问题描述
问题是要找到列表中的最后一个字符,例如一个字符.
The question is to find the last but one character in a list, e.g.
?- last_but_one(X, [a,b,c,d]).
X = c.
我的代码是:
last_but_one(X, [X,_]).
last_but_one(X, [_|T]) :- last_but_one(X, T).
他们提供的代码是:
last_but_one(X, [X,_]).
last_but_one(X, [_,Y|Ys]) :- last_but_one(X, [Y|Ys]).
当我学习Haskell时,我可以回想起,当问题在某个列表中问到第二,第三或第n个字符时,其结构与所提供的答案相同,所以我知道用他们写的方式写已经写了它有一定的意义.但是我似乎仍然能以自己的方式得到正确的答案.
When I was studying Haskell, I can recall that when questions asked for the 2nd, 3rd, or nth character in some list, the structure was the same as the answer that's been supplied, so I know writing it the way they've written it has some significance. But I still seem to get correct answers with the way I have written it.
我写错了吗?做出答案的人写的代码是否更好?—如果这样,怎么做?
Is the way I have written it wrong? Is the code that the guys who made the answer wrote better—and if so, how?
推荐答案
您的原始版本更容易阅读.特别是,递归规则读取-从右到左读取
Your original version is much simpler to read. In particular, the recursive rule reads - reading it right-to-left
last_but_one(X, [_|T]) :- last_but_one(X, T).
^^^^^^^^^^
provided X is the lbo-element in T
^^ then, it follows, that (that's an arrow!)
^^^^^^^^^^^^^^^^^^^^^^
X is also the lbo-element of T with one more element
换句话说:如果给定列表T
中已经有一个lbo元素,那么您可以构建新列表,其中前面还有其他任何元素也具有相同的lbo元素.
In other words: If you have already an lbo-element in a given list T
, then you can construct new lists with any further elements in front that also have the very same lbo-element.
人们可能会争论哪种版本在效率上更可取.如果您真的很喜欢,那就请:
One might debate which version is preferable as to efficiency. If you are really into that, rather take:
last_but_one_f1(E, Es) :-
Es = [_,_|Xs],
xs_es_lbo(Xs, Es, E).
xs_es_lbo([], [E|_], E).
xs_es_lbo([_|Xs], [_|Es], E) :-
xs_es_lbo(Xs, Es, E).
甚至:
last_but_one_f2(E, [F,G|Es]) :-
es_f_g(Es, F, G, E).
es_f_g([], E, _, E).
es_f_g([G|Es], _, F, E) :-
es_f_g(Es, F, G, E).
永远不要忘记常规测试:
Never forget general testing:
| ?- last_but_one(X, Es).
Es = [X,_A] ? ;
Es = [_A,X,_B] ? ;
Es = [_A,_B,X,_C] ? ;
Es = [_A,_B,_C,X,_D] ? ;
Es = [_A,_B,_C,_D,X,_E] ? ;
Es = [_A,_B,_C,_D,_E,X,_F] ? ...
这是我以前的Labtop上的一些基准测试:
And here are some benchmarks on my olde labtop:
SICStus SWI
4.3.2 7.3.20-1
--------------+----------+--------
you 0.850s | 3.616s | 4.25×
they 0.900s | 16.481s | 18.31×
f1 0.160s | 1.625s | 10.16×
f2 0.090s | 1.449s | 16.10×
mat 0.880s | 4.390s | 4.99×
dcg 3.670s | 7.896s | 2.15×
dcgx 1.000s | 7.885s | 7.89×
ap 1.200s | 4.669s | 3.89×
产生较大差异的原因是f1
和f2
都是纯确定性的,而没有任何选择点的创建.
The reason for the big difference is that both f1
and f2
run purely determinate without any creation of a choicepoint.
使用
bench_last :-
\+ ( length(Ls, 10000000),
member(M, [you,they,f1,f2,mat,dcg,dcgx,ap]), write(M), write(' '),
atom_concat(last_but_one_,M,P), \+ time(call(P,L,Ls))
).
这篇关于列表中的Prolog递归,最后一个元素的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!