prctile function vs excel percentile
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Hi,
when I run the below,
A = [ 0.02; 0.01 ; 0.01; 0.01 ; 0.04 ]; B = prctile(A,99.95);
I get B = 0.0400, which seems rounded.
In excel the same percentile function produces a result of 0.03996
Is there a way to show a less rounded result in matlab? Any other function or calculation in matlab e.g. 65/43, will produce a result (1.5116) that doesn't round the decimals like prctile does.
(even if I try "format long" the result will still be 0.040000000...)
Many thanks
Respuesta aceptada
Más respuestas (2)
Chan Dennis
el 23 de Dic. de 2018
Clearly, both percentile alogrithms are the different cases of the linear interpolation between closest ranks method. More details in Percentile - Wikipedia.
- When let C=0.5, the result is equal prctile function in matlab.
- When let C=1, the result is equal excel percentile.
There are two matlab functions below, which are the implementations of the above algorithms.
prctile_one.m (resuls equal excel PERCENTILE and PERCENTILE.INC)
function V_x = prctile_one(V,p)
% The linear interpolation between closest ranks method in the case of `C=1`
% [Percentile - Wikipedia](https://en.wikipedia.org/wiki/Percentile)
if ~isvector(p) || numel(p) == 0 || any(p < 0 | p > 1) || ~isreal(p)
error('Make sure the Second digit within the [0,1] interval');
end
V = sort(V,'ascend');
N = length(V);
x = p*(N-1)+1; % position x
if floor(x) < N
V_x = V(floor(x)) + mod(x,1)*(V(floor(x)+1) - V(floor(x))); % value
else
V_x = V(N); % position N
end
prctile_half.m (resuls equal matlab prctile)
function V_x = prctile_half(V,p)
% The linear interpolation between closest ranks method in the case of `C=0.5`
% [Percentile - Wikipedia](https://en.wikipedia.org/wiki/Percentile)
if ~isvector(p) || numel(p) == 0 || any(p < 0 | p > 1) || ~isreal(p)
error('Make sure the Second digit within the [0,1] interval');
end
V = sort(V,'ascend');
N = length(V);
p_1 = 1/(2*N); % position 1
p_N = 1 - 1/(2*N); % position N
if p_1<=p && p<=p_N
x = N*p + 0.5; % position x
V_x = V(floor(x)) + mod(x,1)*(V(floor(x)+1) - V(floor(x))); % value
else if 0<=p && p<p_1
V_x = V(1); % value 1
else if p_N<p && p<=1
V_x = V(N); % value N
end
end
end
1 comentario
John D'Errico
el 24 de Mzo. de 2015
Personally, I might have chosen the Excel scheme. The highest datapoint seems like it should be 100%, the smallest, 0%. But I can see entirely valid arguments for doing it other ways.
Of course, were I truly the author of the code, I would have offered multiple schemes, and let the user pick among them as options. But then I tend to overdo those things as an author.
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