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可以这样
; k( a) S% P# E, ex = normrnd(0,1,1,1000);
~5 B! r7 X8 g5 y/ x$ p1 Y' W[f, xc] = ecdf(x); % 调用ecdf函数计算xc处的经验分布函数值f
6 F( J+ {" x" Y' Gxx = linspace(min(x),max(x),50);
+ e2 Y. V# r9 ~ H; l) B" q$ Vy = [];( T( I+ A4 b' r/ n1 @" b4 D
for i=1:length(xx)7 H. w2 r& K' \
id = find(xc <= xx(i));
$ c* [$ O: Y( B' o7 Y if ~isempty(id)
$ L. d2 ]0 k! C y(i) = f(id(end));! z4 ]9 J, A; f3 E2 m" x
else
3 Y* n @0 b" U$ z' K' u y(i) = 0;
; q9 {; \* i5 B end
- R: I: L! t0 i" G1 Q1 R9 {4 P& W7 Qend, L5 D' y4 y4 k# k; s, J
stairs(xc,f);; D. J3 J% q4 v8 m5 Q
hold on
! P; O* W, M+ O8 @5 f2 Y- qplot(xx,y,'.r'); |
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