python2.4 generator expression > python2.3 list expression

[snacktime]

I need to convert a generator expression to a list expression so it
will work under python 2.3.

I rewrote this:

for c in range(128):
even_odd = (sum(bool(c & 1<<b) for b in range(8))) & 1

As this:

for c in range(128):
bo = [bool(c & 1<<b) for b in range(8)]
even_odd = sum(bo) & 1
Seems to work, is there a better way to do this?

[Steven Bethard]

snacktime wrote:

I need to convert a generator expression to a list expression so it
will work under python 2.3.

I rewrote this:

for c in range(128):
even_odd = (sum(bool(c & 1<<b) for b in range(8))) & 1

As this:

for c in range(128):
bo = [bool(c & 1<<b) for b in range(8)]
even_odd = sum(bo) & 1

Seems to work, is there a better way to do this?

Well, if you were happy with your generator expression, you can use
almost exactly the same syntax:

for c in range(128):
even_odd = (sum([bool(c & 1<<b) for b in range(8)])) & 1

No need for the 'bo' variable...

STeVe

[Michael Hoffman]

snacktime wrote:

I need to convert a generator expression to a list expression so it
will work under python 2.3.

I rewrote this:

for c in range(128):
even_odd = (sum(bool(c & 1<<b) for b in range(8))) & 1

As this:

for c in range(128):
bo = [bool(c & 1<<b) for b in range(8)]
even_odd = sum(bo) & 1
Seems to work, is there a better way to do this?

If you want to keep it as a generator that doesn't build a list
in memory, you can use itertools:

import itertools

for c in range(128):
def _even_odd_func( b): return bool(c & 1<<b)
even_odd = (sum(itertools. imap(_even_odd_ func, xrange(8)))) & 1

The fact that you used range() instead of xrange() indicates that
you may not care about this, though. ;-)
--
Michael Hoffman

[Peter Otten]

snacktime wrote:

I need to convert a generator expression to a list expression so it
will work under python 2.3.

I rewrote this:

for c in range(128):
even_odd = (sum(bool(c & 1<<b) for b in range(8))) & 1

As this:

for c in range(128):
bo = [bool(c & 1<<b) for b in range(8)]
even_odd = sum(bo) & 1
Seems to work, is there a better way to do this?

Summing over zeros seems pointless, so

for c in range(128):

.... print len([1 for b in range(8) if c & 1 << b]) & 1,
....
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1
1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0
1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0
0 1 0 1 1 0 0 1 1 0 1 0 0 1

The same simplification works for genexps, but you have to use sum() there
instead of len(). Another optimization would be to precalculate the
bitmasks [1 << b for b in range(8)] outside the loop.

Peter

[Dan Sommers]

On Sun, 20 Feb 2005 20:56:52 -0800,
snacktime <sn*******@gmai l.com> wrote:

I need to convert a generator expression to a list expression so it
will work under python 2.3. I rewrote this: for c in range(128):
even_odd = (sum(bool(c & 1<<b) for b in range(8))) & 1 As this: for c in range(128):
bo = [bool(c & 1<<b) for b in range(8)]
even_odd = sum(bo) & 1
Seems to work, is there a better way to do this?

for c in range( 128 ):
even_odd = 0
print '%3d' % c,
while c:
c &= c - 1
even_odd = not even_odd
print int( even_odd )

Okay, so your inner loop is only counting to 8, but IMO this is a good
example of how to use a better algorithm instead of optimizing the code
of a naïve one. My inner loop only iterates over 1-bits.

"Better," of course is all relative. Your algorithm obviously counts
bits in an integer. My algorithm is less clear at first glance (and
even second and third glance), but nearly idiomatic to those of us who
spent lots of time writing embedded assembly code.

If you have the space to spare, a lookup table (pre-calculated or
created during your program's initialization) is probably the best way
to go.

Regards,
Dan

--
Dan Sommers
<http://www.tombstoneze ro.net/dan/>
Never play leapfrog with a unicorn.

[Christos TZOTZIOY Georgiou]

On 21 Feb 2005 06:48:19 -0500, rumours say that Dan Sommers <[e-mail] >
might have written:

[snip: snacktime posts code to count bits]

Seems to work, is there a better way to do this?

[Dan]for c in range( 128 ):
even_odd = 0
print '%3d' % c,
while c:
c &= c - 1
even_odd = not even_odd
print int( even_odd )

Just for the sake of people who haven't messed with bit manipulation in C or
assembly, the effect of

c &= c - 1

is to reset the rightmost (less significant) '1' bit of a number (ie change it
to '0').
--
TZOTZIOY, I speak England very best.
"Be strict when sending and tolerant when receiving." (from RFC1958)
I really should keep that in mind when talking with people, actually...

[Bryan]

Christos TZOTZIOY Georgiou wrote:

On 21 Feb 2005 06:48:19 -0500, rumours say that Dan Sommers <[e-mail] >
might have written:

[snip: snacktime posts code to count bits]

Seems to work, is there a better way to do this?

[Dan]

for c in range( 128 ):
even_odd = 0
print '%3d' % c,
while c:
c &= c - 1
even_odd = not even_odd
print int( even_odd )

Just for the sake of people who haven't messed with bit manipulation in C or
assembly, the effect of

c &= c - 1

is to reset the rightmost (less significant) '1' bit of a number (ie change it
to '0').

i tried c &= c - 1 but i'm not getting the least significant or rightmost bit
reset to zero. am i misunderstandin g something?

2 & 1 # 2 = 0x10; reset right most would be 0x10 0 10 & 9 # 10 = 0x1010; reset right most would be 0x1010

8

bryan

[Duncan Booth]

Bryan wrote:

is to reset the rightmost (less significant) '1' bit of a number (ie
change it to '0').

i tried c &= c - 1 but i'm not getting the least significant or
rightmost bit reset to zero. am i misunderstandin g something?

2 & 1 # 2 = 0x10; reset right most would be 0x10 0 10 & 9 # 10 = 0x1010; reset right most would be 0x1010 8

The difference between the original "reset the rightmost '1' bit", and your
interpretation: "reset the rightmost bit" is the "'1'".

The rightmost bit that is set is reset. So 0x10 -> 0, and 0x1010 -> 0x1000.

If you want to extract the least significant set bit from a number 'x' you
can use (x&-x):

x = 0xab4
while x: print hex(x&-x), hex(x)
x ^= (x&-x)
0x4 0xab4
0x10 0xab0
0x20 0xaa0
0x80 0xa80
0x200 0xa00
0x800 0x800

(but don't try this if x is negative: it works but never terminates).

[Brian Beck]

Duncan Booth wrote:

The difference between the original "reset the rightmost '1' bit", and your
interpretation: "reset the rightmost bit" is the "'1'".

The rightmost bit that is set is reset. So 0x10 -> 0, and 0x1010 -> 0x1000.

If you want to extract the least significant set bit from a number 'x' you
can use (x&-x):

My interpretation of Bryan's (mis?)interpret ation (heh) was that since
in the numbers 2 and 10 (as in his examples), the least significant bit
was already 0, performing an operation that set it to 0 should result in
the number unchanged. As his tests show, this is not the case. This is
because the operation works only if the least significant bit actually
NEEDS to be unset. To zero the least significant bit unconditionally , we
can use:

x &= ~1

--
Brian Beck
Adventurer of the First Order

[Bryan]

Duncan Booth wrote:

Bryan wrote:

is to reset the rightmost (less significant) '1' bit of a number (ie
change it to '0').

i tried c &= c - 1 but i'm not getting the least significant or
rightmost bit reset to zero. am i misunderstandin g something?

>2 & 1 # 2 = 0x10; reset right most would be 0x10

0

>10 & 9 # 10 = 0x1010; reset right most would be 0x1010

8

The difference between the original "reset the rightmost '1' bit", and your
interpretation: "reset the rightmost bit" is the "'1'".

The rightmost bit that is set is reset. So 0x10 -> 0, and 0x1010 -> 0x1000.

If you want to extract the least significant set bit from a number 'x' you
can use (x&-x):

x = 0xab4
while x:

print hex(x&-x), hex(x)
x ^= (x&-x)
0x4 0xab4
0x10 0xab0
0x20 0xaa0
0x80 0xa80
0x200 0xa00
0x800 0x800
(but don't try this if x is negative: it works but never terminates).

thanks duncan... you're right, i did intrepret this as "reset the rightmost bit"
instead of "reset the rightmost '1' bit". and i must have read what christos
wrote 100 times!!!

bryan