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BabyDoragon
Level II

How to distinguish whether there are empty values in a column?

I want to distinguish whether the cells in the column are empty using "is missing". In the JSL below, I hope to differentiate between the "name" and "Apple" fields, but the results are all 1. How should I modify it?

 

 dtTest = Open( "$SAMPLE_DATA/Big Class.jmp" );
 show(IsMissing(:age));
 dtTest << New Column( "Apple", "Character", "Nominal" );
 show(IsMissing(:Apple));
1 ACCEPTED SOLUTION

Accepted Solutions
jthi
Super User

Re: How to distinguish whether there are empty values in a column?

Check for 0 not for 1

Names Default To Here(1);

dt = Open("$SAMPLE_DATA/Big Class.jmp");

dt << New Column("Apple", "Character", "Nominal");
dt << New Column("Orange", Numeric, Continuous);


Show(Col Number(Column(dt, "age")) == 0);
Show(Col Number(Column(dt, "Apple")) == 0);
Show(Col Number(Column(dt, "Orange")) == 0);
-Jarmo

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3 REPLIES 3
jthi
Super User

Re: How to distinguish whether there are empty values in a column?

Do you wish to check if all values are missing? You can for example use Col Number and check if it is 0

Names Default To Here(1);

dt = Open("$SAMPLE_DATA/Big Class.jmp");

dt << New Column("Apple", "Character", "Nominal");
dt << New Column("Orange", Numeric, Continuous);


Show(Col Number(Column(dt, "age")) == 1);
Show(Col Number(Column(dt, "Apple")) == 1);
Show(Col Number(Column(dt, "Orange")) == 1);
-Jarmo
BabyDoragon
Level II

Re: How to distinguish whether there are empty values in a column?

I confirmed that all three results are displayed as 0, but in reality, the age field has a value. This makes it impossible to distinguish between the Apple and Orange fields.
jthi
Super User

Re: How to distinguish whether there are empty values in a column?

Check for 0 not for 1

Names Default To Here(1);

dt = Open("$SAMPLE_DATA/Big Class.jmp");

dt << New Column("Apple", "Character", "Nominal");
dt << New Column("Orange", Numeric, Continuous);


Show(Col Number(Column(dt, "age")) == 0);
Show(Col Number(Column(dt, "Apple")) == 0);
Show(Col Number(Column(dt, "Orange")) == 0);
-Jarmo

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