Simple Recursive Generator Question

[MetalOne]

I am trying to write a generator function that yields the index position
of each set bit in a mask.
e.g.
for x in bitIndexGenerator(0x16): #10110
print x
--> 1 2 4


This is what I have, but it does not work.
Changing yield to print, shows that the recursion works correctly.

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
bitIndexGenerator(mask >> 1, index+1)

What am I missing?

 

[Peter Otten]

I am trying to write a generator function that yields the index position
of each set bit in a mask.
e.g.
for x in bitIndexGenerator(0x16): #10110
print x
--> 1 2 4


This is what I have, but it does not work.
Changing yield to print, shows that the recursion works correctly.

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
bitIndexGenerator(mask >> 1, index+1)

What am I missing?

The bitIndexGenerator(mask>>1, index+1) just returns a new generator which
is immediately discarded. For a generator to do anything, you must invoke
its next() method, and one way to do it is a for loop, e. g (not mask<0
proof):
.... if mask != 0:
.... if mask & 1:
.... yield index
.... for result in big(mask>>1, index+1):
.... yield result
........ print x,
....
1 2 4

Peter

 

[Francis Avila]

MetalOne wrote in message

What am I missing?

The difference between a generator and a generator function.

Also, this isn't a recursive problem. Use a loop.

 

[Christian Tismer]

I am trying to write a generator function that yields the index position
of each set bit in a mask.
e.g.
for x in bitIndexGenerator(0x16): #10110
print x
--> 1 2 4


This is what I have, but it does not work.
Changing yield to print, shows that the recursion works correctly.

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
bitIndexGenerator(mask >> 1, index+1)

What am I missing?

Your recursive generator is yielding to itself,
but not taking its result. Recursion does not
make sense with generators, since they can yield
only one level up.

The above would work pretty fine with a Stackless
tasklet.

Here is a recursion solution, but be warned, this
is really really inefficient, since it has quadratic
behavior due to the repeated recursion activation:

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
for each in bitIndexGenerator(mask >> 1, index+1):
yield each

An interative version is easy and adequate here.

def bitIndexGenerator(mask):
index = 0
while mask:
if mask & 0x1: yield index
mask >>= 1
index += 1

--
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[David Eppstein]

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
bitIndexGenerator(mask >> 1, index+1)

The actual answer to your question is that when you recursively call a
generator, its return value (the iterator of its results) needs to be
used not ignored.

But the reason I'm posting is to suggest an alternative approach:


bitIndices = dict([(1L<<i,i) for i in range(32)])

def bitIndexGenerator(mask):
while mask:
bit = mask &~ (mask - 1)
yield bitIndices[bit]
mask &=~ bit

 

[Samuel Bronson]

> This is what I have, but it does not work.
> Changing yield to print, shows that the recursion works correctly.
>
> def bitIndexGenerator(mask, index=0):
> if mask == 0: return
> elif mask & 0x1: yield index
> bitIndexGenerator(mask >> 1, index+1)
>
> What am I missing?

Everything needs to be yielded from the outermost generator, like

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
for i in bitIndexGenerator(mask >> 1, index+1):
yield i

However, this is ugly ;-), and kind of defeats the purpose of
generators. Why do you want to use recursion like this?

 

[Skip Montanaro]

jcb> This is what I have, but it does not work.

jcb> def bitIndexGenerator(mask, index=0):
jcb> if mask == 0: return
jcb> elif mask & 0x1: yield index
jcb> bitIndexGenerator(mask >> 1, index+1)

Try:

def bitIndexGenerator(mask, index=0):
if mask == 0:
return
elif mask & 0x1:
yield index
for index in bitIndexGenerator(mask >> 1, index+1):
yield index

Skip

 

[Bengt Richter]

I am trying to write a generator function that yields the index position
of each set bit in a mask.
e.g.
for x in bitIndexGenerator(0x16): #10110
print x
--> 1 2 4


This is what I have, but it does not work.
Changing yield to print, shows that the recursion works correctly.

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
bitIndexGenerator(mask >> 1, index+1)

What am I missing?

Here is one that works also for negative numbers (includes the least significant
of the arbitrarily extended sign bits):
... """Little-endian bit number generator"""
... bits = long(self)
... sign = bits<0
... bitno = 0
... while bits>0 or sign and bits!=-1L:
... if bits&1: yield bitno
... bitno += 1
... bits >>= 1
... if sign: yield bitno
...

(I'll use a subclass of long I recently posted (with missing ~ operator in first version, but
fix followup posted) to show bits) The above is a mod of the bit list generator from the latter.

>>> from lbits import LBits
>>>
>>> for i in range(-3,4)+[0x16, -0x16]:

... print '%3s %8r %s' %(i, LBits(i), [bit for bit in bitnos(i)])
...
-3 101b [0, 2]
-2 110b [1]
-1 11b [0]
0 0b []
1 01b [0]
2 010b [1]
3 011b [0, 1]
22 010110b [1, 2, 4]
-22 101010b [1, 3, 5]

Regards,
Bengt Richter

 

[David Eppstein]

Here is one that works also for negative numbers (includes the least
significant
of the arbitrarily extended sign bits):

... """Little-endian bit number generator"""
... bits = long(self)
... sign = bits<0
... bitno = 0
... while bits>0 or sign and bits!=-1L:
... if bits&1: yield bitno
... bitno += 1
... bits >>= 1
... if sign: yield bitno
...

I'm not sure I would call that working -- what I'd expect for a negative
number is to generate an infinite sequence.

 

[MetalOne]

Thanks.

 

[Bengt Richter]

I'm not sure I would call that working -- what I'd expect for a negative
number is to generate an infinite sequence.

Hm, yeah, if you don't know the sign, there's obviously an ambiguity.
Maybe I should flag by making the last line

if sign: yield -bitno.

With that change we can get the number back:

>>> for i in range(-3,4)+[0x16,-0x16]:

... print '%3s => %s' % (i, sum([p>=0 and 2**p or -2**-p for p in bitnos(i)]))
...
-3 => -3
-2 => -2
-1 => 1
0 => 0
1 => 1
2 => 2
3 => 3
22 => 22
-22 => -22


The repr of my lbits.LBits class doesn't have that problem, since the msb
is always the lsb sign bit (except that I used '11b' for -1 for symmetry)

BTW, what would you think of having signed binary literals in python, so you
could write natively

a = 010110b # not legal now
b = 101010b # ditto

like the strings my LBits constructor accepts
101010b

>>> a+b 0b
>>> a+b, a==-b (0b, True)
>>> map(int, [a,b])

[22, -22]

Sometimes it's nice to see bits, e.g., the bit isolation of your algo, shown step by step:
0b

Regards,
Bengt Richter

 

[Andrew Koenig]

I am trying to write a generator function that yields the index position
of each set bit in a mask.
e.g.
for x in bitIndexGenerator(0x16): #10110
print x
--> 1 2 4


This is what I have, but it does not work.
Changing yield to print, shows that the recursion works correctly.

def bitIndexGenerator(mask, index=0):
if mask == 0: return
elif mask & 0x1: yield index
bitIndexGenerator(mask >> 1, index+1)

What am I missing?

If you really want to make your program work the way you apparently intend,
then you need to activate the generator you create recursively and yield
each value that it yields. For example:

def bitIndexGenerator(mas, index=0):
if mask == 0: return
elif mask & 0x1: yield index
for n in bitIndexGenerator(mask >> 1, index + 1): yield n

However, unless your purpose in writing this program is to improve your
understanding of recursion, this technique uses much more mechanism that
needed to solve the problem--as other replies have already pointed out.