在OPL中查找下一组编号 [英] Find next number for sets in OPL
问题描述
我有2套如下
jks - {<1 1> <1 2> <1 3> <2 1> <2 2> <2 3> <3 1> <3 2> <3 3>}
benchPerjk - [{1 3} {1 3} {1 3} {1 3} {1 3} {1 3} {1 2 3} {1 2 3} {1 2 3}]
我现在想在BenchPerjk中查找某个jks的下一个数字
I now want to find the next number in the benchPerjk for a certain jks
代码如下
tuple jk {
int j;
int k;
}
{jk} jks = { <t.j, t.k> | t in PitBlocksType };
{int} BenchPerjk[v in jks] = { t.i | t in PitBlocksType : t.j == v.j && t.k == v.k };
其中PitBlocksType来自以下数据：
Where the PitBlocksType come from the following data :
tuple blockType {
string id;
int i;
int j;
int k;
};
{blockType} PitBlocksType = ...; // Read from excel table which contains several rows, a short example below
/*
Example below
Block Id (i) (j) (k)
P1 1 1 1
P2 1 1 2
P3 1 1 3
P7 3 1 1
P8 3 1 2
P9 3 1 3
P10 1 2 1
P11 1 2 2
P12 1 2 3
P16 3 2 1
P17 3 2 2
P18 3 2 3
P19 1 3 1
P20 1 3 2
P21 1 3 3
P22 2 3 1
P23 2 3 2
P24 2 3 3
P25 3 3 1
P26 3 3 2
P27 3 3 3
*/
我之所以这样做，是因为我想使用BenchPerjk查找下一个i。用于在下面的代码中代替b.i + 1
The reason why I am lookng for this is I want to use the BenchPerjk to find the next i. for using in the code below in place of the b.i+1
如果i是连续的并且所有j存在于所有i中，则b.i + 1可以很好地工作， k。这意味着Benchperjk是否会
The b.i+1 works well if the i is continuous and all i is present for any set of j,k. Which means if Benchperjk would be
benchPerjk-[{1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3}]
benchPerjk - [{1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3} {1 2 3}]
但这不是我的数据。
{blockType} OntopPit[b1 in PitBlocksType] =
{b | b in PitBlocksType: b1.i == b.i +1 &&
((b1.k == b.k-1 ) ||
(b1.k == b.k+1 ) ||
(b1.k == b.k ) ) &&
((b1.j == b.j-1 ) ||
(b1.j == b.j+1 ) ||
(b1.j == b.j ) ) };
建议。
我尝试了以下代码，但是它附带了错误：
I have tried the following code but it comes up with errors :
{blockType} OntopPit[b1 in PitBlocksType] =
{b | b in PitBlocksType: b1.i == next(BenchPerjk[< b.j , b.k>],b.i) &&
((b1.k == b.k-1 ) ||
(b1.k == b.k+1 ) ||
(b1.k == b.k ) ) &&
((b1.j == b.j-1 ) ||
(b1.j == b.j+1 ) ||
(b1.j == b.j ) ) };
，但出现了两行错误
b.i ：next（）元素不存在。
数组项的无效初始化表达式：OntopPit [< P1，1,1,1>]
but came up with 2 lines of error "b.i" : next() element does not exist. Invalid initialization expression for array item: OntopPit[<"P1",1,1,1>]
推荐答案
您可以使用下一个：
tuple jk {
int j;
int k;
}
{jk} jks = {<1 ,1> ,<1 ,2>, <1 ,3>, <2, 1> ,<2 ,2> ,<2, 3>, <3, 1> ,<3, 2>, <3, 3>};
{int} BenchPerjk[jks] = [{1 ,3}, {1, 3}, {1, 3} ,{1 ,3}, {1, 3}, {1, 3}, {1 ,2 ,3} ,{1 ,2, 3} ,{1 ,2, 3}];
// Now suppose you want to get the next after 1 for <j,k> = <2,1>
int res=next(BenchPerjk[<2,1>],1);
execute
{
writeln(res);
}
提供3
后来需要一些通用的东西：
Something more generic was later required:
tuple jk {
int j;
int k;
}
{jk} jks = {<1 ,1> ,<1 ,2>, <1 ,3>, <2, 1> ,<2 ,2> ,<2, 3>, <3, 1> ,<3, 2>, <3, 3>};
{int} BenchPerjk[jks] = [{1 ,3}, {1, 3}, {1, 3} ,{1 ,3}, {1, 3}, {1, 3}, {1 ,2 ,3} ,{1 ,2, 3} ,{1 ,2, 3}];
// Now suppose you want to get the next after 1 for <j,k> = <2,1>
int res=next(BenchPerjk[<2,1>],1);
execute
{
writeln(res);
}
// And now as a generic function
{int} benches=union (jk in jks) BenchPerjk[jk];
int nextone[jk in jks][i in benches]=(i in BenchPerjk[jk])?((i!=last(BenchPerjk[jk]))?next(BenchPerjk[jk],i):-2):-1;
int res2=nextone[<2,1>][1];
execute
{
writeln(res2);
}
给予
3
3
3 3
这篇关于在OPL中查找下一组编号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IT屋！
