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Oct 7, 2014 6:35 AM
(18814 views)

Hello JMP Community,

I am trying to create a column that indexes rows by an ID column.

In other scripting language, this is accomplished with the following lines.

By ID: gen IDIndex = _n

By ID: gen IDCount = _N

I can do the following, but I index and count the whole table

```
New Column( "IDIndex", formula( Row() ) );
New Column( "IDCount", formula( N Row() ) );
```

Advice on this would be great!

Very Sincerely,

ID |
IDIndex |
IDCount |
---|---|---|

444 | 1 | 4 |

444 | 2 | 4 |

444 | 3 | 4 |

444 | 4 | 4 |

333 | 1 | 3 |

333 | 2 | 3 |

333 | 3 | 3 |

1 ACCEPTED SOLUTION

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In your formula you need to check the ID value to see if it has changed from the previous row and reset your index if it has.

Try this formula:

If(:ID != Lag(:ID, 1) | Row() == 1, 1, Lag(:IDIndex, 1) + 1)

Your ID Count column has use IDIndex to get its value:

Col Maximum( :IDIndex, :ID )

The second argument to the Col Maximum function is the By variable.

I hope that helps!

-Jeff

-Jeff

5 REPLIES 5

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In your formula you need to check the ID value to see if it has changed from the previous row and reset your index if it has.

Try this formula:

If(:ID != Lag(:ID, 1) | Row() == 1, 1, Lag(:IDIndex, 1) + 1)

Your ID Count column has use IDIndex to get its value:

Col Maximum( :IDIndex, :ID )

The second argument to the Col Maximum function is the By variable.

I hope that helps!

-Jeff

-Jeff

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Re: Syntax: How do I create an index column by ID

Great, that worked perfectly!

Thanks for quick reply.

Have a great day,

Matt

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Re: Syntax: How do I create an index column by ID

Jeff is correct and i will only add that the formula for IDIndex depends on previous sorting.

If the table needs to be re-sorted later you would need to suppress evaluation or delete it.

column (dt, "IDIndex") << suppress eval ( true );

or:

column (dt, "IDIndex") << delete formula;

to produce the column IDCount without sorting you can use the table summery option:

dt << Summary (Group ( :ID );

ron

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Re: Syntax: How do I create an index column by ID

I think the below formula will work for IDIndex independently of the initial sorting of the ID column:

Sum( :ID[Index( 1, Row() )] == :ID )

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Re: Syntax: How do I create an index column by ID

Thanks MS, the command you shared does work independently of the sorting.

Have a great day!