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可以这样' e/ x' I7 J4 R
x = normrnd(0,1,1,1000);6 _ E6 r2 \7 a/ j# L( j
[f, xc] = ecdf(x); % 调用ecdf函数计算xc处的经验分布函数值f
$ w1 y( j3 D' ]) t, ? ?' D* a7 Axx = linspace(min(x),max(x),50);, e, L1 g8 u( V
y = [];
4 x% `3 x3 b+ {$ Q& q' }) H8 X" F5 afor i=1:length(xx)5 W2 \$ L# I" w( s& @
id = find(xc <= xx(i));
+ W: P8 C* o5 ]7 B; d8 E" K/ i if ~isempty(id)# N0 f7 y4 e) h' J; B4 b
y(i) = f(id(end));* ~7 K! l. P2 r1 V6 A* B' _, A) R' O
else
' z* q" `/ Q- |; A- [ y(i) = 0;7 a8 q* f- v# Y6 g( c6 Z- a7 g
end* p6 Y+ V5 V0 q" y' f3 [3 j/ R/ |; e
end% s, o" X! ^, L; h. q, |. p
stairs(xc,f);2 @0 ~7 E& k* o' |" b
hold on
5 [& c7 y! K/ e) l$ k' kplot(xx,y,'.r'); |
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