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Barmar's answer
,并且它很好地处理任意权重。但是,它确实需要两次调用
随机的
,这可能是不可取的。另一种方法是创建一个向量,其中包含根据其预期频率出现的元素。E、 例如,如果元素a和b的选择概率为1/3和2/3,那么可以创建一个数组
(a b b)
并从中随机选择。
(defun biased-generator (values weights)
(multiple-value-bind (total values)
(loop for v in values
for w in weights
nconc (make-list w :initial-element v) into vs
sum w into total
finally (return (values total (coerce vs 'vector))))
(lambda ()
(aref values (random total)))))
CL-USER> (defparameter *gen* (biased-generator '(a b) '(1 2)))
*GEN*
CL-USER> (loop for i from 1 to 100 collect (funcall *gen*))
(A A B A B A A B B A B B A A A B A A B A A A B A A A B B B B B A B B B B A A B
A B B A A A A B B B A A A A B A A B B B A A B B B A B B B B B B B B B B A B A
A A A B B B B A B A A B B A B A A B B B B B)
CL-USER> (let ((abs (loop for i from 1 to 10000 collect (funcall *gen*))))
(list (count 'a abs)
(count 'b abs)))
(3293 6707)