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Re: creating new variables

by Melissa Ives :: Rate this Message:

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One way would be to use the sort function so that the 209 record is last. (it may help to create a dichotomy for rep209=1 or 0 if not--something like this would be needed if there are >2 repairs in a day)

Then do a lag function to create morejobs and moremoney.

THE FOLLOWING ASSUMES that no client has >2 repairs in a day AND 209 is the highest repair value.

sort cases by id repdate reptype.

if (id=lag(id) and repdate=lag(repdate) and reptype=209) morejobs=1.

if ((id ne lag(id) or repdate ne lag(repdate)) and reptype=209) morejobs=0.

if (id=lag(id) and repdate=lag(repdate) and reptype=209) moremoney=lag(money).

The actual syntax will differ if there are more than one possible extra repairs on the same date (i.e. one id could have 3+ per day), but similar logic should work.

HTH,

Melissa

From: SPSSX(r) Discussion [mailto:SPSSX-L@...] On Behalf Of abdelrhman elmubarak
Sent: Thursday, May 07, 2009 3:36 AM
To: SPSSX-L@...
Subject: [SPSSX-L] creating new variables

my data looks as below

id	Repair Date	Repair Type	money
16	20090430	115	23
30	20090325	209	45
30	20090124	103	62
47	20090409	209	78
47	20090409	101	69

I want to create two more variables for those who did repair type 209…the purpose is to know did they did more jobs on the same day of their visit and what the money generated from that ….at the final my data should be look as below

id	Repair Date	Repair Type	money	more jobs plus	more money plus
16	20090430	115	23
30	20090325	209	45	0
30	20090124	103	62
47	20090409	209	78	1	69
47	20090409	101	69

I can explain more if it is required

Thanks in advance

Abdulrahman

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