[Python-3000] could range be smarter?

[Vitor Bosshard]

Hello,

You're using range objects in ways that were hardly intended. If range objects had to be magically smart about everything, the language would be slower overall just to enable a questionable use case.

Plain comparison operators and isinstance or type checks against int will make your life a lot easier in this case:

>>> a = .1
>>> type(a) is int and 0 <= a < 20**30
False
 
 
Vitor




----- Mensaje original ----
> De: Owolabi Abdulkareem <adigunoble at gmail.com>
> Para: python-3000 at python.org
> Enviado: martes, 18 de noviembre, 2008 12:01:35
> Asunto: [Python-3000] could range be smarter?
> 
> Hello ,
> I'm not really a programmer but I am working to be one
> 
> I just thought may be this is worth considering for future enhancement
> with python's range
> 
> I've you considered trying out something of these nature with range:
> 
> />>> 9000000 in range(20**30)/
> 
> you'll have to wait for sometime to get a response but if you try
> something like this
> 
> />>> 900000.7 in range(20**30)
> 
> /or even
> 
> />>> 0.1 in range(20**30)/
> 
> range iterates through till  the end before it gives you a response. It
> is clear that a float could not be in the above range and /0.1/ is less
> than a unit integer, why do the function /range  /*have to iterate
> through till the end* wasting CPU resource when it is clear that it is
> not in the range.
> 
> To avoid this ,I have to /type check /and test if the number is an
> integer and not a float .
> 
> Failure to do this would lead to my program to freeze until the
> iteration is done. This makes range inefficient and one would have to
> remember this for it does not do the job of checking for you or/ and if
> a step is included in the range I have to test it with % operator:
> 
> Putting this in a function:
> /
> def Inrange(begin,end,step,number):
>     index=0 # Here is where the index of the number is placed if is in range
>     for i in begin,end,step,number:
>         if type(i) == type(int(i)): #Check to see all that parameters  
> are all integers
>             pass
>         else:
>             return 'All parameters must be integers'
>         try:                                                    #check
> to see if parameters are  appropriate to produce a true range
>             if (begin < end and step>0):
>                 assert(begin <= number < end)
>             elif begin > end and step < 0:
>                 assert(begin >= number > end)
>             else:
>                 raise AssertionError
> 
>         except AssertionError:
> 
>             return 'Check your parameters ;range between them is not
> executable'
>         if (number - begin)%step != 0:
>             return 'Your number is not in range'
>         else:
>             index=(number - begin) // step
> 
>             return (True,index)
> /
> (I'm sorry about my code; any correction will be educating)
> 
> Ruby's range does the comparing check but gives an erroneous 'True'  with
> 
> /(0 .. 20*30) include? 7.7482/
> 
> Where 7.7482 is a float ( and should not be found in a range of integers)
> 
> I think it would be nice (and more efficient for there won't be a need
> to iterate through for this) if we could do this:
> /
> >>> 7.7483 in range(20**30)
> Traceback (most recent call last):
>   File "", line 1, in 
>     7.7483 range(20**30)
> TypeError: 'float' object cannot be in a range of integers
> 
> >>> 622168 in range(20**30,8)
> (True, 77770)
> 
> >>> 1219 in range(-181,20**30,8)
> (True, 175)
> 
> >>> range(-181,20**30,8)[175]
> 1219
> 
> >>> list(range(-181,20**30,8)).index[1219]  # as in  index = (number -
> begin) // step
> 175/
> 
> In python 3.0 range(-181,20**30,8) gives:
> /
> Traceback (most recent call last):
>   File "", line 1, in 
>     list(range(20**20)).index(20)
> OverflowError: Python int too large to convert to C ssize_t/
> 
> I believe this simple mathematical expression in the function above 
> could be used to avoid iteration for this and
> go past this OverflowError (with some  initial test similar to that in
> the function 'Inrange' above):
> /
> number = index*step + begin
> index = (number - begin) // step #making index the subject of formul/a
> 
> I don't know if something could be done about this now ( or if it is
> really a good idea) especially when python 3000 is in its rc2 but I
> thought I should say something on this.
> 
> 
> Yours
> 
> Abdulkareem Owolabi
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