对嵌套列表降序排序



我有此嵌套列表:

Xs =  [ [Joe, Pilot, 100], [Stan, Co-Pilot, 300], [Steve, Pilot, 150],  ].

我如何对此嵌套列表进行排序以执行以下操作:(以降序使用第三个元素)

Xs = [ [Stan, Co-Pilot, 300], [Steve, Pilot, 150], [Joe, Pilot, 100] ]. ?

定义:

criteria(R,[_,_,N1],[_,_,N2]) :- compare(R,N2,N1).

并使用" predsort/3"喜欢:

?- predsort(criteria,[ [Joe, Pilot, 100], [Stan, Co-Pilot, 300], [Steve, Pilot, 150]  ], Xs).
Xs = [[Stan, Co-Pilot, 300], [Steve, Pilot, 150], [Joe, Pilot, 100]].

如果存在重复的第三个要素,则必须更改"标准"。例如:

criteria(R,[_,_,N1],[_,_,N2]) :- N1==N2, !, compare(R,N2,N1).
criteria(R,E1,E2) :- compare(R,E1,E2).

如果将数据表示为结构(无论如何是一个好主意),并且您的Prolog System具有sort/4(例如Eclipse,SWI),则可以简单地做:

?- Xs = [emp(joe, pilot, 100), emp(stan, copilot, 300), emp(steve, pilot, 150)],
   sort(3, >=, Xs, Ys).
Xs = [emp(joe, pilot, 100), emp(stan, copilot, 300), emp(steve, pilot, 150)]
Ys = [emp(stan, copilot, 300), emp(steve, pilot, 150), emp(joe, pilot, 100)]
Yes (0.00s cpu)

一个更便携的(ISO)方法是将每个列表元素与排序键配对,然后使用Keysort/2,然后再次剥离键:

?- Xs = [emp(joe, pilot, 100), emp(stan, copilot, 300), emp(steve, pilot, 150)],
   add_keys(Xs, KXs),
   keysort(KXs, KYs),
   strip_keys(KYs, Ys).
Xs = [emp(joe, pilot, 100), emp(stan, copilot, 300), emp(steve, pilot, 150)]
KXs = [-100 - emp(joe, pilot, 100), -300 - emp(stan, copilot, 300), -150 - emp(steve, pilot, 150)]
KYs = [-300 - emp(stan, copilot, 300), -150 - emp(steve, pilot, 150), -100 - emp(joe, pilot, 100)]
Ys = [emp(stan, copilot, 300), emp(steve, pilot, 150), emp(joe, pilot, 100)]
Yes (0.00s cpu)

使用辅助谓词

add_keys([], []).
add_keys([Emp|Emps], [Key-Emp|SEmps]) :-
    Emp = emp(_Name,_Job,Salary),
    Key is -Salary,
    add_keys(Emps, SEmps).
strip_keys([], []).
strip_keys([_-V|KVs], [V|Vs]) :-
    strip_keys(KVs, Vs).

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