
From  Nikolay Markov <nmarkov@ibn.bg> 
To  statalist@hsphsun2.harvard.edu 
Subject  Re: st: a question about number precision 
Date  Mon, 03 Apr 2006 12:37:35 +0300 
Nick Cox wrote:
Phil gave an excellent answer, and I just want to add one more detail.
If all you want to do is to extract the last three digits of a numeric ID, you can use
mod(ID, 1000)
You may have been taught about the modulus function under the name of remainder (or the equivalent in your first language).
mod(21557127, 1000)
is the remainder (what is left over) after dividing 21557127 by 1000, namely 127.
Nick n.j.cox@durham.ac.uk
Phil Schumm replied to Jian Zhang
I have a problem about number precision. I cann't figure out what it happened. Hope that you can help me out. Thanks.
Here is the data:
ID
21557127
then i run the following do file trying to extract the last three digits from the ID:
gen double temxxx=(ID/1000)
gen temyyy=int(temxxx)
gen temzzz=temxxxtemyyy
gen areaxxx=(temzzz*1000)
drop temxxx temyyy temzzz
the generated data looks like the following:
ID areaxxx
21557127 127
However, when I typed: list if areaxxx==127, stata in fact listed nothing!
First I thought it may be because areaxxx is a floatingpoint variable, so I type: list if areaxxx=float(127). However, Stata listed nothing again.
First, let me say that if all you want to do is to extract the last three digits of the ID, here is the way to do it:
. di real(substr(string(ID,"%12.0g"),3,.))
127
Note that if you just use string(ID) this will not work, as string() uses a default format which is not wide enough for your ID (%12.0g is the default format for the long storage type, which I presume is how your ID variable is stored).
Second, this is exactly the reason why you should not store IDs as numbers  you should store them as strings instead. For example, if ID were a string variable, then extracting the last three digits would be even simpler:
. di substr(ID,3,.)
127
and would be guaranteed to work no matter how long your IDs are (provided they are no longer than 244 characters).
Finally, what happened above? The problem was indeed due to the error inherent in floatingpoint arithmetic. For example, here is the calculation you performed:
. di %24.18f float( 1000 * float( (21557127/1000)  float( int (21557127/1000) ) ) )
127.000007629394531250
which, as you can see is not equal to 127. Let's take a closer look:
float( 1000 * float( (21557127/1000)  float( int (21557127/1000) ) ) )
 temxxx   temxxx 
 temyyy 
 temzzz 
 areaxxx 
Notice how I am using the float() function to mimic the fact that, although you created temxxx as a double, you did not do so for the other intermediate variables. Now in this case, had you also created temzzz as a double, you would have gotten what you wanted:
. assert float( 1000 * ( (21557127/1000)  float( int (21557127/1000) ) ) ) == 127
However, as I said above, it is nearly always better to store IDs such as these as string variables.
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