[Tutor] Workaround for limitation in xrange()?

[Dick Moores]

>
>> >>> for x in xrange(2**31):
>>     pass
>>
>>Traceback (most recent call last):
>>   File "<pyshell#16>", line 1, in <module>
>>     for x in xrange(2**31):
>>OverflowError: long int too large to convert to int
>

Here are the suggestions I've received:

Danny's
>#################
>def myxrange(m, n=None, skip=1):
>     """An xrange-like function that can deal with long ints.
>     Note: doesn't deal with negative ranges particularly well."""
>     if n is None:
>        m, n = 0, m
>     i = m
>     while i < n:
>         yield i
>         i = i + skip
>#################

Kent's
Encapsulate the while loop in a generator:
def count(limit):
   n=0
   while n<limit:
     yield n
     n += 1

Andrei's
Write your own iterator:
 >>> def hugerange(minval, maxval):
...     val = minval
...     while val < maxval:
...         yield val
...         val += 1

All 3 are essentially the same, aren't they. Which makes me feel even 
dumber, because I don't understand any of them. I've consulted 3 
books, and still don't understand the use of yield.

Let me show you what I was doing when I ran into the 2**31-1 
limitation: <http://www.rcblue.com/Python/1histogram-5a-ForWeb.txt>

This is one of my attempts to test the degree of uniformity of 
random(). The others are seeing how close the average random() return 
comes to 0.5, and using random() in an approximation of pi.

I've realized now that all 3 scripts require an immense number of 
calls to random(), so many that I wouldn't be able to use my computer 
for anything else for weeks, maybe months. But be that as it may, I'd 
still like to know how to use yield in my 1histogram-5a.

For your convenience, here's the list h, and the while loop I used:

h = 100*[0]

c = 0
while c < k:
     r = random()
     i = int(math.floor(100*r))
     h[i] += 1
     c += 1

So, for a random() return of 0.27862506717494118, i would be 27, and 
h[27] would be incremented by 1.
Thanks,

Dick