通用Lisp序列函数remove-duplicates留下了每个多重性的一个元素。以下类似函数remove-equals的目标是消除所有的多重性。
但是,我希望使用内置函数remove-if (非迭代)和SBCL的哈希表功能,以便使测试函数的时间复杂度保持在O(n)。当前的问题是SBCL等式测试需要是全局的,但是测试也需要依赖于key参数到remove-equals。它能被写成满足这两种要求吗?
(defun remove-equals (sequence &key (test #'eql) (start 0) end (key #'identity))
"Removes all repetitive sequence elements based on equality test."
#.(defun equality-test (x y)
(funcall test (funcall key x) (funcall key y)))
#.(sb-ext:define-hash-table-test equality-test sxhash)
(let ((ht (make-hash-table :test #'equality-test)))
(iterate (for elt in-sequence (subseq sequence start end))
(incf (gethash (funcall key elt) ht 0)))
(remove-if (lambda (elt)
(/= 1 (gethash elt ht)))
sequence :start start :end end :key key)))发布于 2019-06-10 11:47:01
define-hash-table-test的第三个参数将测试与哈希函数关联起来。使用sxhash并不能达到这个目的,因为它应该适合test函数。(equal x y)暗示着(= (sxhash x) (sxhash))。因此,第二个参数应该是一个函数test-hash,这样(funcall test x y)就意味着(= (test-hash x) (test-hash y))。仅仅拥有测试函数是不可能做到的。也许最好是通过记录它需要有散列支持来绕过整个事情:
(defun remove-duplicated (sequence &key (test #'eql) (start 0) end (key #'identity))
"Removes all repetitive sequence elements based on equality test.
equalily tests other than eq, eql, equal and equalp requires you
add it to be allowed in a hash table eg. sb-ext:define-hash-table-test in SBCL"
(let ((ht (make-hash-table :test test)))
(iterate (for elt in-sequence (subseq sequence start end))
(incf (gethash (funcall key elt) ht 0)))
(remove-if (lambda (elt)
(/= 1 (gethash elt ht)))
sequence :start start :end end :key key)))现在,如果用户需要一个自定义测试,那么他们需要自己对它进行测试:
(defun car-equals (a b)
(equal (car a) (car b)))
(defun car-equals-hash (p)
(sxhash (car p)))
(sb-ext:define-hash-table-test car-equals car-equals-hash)
(car-equals '(1 2 3 4) '(1 3 5 7)) ; ==> t
(defparameter *ht* (make-hash-table :test 'car-equals))
(setf (gethash '(1 2 3 4) *ht*) 'found)
(gethash '(1 3 5 7) *ht*) ; ==> found
(remove-duplicated '((5 0 1 2) (5 1 2 3) (5 1 3 2) (5 2 3 4))
:test #'car-equals
:key #'cdr)
; ==> ((5 0 1 2) (5 2 3 4))发布于 2019-06-10 08:28:08
像这样的读取时间计算函数不会像你想的那样做。从代码中简化:
(defun foo (a b test)
#.(defun equality-test (x y)
(funcall test x y))
(funcall #'equality-test a b)),这是不可能的.
原因1:读取时间创建的函数无法访问来自周围代码的词法变量(这里没有方法引用test,因为在读取过程中不存在带有函数foo的环境)
test变量在equality-test中并不是指词法变量。它是未定义的/未声明的。
原因2:DEFUN计算结果为符号
在读取和评估读取时代码后,代码如下所示:
(defun foo (a b test)
equality-test
(funcall #'equality-test a b))equality-test是一个未绑定变量。这是运行时的一个错误。
原因3:函数equality-test可能不存在
如果我们用文件编译器编译代码,函数equality-test是在读取表单时在编译时环境中创建的,但它不会是编译代码的一部分。
发布于 2019-06-10 06:21:04
免责声明:我发现@Sylwester的答案更清晰、更干净--只是更好(没有宏)。
然而,这只是假设(但不是一个好做法):
(ql:quickload :iterate) ;; you forgot these - but they are necessary
(use-package :iterate) ;; for your code to run - at least my definition
(ql:quickload :alexandria) ;; of 'minimal working example' is to include imports.
(defmacro remove-equals (sequence &key (test #'eql) (start 0) end (key #'identity))
"Remove all repetitive sequence alements based on equality test."
(alexandria:once-only (sequence test start end key) ; as hygyenic macro
`(progn
(defun equality-test (x y)
(funcall ,test (funcall ,key x) (funcall ,key y)))
(sb-ext:define-hash-table-test equality-test sxhash)
(let ((ht (make-hash-table :test #'equality-test)))
(iterate (for elt in-sequence (subseq ,sequence ,start ,end))
(incf (gethash (funcall ,key elt) ht 0)))
(remove-if (lambda (elt)
(/= 1 (gethash (funcall ,key elt) ht)))
,sequence :start ,start :end ,end :key ,key)))))
(remove-equals '(1 2 3 1 4 5 3) :test #'= :end 6)
;; WARNING: redefining COMMON-LISP-USER::EQUALITY-TEST in DEFUN
;;
;; (2 3 4 5 3)
(describe 'equality-test) ;; shows new definition
;; COMMON-LISP-USER::EQUALITY-TEST
;; [symbol]
;;
;; EQUALITY-TEST names a compiled function:
;; Lambda-list: (X Y)
;; Derived type: (FUNCTION (T T) (VALUES BOOLEAN &OPTIONAL))
;; Source form:
;; (SB-INT:NAMED-LAMBDA EQUALITY-TEST
;; (X Y)
;; (BLOCK EQUALITY-TEST
;; (FUNCALL #'= (FUNCALL #1=#<FUNCTION IDENTITY> X)
;; (FUNCALL #1# Y))))警告总是会发生的--如果您使用的不仅仅是一个哈希表,这肯定会干扰并导致错误。所以我不建议你这么做!
https://stackoverflow.com/questions/56519209
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