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? 10 1800
MetalOne wrote: 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): def big(mask, index=0):
.... if mask != 0:
.... if mask & 1:
.... yield index
.... for result in big(mask>>1, index+1):
.... yield result
.... for x in big(0x16):
.... print x,
....
1 2 4
Peter
MetalOne wrote in message
<92**************************@posting.google.com>. .. What am I missing?
The difference between a generator and a generator function. def gen_func(): yield None
.... gen_func() # what does this return?
Also, this isn't a recursive problem. Use a loop.

Francis Avila
In article <92**************************@posting.google.com >, jc*@iteris.com (MetalOne) wrote: 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

David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer Science
MetalOne wrote: 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?
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
On 19 Dec 2003 11:13:39 0800, jc*@iteris.com (MetalOne) wrote: 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): def bitnos(self):
... """Littleendian 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
In article <br**********@216.39.172.122>, bo**@oz.net (Bengt Richter)
wrote: Here is one that works also for negative numbers (includes the least significant of the arbitrarily extended sign bits):
>>> def bitnos(self):
... """Littleendian 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.

David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer Science
On Fri, 19 Dec 2003 14:49:10 0800, David Eppstein <ep******@ics.uci.edu> wrote: In article <br**********@216.39.172.122>, bo**@oz.net (Bengt Richter) wrote:
Here is one that works also for negative numbers (includes the least significant of the arbitrarily extended sign bits):
>>> def bitnos(self): ... """Littleendian 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.
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
a = LBits('010110b') a
010110b b = LBits('101010b') b
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:
a
010110b a &~(a1)
010b a ^= a &~(a1) a
010100b a &~(a1)
0100b a ^= a &~(a1) a
010000b a &~(a1)
010000b a ^= a &~(a1) a
0b
Regards,
Bengt Richter
MetalOne wrote: 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

Christian Tismer :^) <mailto:ti****@tismer.com>
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