[Tutor] A faster x in S

[bob gailer]

Dinesh B Vadhia wrote:
> For some significant data pre-processing we have to perform the 
> following simple process:
>  
> Is the integer x in a list of 13K sorted integers.  That's it except 
> this has to be done >100m times with different x's (multiple times).  
> Yep, a real pain! 
>  
> I've put the 13K integers in a list S and am using the is 'x in S' 
> function.
>  
> I was wondering if there is anything faster?
I agree with Kent.

 >>> l = range(13000)
 >>> s=set(l)
 >>> d=dict(enumerate(l))
 >>> import time
 >>> def f(lookupVal, times, values):
...     st=time.time()
...     for i in range(times):
...         z = lookupVal in values
...     return time.time()-st   
 >>> f(6499,1000,l)
0.31299996376037598
 >>> f(6499,1000000,s)
0.31200003623962402

So set is 1000 times faster than list!

 >>> f(6499,1000000,d)
0.31300020217895508

And dict is (as expected) about the same as set.

So 100,000,000 lookups should take about 30 seconds. Not bad, eh?

Let's explore another angle. What range are the integers in (min and max)?

Bob