Iterators and generators.

Dev Shed Forums Sponsor:

January 8th, 2013, 12:01 PM

Nightmareix35

Contributing User

Join Date: Nov 2012

Posts: 32

Time spent in forums: 8 h 42 m 41 sec
Reputation Power: 1

Iterators and generators.

I have this generator:

PHP Code:


		
			
 def takeOnly(it,pred,f):

    count = 0

    for i in it:

        if not count == f:

            if pred(i):

                count = 0

                yield i

            else:

                count += 1

        else:

            break

it receives an iterator, a predicate and a limit "f". The generator will yield all the items that meet the requirement of the predicate. As soon as "f" consecutive items in the iterator don't meet the requirement, the operation is terminated.

the generator works fine but I noticed that it has a little problem handling "infinite" iterators (the generator in this case doesn't stop, the memory is filled and the python shell crashes). How can I take care of this issue?

January 8th, 2013, 01:04 PM

b49P23TIvg

Contributing User

Join Date: Aug 2011

Posts: 3,348

Time spent in forums: 1 Month 2 Weeks 3 Days 7 h 35 m 46 sec
Reputation Power: 383

please show failure, works fine in python3 for several test cases.

Code:

>>> def takeOnly(it,pred,f):
    count = 0
    for i in it:
        if not count == f:
            if pred(i):
                count = 0
                yield i
            else:
                count += 1
        else:
            break

def odd(a):
    return a & 1

... ... ... ... ... ... ... ... ... ... ... >>> ... ... >>> 
>>> for i in takeOnly([1,2,3,4,5,6,6,6,6,6],odd,3):
...  print(i)
... 
1
3
5
>>> for i in takeOnly([1,2,2,2,2,3,4,5,6,6,6,6,6],odd,3):
...  print(i)
... 
1
>>> for i in takeOnly([1,2,3],odd,8):
...  print(i)
... 
1
3
>>> for i in takeOnly(itertools.repeat(1),odd,8):
...  print(i)
... 
1   (many 1's removed)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in takeOnly
KeyboardInterrupt
>>>

__________________
[code]Code tags[/code] are essential for python code!

January 8th, 2013, 01:22 PM

Nightmareix35

Contributing User

Join Date: Nov 2012

Posts: 32

Time spent in forums: 8 h 42 m 41 sec
Reputation Power: 1

Quote:

Originally Posted by b49P23TIvg

please show failure, works fine in python3 for several test cases.

Code:

>>> def takeOnly(it,pred,f):
    count = 0
    for i in it:
        if not count == f:
            if pred(i):
                count = 0
                yield i
            else:
                count += 1
        else:
            break

def odd(a):
    return a & 1

... ... ... ... ... ... ... ... ... ... ... >>> ... ... >>> 
>>> for i in takeOnly([1,2,3,4,5,6,6,6,6,6],odd,3):
...  print(i)
... 
1
3
5
>>> for i in takeOnly([1,2,2,2,2,3,4,5,6,6,6,6,6],odd,3):
...  print(i)
... 
1
>>> for i in takeOnly([1,2,3],odd,8):
...  print(i)
... 
1
3
>>> for i in takeOnly(itertools.repeat(1),odd,8):
...  print(i)
... 
1   (many 1's removed)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in takeOnly
KeyboardInterrupt
>>>

Yes, take the last test you tried for example. That's exactly where my program gets stuck, I need to find an appropriate halting condition that blocks it from entering in an infinite loop. As I noticed, you too had to stop it with a keyboardInterrupt order. Rather, I would like to raise a StopIteration to exit the loop. How could this be done so it can work with "any" infinite iterator?

January 8th, 2013, 01:36 PM

b49P23TIvg

Contributing User

Join Date: Aug 2011

Posts: 3,348

Time spent in forums: 1 Month 2 Weeks 3 Days 7 h 35 m 46 sec
Reputation Power: 383

use another argument

Code:

def takeOnly(it,pred,f,at_most=8888888888):
    count = 0
    yielded = 0
    for i in it:
        if (count < f) and (yielded < at_most):
            if pred(i):
                count = 0
                yielded += 1
                yield i
            else:
                count += 1
        else:
            break

def odd(a):
    return a & 1


import itertools

for i in takeOnly(itertools.repeat(1),odd,8,at_most=20):
    print(i)

Iterators and generators.

Developer Shed Advertisers and Affiliates