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- How to use Dif/Lag function by Column group

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Jul 29, 2016 12:31 PM
(4470 views)

Hi JMP community folks,

I have a data table with four columns as Box, iClass, BinPos, and Slot. iClass means 24 level of products, BinPos means the Box location in the sorting equipment and Slot indicated which slot position the product be placed in the box. The Box is designed to hold a max quantity of 100 product, after that a new empty box would be placed in. In order to check the product is be placed in the right box and the Slot increment is exactly by "1" (because sometime I do manually observed the slot increment increase >1 or increment=0, which meant the placing machine having issues),I tried to use the dif/lag function to check the increment. However, the dif/lag formula is not allowing me the add a function to group by column of Box. The Box is actually the unique identifier.

A formula like Dif(:Slot,1) can tell me the difference b/w a certain with respect to the row before it. However, the rows are mixed by with different Box because of difference level of product belong to iClass column.

I am hoping to have a formula like Dif(:Slot, 1, :Box) to work out. Can any expert provide some clues to work this out? A example table is attached for investigation.

Many Thanks in advance

Box | iClass | BinPos | Slot |

72813_2 | 11 | 11 | 64 |

73213_2 | 15 | 15 | 2 |

73213_2 | 15 | 15 | 3 |

73213_2 | 15 | 15 | 4 |

73213_2 | 15 | 15 | 5 |

73213_2 | 15 | 15 | 6 |

73213_2 | 15 | 15 | 7 |

73213_2 | 15 | 15 | 8 |

73214_2 | 14 | 14 | 1 |

73214_2 | 14 | 14 | 2 |

73214_2 | 14 | 14 | 3 |

73214_2 | 14 | 14 | 4 |

73214_2 | 14 | 14 | 5 |

73214_2 | 14 | 14 | 6 |

73213_2 | 15 | 15 | 9 |

73214_2 | 14 | 14 | 7 |

73213_2 | 15 | 15 | 10 |

73213_2 | 15 | 15 | 11 |

73213_2 | 15 | 15 | 12 |

73213_2 | 15 | 15 | 13 |

73213_2 | 15 | 15 | 14 |

73213_2 | 15 | 15 | 15 |

73214_2 | 14 | 14 | 8 |

Solved! Go to Solution.

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Jul 29, 2016 5:32 PM
(8293 views)

Solution

The example is a formula for a new table, not a JSL standalone script. Create a new column, and paste the formula into the formula for the column, and it will generate the incremental values you want.

Jim

11 REPLIES

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Jul 29, 2016 1:35 PM
(4310 views)

Can you sort by :Box and then apply the formula after checking that :Box==Lag(:Box,1)?

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Jul 29, 2016 4:08 PM
(4310 views)

If the absolute order of the original data is not important, sorting the data and using the Dif function will work. However, here is a formula that you can use that will calculate the increment without having to sort the data.

i = Row**()**;

allrows = Current Data Table**()** << **get rows where(** :Box == :Box**[**i**]** **)**;

If**(** Loc**(** As List**(** allrows **)**, i **)[****1]** == **1**,

**1**,

slot**[**allrows**[**Loc**(** As List**(** allrows **)**, i **)[****1]]]** - slot**[**allrows**[(**Loc**(** As List**(** allrows **)**, i **)[****1])**-**1]]**

**)**;

Jim

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Jul 29, 2016 4:28 PM
(4310 views)

Hello Jim,

I tried to run the script you made (just copy and pasted the entire script), however, I cannot see anything change in the data table and no New Column to generate.

I tried to make a new column with your script but nothing work either.

This is what the script I tried to run,

i = Row();

allrows = Current Data Table() << get rows where( :Box_ID == :Box_ID* );*

allrows<< New Column("Test", Numeric, Continuous, Formula(If( Loc( As List( allrows ), i )[1] == 1,

1,

slot[allrows[Loc( As List( allrows ), i )[1]]] - slot[allrows[(Loc( As List( allrows ), i )[1])-1]]

)));

Regards,

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Jul 29, 2016 5:32 PM
(8294 views)

The example is a formula for a new table, not a JSL standalone script. Create a new column, and paste the formula into the formula for the column, and it will generate the incremental values you want.

Jim

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Jul 29, 2016 8:47 PM
(4310 views)

Hi Jim,

Thanks for the help. Actually, I have more than 60,000+ rows in a single daily table if I run a formula you provide it would take a fairly long time.

Would it be quicker if I sort the data first then run the formula?

Btw, is there a way for jmp a count the time it run for a script or formula?

Regards,

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Jul 30, 2016 6:34 AM
(4310 views)

With that many rows, using the sort method would be faster. Below is the method that I use to find the time a script takes:

start = today();

the script or part of a script you want to find the timings on

show( today() - start );

Jim

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Aug 1, 2016 7:42 AM
(4310 views)

If you need more accuracy in the timing, @msharp noted in a recent post:

*Today() is only accurate to the second, Tick Seconds() is accurate to 1/60 of a second, and HP TIME() is accurate to the microsecond.*

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Jun 11, 2018 8:35 PM
(1117 views)

Here is a rework of the previous formula. This formula works with JMP 14

```
If( Row() == 1,
dt = Current Data Table()
);
i = Row();
holdBox = :Box;
allrows = dt << get rows where( holdBox == :Box[i] );
If( Loc( As List( allrows ), i )[1] == 1,
1,
:Slot[allrows[Loc( As List( allrows ), i )[1]]] - :Slot[allrows[Loc( As List( allrows ), i )[1] - 1]]
);
```

Jim

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Jun 11, 2018 4:40 PM
(1132 views)

Jim,

This is a very valuable workaround since it provides a functionality of a non-existen formula ColLag(); :)

However, I see it's been quite a while since you posted this workaround. JMP 14 nowadays doesn't like how it's written. It says Get Rows Where is an unknown function and it says that using Current Data Table() in formulas is not stable.

Is there any newer version of this workaround?

Thanks,

M.