Re: python2.4 generator expression python2.3 list expression
On Mon, 21 Feb 2005 10:55:05 -0800, rumours say that Bryan [EMAIL PROTECTED] might have written: [I] is to reset the rightmost (less significant) '1' bit of a number (ie change it to '0'). [bryan] i tried c = c - 1 but i'm not getting the least significant or rightmost bit reset to zero. am i misunderstanding something? 2 1 # 2 = 0x10; reset right most would be 0x10 snip [Duncan] 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. snip 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!!! Don't worry, Bryan, I'm probably more to blame, since I have this tendency to interject parenthesized sub-sentences all over my paragraphs, that probably confuse more than clarify things ( self.remind(prose is not code) :). Perhaps I should crosspost my replies (esp. the ones with nested parentheses) to comp.lang.lisp ... -- 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... -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
Dan Sommers wrote: 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. Here's yet another way to achieve the same results. This version doesn't iterate over any bits at all: import operator parity = [ False ] for i in range(7): parity += map(operator.not_, parity) And if you want the same output: for even_odd in parity: print int(even_odd) -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
On 22 Feb 2005 09:14:50 GMT, Duncan Booth [EMAIL PROTECTED] wrote: Here's yet another way to achieve the same results. This version doesn't iterate over any bits at all: import operator parity = [ False ] for i in range(7): parity += map(operator.not_, parity) Very clever! :-) Picking a nit, that version iterates over *two* sets of bits. The for loop over each possible bit in the input values. The map function over the parity bits accumulated up to that point. And the += operator over those same bits again. Make that *three* sets of bits. I stand humbled. Regards, Dan -- Dan Sommers http://www.tombstonezero.net/dan/ c = 1 -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
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 1b) for b in range(8))) 1 As this: for c in range(128): bo = [bool(c 1b) 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
On Sun, 20 Feb 2005 20:56:52 -0800, snacktime [EMAIL PROTECTED] 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 1b) for b in range(8))) 1 As this: for c in range(128): bo = [bool(c 1b) 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 nave 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.tombstonezero.net/dan/ Never play leapfrog with a unicorn. -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
On 21 Feb 2005 06:48:19 -0500, rumours say that Dan Sommers [EMAIL PROTECTED] 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... -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
Christos TZOTZIOY Georgiou wrote: On 21 Feb 2005 06:48:19 -0500, rumours say that Dan Sommers [EMAIL PROTECTED] 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 misunderstanding something? 2 1 # 2 = 0x10; reset right most would be 0x10 0 10 9 # 10 = 0x1010; reset right most would be 0x1010 8 bryan -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
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?)interpretation (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 -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
Christos TZOTZIOY Georgiou [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] On 21 Feb 2005 06:48:19 -0500, rumours say that Dan Sommers [EMAIL PROTECTED] 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'). Cute. I tried it a few times until I saw why it works. But it is also dangerous (within a loop like the above) in a language like current Python (and unlike C/assembler) in which the binary representation of -1 is effectively a left infinite string of '1's: ...11 Terry J. Reedy -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
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 1b) for b in range(8))) 1 As this: for c in range(128): bo = [bool(c 1b) 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 1b) for b in range(8)])) 1 No need for the 'bo' variable... STeVe -- http://mail.python.org/mailman/listinfo/python-list
Re: python2.4 generator expression python2.3 list expression
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 1b) for b in range(8))) 1 As this: for c in range(128): bo = [bool(c 1b) 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 1b) 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 -- http://mail.python.org/mailman/listinfo/python-list
