7 Applicative Programming
7.2 funcall
(funcall #'cons 'a 'b) -> (a . b)
funcall可以通过函数名字来调用函数
lisp里面引用函数使用符号#'
再看一个例子，
> (setf fn #'cons)
#<compilied-function cons {6041410}>
> (funcall fn 'c 'd)
(c . d)
7.3 mapcar
mapcar可以将一个函数应用到一个list中的每一个元素中，然后将每次返回的结果全部放到一个list中返回
(defun square (n) (* n n)) (square 3) -> 9 (square '(1 2 3)) -> error (mapcar #'square '(1 2 3) -> (1 4 9)
7.4 manipulating tables with mapcar
(mapcar #'function-name list)
现在有一张英语和法语的转换表
(setf words
`((one un)
(two deux)
(three trois)
(four quatre)
(five cinq)))
获取所有list的第一个item:
> (mapcar #'first words) (one two three four five)
ex 7.1
(defun add1 (x) (+ x 1)) > (mapcar #'add1 '(1 3 5 7 9)
7.5 lambda expressions
(lambda (n) (* n n))
在使用mapcar的时候如果不想另外定义一个函数，可以直接在mapcar函数里面定义匿名函数lambda
> (mapcar #'(lambda (n) (* n n)) '(1 2 3) (1 4 9)
ex 7.7
(defun flips (x) (mapcar #'(lambda (n) (if (equal n 'up) 'down 'up)) x))
7.6 find-if
(find-if #'predicate list)
对list中的每个item调用predicate，返回第一个返回值是t的item，没有返回t的就返回nil
> (find-if #'oddp '(2 4 6 7 8)) 7
7.7 my-assoc
assoc函数有2个参数(key table),返回在table中第一个元素是key的那个list
现在可以用find-if来写一个assoc功能的函数
(defun my-assoc (key table) (find-if #'(lambda (entry) (equal key (first entry))) table))
对于table中的每个list,我们定义的lambda函数要判断key是否等于这个list的第一个元素，如果想等就返回t,而find-if函数只要lambda函数返回t就将那个list返回，刚好可以实现assoc函数的功能
要注意的一点，例子中lambda函数可以访问定义在my-assoc函数中的局部变量key，说明lambda函数可以包含这个lambda函数的外层函数的local vars
ex 7.8
(defun foo (x k) (find-if #'(lambda (n) (if (< n (- k 10)) nil (<= n (+ k 10)))) x))
ex 7.9
(defun find-nested (x) (find-if #'(lambda (entry) (and (listp entry) (not (equal entry nil)))) x))
7.8 remove-if, remove-if-not
(remove-if #'function-name list)
对于list中的每个元素都调用函数进行判断，如果为t就将其删除，返回剩下的list
> (remove-if #'numberp '(2 for 1 sale) (for sale)
remove-if-not则相反，如果函数返回为nil就将该元素删除
> (remove-if-not #'(lambda (x) (> x 3)) '(2 3 4 5 6)) (4 5 6)
ex 7.15
(defun rank (x) (first x)) (defun suit (x) (second x)) (setf my-hand `((3 hearts) (5 clubs) (2 diamonds) (4 diamonds) (ace spades))) (defun count-suit (suit hand) (length (remove-if-not #'(lambda (entry) (equal suit (second entry))) hand))) (setf colors '((clubs black) (diamonds red) (hearts red) (spades black))) (defun color-of (x) (second (assoc (second x) colors))) (defun first-red (hand) (find-if #'(lambda (card) (equal 'red (color-of card))) hand)) (defun black-cards (hand) (remove-if-not #'(lambda (card) (equal 'black (color-of card))) hand)) (defun what-ranks (suit hand) (mapcar #'first (remove-if-not #'(lambda (card) (equal suit (second card))) hand))) (setf all-ranks '(2 3 4 5 6 7 8 9 10 jack queen king ace)) (defun higher-rank-p (card1 card2) (member (first card1) (rest (member (first card2) all-ranks)))) (defun high-card (hand) ;;建立一个rank的list(ace king queen jack 10 ... 2),使用find-if对这个list的rank逐个查找是否出现在hand中 ;;(mapcar #'first hand)可以得到hand中所有的rank的list (assoc (find-if #'(lambda (rank) (member rank (mapcar #'first hand))) (reverse all-ranks)) hand))
7.9 reduce
(reduce #'function list)
reduce函数对list不断进行操作直到只剩一个结果，如对一个list使用加法相当与对list中的所有因素求和
比如list是(a b c d)，则先计算a b的结果，再用这个结果与c运算，得到的结果再与d运算，最后剩下一个结果
其中function必须是有两个参数的，eg,
(reduce #'+ '(1 2 3)) -> 6 (reduce #'* '(2 3 4)) -> 24
reduce可以将嵌套的list变为一个list，eg,
> (reduce #'append '((one un) (two deux) (three trois))) (one un two deux three trois)
ex 7.17
(defun total-length (x) (reduce #'+ (mapcar #'length x)))
7.10 every
(every #'predicate list)
如果list中的所有元素都使得predicate返回t则函数返回t，意思就是对于每一个list中的因素都成立,否则返回nil,eg,
> (every #'numberp '(1 2 3)) t
如果第二个参数是nil，则every返回t
every后面也可以给多个list,eg,
> (every #'> '(10 20 30 40) '(1 4 11 23)) t
>函数需要两个参数，every每次从后面的两个list中各取出一个作为参数，如例子中10>1,20>4,30>11,40>24都成立，所以返回t
debug tool: trace
- (trace func1 func2 ...)
- (trace) : 查看正在trace的函数
- (untrace func1 func2 ...)
- (untrace) : untrace所有函数
(defun half (n) (* n 0.5)) (defun average (x y) (+ (half x) (half y))) > (trace half average) (half average) > (average 3 7) 1. Trace: (AVERAGE '3 '7) 2. Trace: (HALF '3) 2. Trace: HALF ==> 1.5 2. Trace: (HALF '7) 2. Trace: HALF ==> 3.5 1. Trace: AVERAGE ==> 5.0 5.0
ex 7.29
(setf database
'((b1 shape brick)
(b1 color green)
(b1 size small)
(b1 supported-by b2)
(b1 supported-by b3)
(b2 shape brick)
(b2 color red)
(b2 size small)
(b2 supports b1)
(b2 left-of b3)
(b3 shape brick)
(b3 color red)
(b3 size small)
(b3 supports b1)
(b3 right-of b2)
(b4 shape pyramid)
(b4 color blue)
(b4 size large)
(b4 supported-by b5)
(b5 shape cube)
(b5 color green)
(b5 size large)
(b5 supports b4)
(b6 shape brick)
(b6 color purple)
(b6 size large)))
(defun match-element (x y)
(or (equal x y) (equal y '?)))
(defun match-triple (assertion pattern)
(every #'match-element assertion pattern))
(defun fetch (pattern)
(remove-if-not #'(lambda (entry) (match-triple entry pattern)) database))
> (fetch '(b4 shape ?))
> (fetch '(? shape brick))
> (fetch '(b2 ? b3))
> (fetch '(? color ?))
> (fetch '(b4 ? ?))
(defun ask-color (block)
(list block 'color '?))
(defun supporters (block)
(mapcar #'first (fetch (list '? 'supports block))))
(defun supp-cube (block)
(find-if #'(lambda (entry) (fetch (list entry 'shape 'cube))) (supporters block)))
(defun desc1 (block)
(remove-if-not #'(lambda (entry) (equal block (first entry))) database))
(defun desc2 (block)
(mapcar #'rest (desc1 block)))
(defun description (block)
(reduce #'append (desc2 block)))
(setf (cdr (last database)) '((b1 material wood) (b2 material plastic)))
7.11 operating on multiple lists
之前我们使用mapcar是将一个list的每个item作为参数调用一个函数，这个函数只接受一个参数。其实mapcar也可以对多参数的函数使用，eg,
> (mapcar #'(lambda (x y) (list x 'get y)) '(fred wilma) '(job1 job2)) ((fred gets job1) (wilma gets job2)) > (mapcar #'+ '(1 2 3) '(10 20 30 40)) (11 22 33)
这种方式类似于前面的every后面跟多个list，如果两个list的长度不一样，则遍历完较短那个就会停止
ex 7.30
(mapcar #'append words (mapcar #'list '(uno dos tres quatro cinco)))
7.12 function函数
'代表函数quote，#'则代表函数function,#'cons等价与(function cons)
> 'cons cons > #'cons #<compiled-function cons ...> > #'(lambda (x) (...)) #<lexical-closure ...>
#'返回的是一个函数对象，可以用funcall来进行调用
> (setf g #'(lambda (x) (* x 10))) #<lexical...> > (funcall g 12) 120
7.13 kwargs for applicative operators
:from-end等关键字参数也可以应用在find-if,remove-if,reduce这些函数
> (reduce #'cons '(a b c d e)) ((((a . b) . c) . d) . e) > (reduce #'cons '(a b c d e) :from-end t) (a b c d . e)