Re: st: a question about number precision

[Nikolay Markov <nmarkov@ibn.bg>]

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=temxxx-temyyy
> > > 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 floating-point  
> > > 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 floating-point 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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