list 计算一个数字在列表中出现的次数
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Count the number of occurrences of a number in a list
提问by Ratzo
I'm writing a program in prolog that count the number of occurrences of a number in a list
我正在用 prolog 编写一个程序，它计算列表中某个数字的出现次数
count([],X,0).
count([X|T],X,Y):- count(T,X,Z), Y is 1+Z.
count([_|T],X,Z):- count(T,X,Z).
and this is the output
这是输出
?- count([2,23,3,45,23,44,-20],X,Y).
X = 2,
Y = 1 ;
X = 23,
Y = 2 ;
X = 23,
Y = 1 ;
X = 3,
Y = 1 ;
X = 45,
Y = 1 ;
X = 23,
Y = 1 ;
X = 44,
Y = 1 ;
X = -20,
Y = 1 ;
false.
it's count the same number several times
它多次计算相同的数字
Any help is appreciated
任何帮助表示赞赏
回答by sumx
Instead of the dummy variable _ just use another variable X1 and ensure it does not unify with X.
而不是虚拟变量 _ 只需使用另一个变量 X1 并确保它不与 X 统一。
count([],X,0).
count([X|T],X,Y):- count(T,X,Z), Y is 1+Z.
count([X1|T],X,Z):- X1\=X,count(T,X,Z).
However note that the second argument X is supposed to be instantiated. So e.g. count([2,23,3,45,23,44,-20],23,C) will unify C with 2. If you want the count for every element use
但是请注意，第二个参数 X 应该被实例化。因此，例如 count([2,23,3,45,23,44,-20],23,C) 会将 C 与 2 统一。如果您想要每个元素的计数，请使用
:- use_module(library(lists)).
count([],X,0).
count([X|T],X,Y):- count(T,X,Z), Y is 1+Z.
count([X1|T],X,Z):- X1\=X,count(T,X,Z).
countall(List,X,C) :-
sort(List,List1),
member(X,List1),
count(List,X,C).
Then you get
然后你得到
?- countall([2,23,3,45,23,44,-20],X,Y).
X = -20,
Y = 1 ? ;
X = 2,
Y = 1 ? ;
X = 3,
Y = 1 ? ;
X = 23,
Y = 2 ? ;
X = 44,
Y = 1 ? ;
X = 45,
Y = 1 ? ;
no
回答by Bj?rn Lindqvist
You can also use the includepredicate:
您还可以使用include谓词：
count(L, E, N) :-
include(=(E), L, L2), length(L2, N).
回答by Andrei
ocr(X,[],0):- !.
ocr(X,[Element|Rs],V):- X = Element -> ocr(X,Rs,Ocr), V is 1+ Ocr; ocr(X,Rs,V).
I did it like that. That gives you only one answer and finishes.
我就是这样做的。那只给你一个答案并完成。
