thus, you observe consistent results.
To obtain accurate results for partial rank correlation using the 'Spearman' type, it is recommended to use double-precision data instead of single-precision. The 'Spearman' type relies on the ranks of the data, and the limited precision of single-precision numbers can introduce inconsistencies in the ranking process.
Or you can convert your data to double precision to get the result as shown below:-
% generate data
rng(1);
y = randn(10000,1);
x = y.*randn(10000,1);
z = x + randn(10000,1);
partialcorr(double(y),double(x),double(z),'Type','Spearman') % rho output = 0.2227
partialcorr(y,x,z,'Type','Spearman') % rho output = 0.0155
partialcorr(single(y),single(x),single(z),'Type','Pearson') % rho output = 0.0255
partialcorr(y,x,z,'Type','Pearson') % rho output = 0.0255
Or else if you want to use a single-precision data, partialcorri would be a better choice with 'Spearman'.
% generate data
rng(1);
y = randn(10000,1);
x = y.*randn(10000,1);
z = x + randn(10000,1);
partialcorri(single(y),single(x),single(z),'Type','Spearman') % rho output = 0.2227
partialcorr(y,x,z,'Type','Spearman') % rho output = 0.0155
partialcorr(single(y),single(x),single(z),'Type','Pearson') % rho output = 0.0255
partialcorr(y,x,z,'Type','Pearson') % rho output = 0.0255
Also kindly refer to the links mentioned below for better understanding:-
Hope this helps!
