你好，
我认为您需要在最后一个 New Column 语句中指定矩阵（见下文）。 我无法测试它，因为您的脚本还有其他错误。
nrun=10;
nTraining = 0.75;
nVal = 0.25;
mat = J(1,3);
mat [1,1] = nTraining;
mat [1,2] = nVal;
mat [1,3] = 0;
//getting this error
//Invalid matrix token.
//Line 232 Column 38: ...ake Validation Formula( [►nTraining, nVal, 0], <<ST... for=""></ST...>
很抱歉 JSL 脚本中的拼写错误。 这是正确的语法。
nrun=10;
nTraining = 0.75;
nVal = 0.25;
mat = J(1,3);
mat [1,1] = nTraining;
mat [1,2] = nVal;
mat [1,3] = 0;
//getting this error
//Invalid matrix token.
//Line 232 Column 38: ...ake Validation Formula( [►nTraining, nVal, 0], <<ST... for=""></ST...>
再次抱歉，我试图回答得太快了。 我没有合适的 JMP 版本来测试这个问题，但我想我解决了这个问题。
nrun=10;
nTraining = 0.75;
nVal = 0.25;
mat = J(1,3);
mat [1,1] = nTraining;
mat [1,2] = nVal;
mat [1,3] = 0;
//getting this error
//Invalid matrix token.
//Line 232 Column 38: ...ake Validation Formula( [►nTraining, nVal, 0], <<ST... for=""></ST...>
你好，
为了探索所有途径，您是否尝试过生成转置矩阵？
nTraining = 0.75;
nVal = 0.25;
mat = J(3,1);
show (mat);
mat [1,1] = nTraining;
mat [2,1] = nVal;
mat [3,1] = 0;
show (mat);
//getting this error
//Invalid matrix token.
//Line 232 Column 38: ...ake Validation Formula( [►nTraining, nVal, 0], <<ST... for=""></ST...>
我无法判断该公式是否会被接受，因为我没有可用的 JMP Pro，但至少，它会在公式中生成一个以逗号分隔的矩阵。
很抱歉没有提供更大的帮助。
最好的，
TS
嗨蒂埃里，
你的建议实际上非常接近。 您建议在需要 3x1 矩阵时构建 1x3 矩阵
dtstub= data table("test");
nrun=3;
nTraining = 0.75;
nVal = 0.25;
// Create the matrix
mat = J(3,1, {.}); // 1x3 matrix with missing values
mat [1,1] = nTraining;
mat [2,1] = nVal;
mat [3,1] = 0;
show (mat);
For( i = 1, i <= nrun, i++,
data table("test") <<
New Column( "Val",
Numeric,
"Nominal",
Format( "Best", 12 ),
Formula(
EVal (Eval Expr (Make Validation Formula(
expr(mat),
// [0.75, 0.25, 0],
<<STRATIFICATION columns=""></STRATIFICATION>在评估 ML 模型性能的过程中，我尝试创建一个 UI，用于指定模型将执行的运行次数以及指定训练和验证数据的比例。
我试图将该信息传递到一个 for 循环中，该循环将根据指定的运行次数创建列，并传递比例。 当我指定比例（第二个示例）时，循环工作正常，但在使用变量（nTraining、nVal）时则不然。
一些建议将不胜感激。
谢谢。
小号
dtsub=New Table( "test",
Add Rows( 34 ),
New Column( "test_article",
Character( 7 ),
"Nominal",
Set Values(
{"1269442", "1100729", "1269449", "1269448", "1100723", "1100728",
"1269456", "1269457", "1269458", "1269454", "1269468", "1269469",
"1269466", "1269471", "1100397", "1100405", "1100406", "1100407",
"1100528", "1100534", "1100535", "1100540", "1100542", "1100566",
"1100585", "1100590", "1100725", "1100873", "1100914", "1100915",
"1100920", "1100990", "1100991", "1101101"}
),
Set Display Width( 61 )
),
New Column( "basic_class",
Character,
"Nominal",
Set Values(
{"0_Active", "0_Active", "0_Active", "0_Active", "0_Active", "0_Active",
"0_Active", "0_Active", "0_Active", "0_Active", "0_Active", "0_Active",
"1_MarginalActive", "1_MarginalActive", "1_MarginalActive",
"1_MarginalActive", "1_MarginalActive", "1_MarginalActive", "2_Inactive",
"1_MarginalActive", "2_Inactive", "1_MarginalActive", "2_Inactive",
"1_MarginalActive", "1_MarginalActive", "1_MarginalActive", "0_Active",
"1_MarginalActive", "0_Active", "0_Active", "2_Inactive", "0_Active",
"0_Active", "0_Active"}
),
Set Display Width( 90 )
),
Set Row States(
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1,
1, 0, 1, 0, 0, 0, 0, 0, 0]
)
);
nrun=10;
//the proportion forthe valdation column are specified here
nTraining = 0.75;
nVal = 0.25;
//getting this error
//Invalid matrix token.
//Line 232 Column 38: ...ake Validation Formula( [►nTraining, nVal, 0], <<ST... for=""></ST...>