Re: Noob questions about Python
On 2007-10-19, Hendrik van Rooyen [EMAIL PROTECTED] wrote: Arnaud Delobelle arnoemail.com wrote: In binary 2 is 10. When you multiply by 10, you shift all your digits left by 1 place. When you multiply by 10**n (which is 1 followed by n zeroes), you shift all your digits left by n places. I read somewhere: Only 1 person in 1000 understands binary. The other 111 don't have a clue. There are only 10 kinds of people: those who understand binary and those who don't. -- Grant Edwards grante Yow! Will this never-ending at series of PLEASURABLE visi.comEVENTS never cease? -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
Arnaud Delobelle arnoemail.com wrote: In binary 2 is 10. When you multiply by 10, you shift all your digits left by 1 place. When you multiply by 10**n (which is 1 followed by n zeroes), you shift all your digits left by n places. I read somewhere: Only 1 person in 1000 understands binary. The other 111 don't have a clue. - Hendrik -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On 2007-10-19, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: On Oct 19, 1:44 am, MRAB [EMAIL PROTECTED] wrote: On Oct 18, 7:05 am, Michele Simionato [EMAIL PROTECTED] if number == 0: return 0 Hey, Isn't if not number: return 0 faster? Depends on who is parsing it. If a computer, possibly a few micro-seconds (if the Python byte-code generator doesn't optimize it away). If a human, then it is slower by at least a second. Since I prefer to optimize on my time rather than CPU time, I'd prefer the first version Albert -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 19, 1:44 am, MRAB [EMAIL PROTECTED] wrote: On Oct 18, 7:05 am, Michele Simionato [EMAIL PROTECTED] wrote: On Oct 17, 5:58 pm, Ixiaus [EMAIL PROTECTED] wrote: def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] print sum(res) It basically does the same thing int(string, 2) does. Thank you for the responses! BTW, here is the reverse function dec2bin, so that you can bang your head on it for a while ;) It returns '' when number == 0, so you need to test for that case: def baseN(number, N=2): baseN(9, 2) '1001' assert 2 = N = 10 assert isinstance(number, int) and number = 0 if number == 0: return 0 Hey, Isn't if not number: return 0 faster? -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 17, 5:58 pm, Ixiaus [EMAIL PROTECTED] wrote: def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] print sum(res) It basically does the same thing int(string, 2) does. Thank you for the responses! BTW, here is the reverse function dec2bin, so that you can bang your head on it for a while ;) def baseN(number, N=2): baseN(9, 2) '1001' assert 2 = N = 10 assert isinstance(number, int) and number = 0 b = [] while number: b.append(str(number % N)) number /= N return ''.join(reversed(b)) Michele Simionato -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
Right idea: now to remove all those intermediate lists you construct. 1. reversed(val) creates an iterator that runs over the elements (here of a string) in reverse order. 2. enumerate() is usually better than using an explicit list index. 3. You can use a generator in your sum to avoid constructing the final list. Applying these to your function, and noting that n k is nicer than n * 2 ** k, we get a one-liner: def bin2dec(val): return sum(int(i) k for k, i in enumerate(reversed(val))) Or a slightly nicer alternative is to filter the generator using 'if': def bin2dec(val): return sum(1 k for k, i in enumerate(reversed(val)) if int(i)) -- Paul Hankin Thank you for this reply, I only wish I could come up with functions so elegantly refined! I know '' is shifting x over by n bits; but could you point me to some literature that would explain why it is the same as x*2**n? (Googling only returns bit shift, but doesn't necessarily explain the manner in which you are using it) I will have to read up more on Generators, but maybe you can give me a lowdown on why sum([1 k for k, i in enumerate(reversed(val)) if int(i)]) is less efficient than using a Generator (is the generator a 'temporary' list?) sum(1 k for k, i in enumerate(reversed(val)) if int(i)) -- Parnell Springmeyer -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Wed, 17 Oct 2007 22:05:36 +0200, Bruno Desthuilliers wrote:
[snip]
Note that there's also the reverse() function that returns a reverse
iterator over any sequence, so you could also do:
li = list('allo')
print ''.join(reverse(li))
Note this certainly should've been `reversed()`, with a trailing 'd'.
--
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Re: Noob questions about Python
Stargaming a écrit :
On Wed, 17 Oct 2007 22:05:36 +0200, Bruno Desthuilliers wrote:
[snip]
Note that there's also the reverse() function that returns a reverse
iterator over any sequence, so you could also do:
li = list('allo')
print ''.join(reverse(li))
Note this certainly should've been `reversed()`, with a trailing 'd'.
2-0 for Stargaming. I'll have to either change glasses, buy a new
keyboard (this one is obviously broken), or learn to both use a keyboard
and re-read what I post.
--
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Re: Noob questions about Python
On Oct 18, 7:06 am, Ixiaus [EMAIL PROTECTED] wrote: [...] I know '' is shifting x over by n bits; but could you point me to some literature that would explain why it is the same as x*2**n? I haven't got literature but I've got a (hopefully straightforward) explanation: In binary 2 is 10. When you multiply by 10, you shift all your digits left by 1 place. When you multiply by 10**n (which is 1 followed by n zeroes), you shift all your digits left by n places. HTH -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 18, 7:05 am, Michele Simionato [EMAIL PROTECTED] wrote: On Oct 17, 5:58 pm, Ixiaus [EMAIL PROTECTED] wrote: def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] print sum(res) It basically does the same thing int(string, 2) does. Thank you for the responses! BTW, here is the reverse function dec2bin, so that you can bang your head on it for a while ;) It returns '' when number == 0, so you need to test for that case: def baseN(number, N=2): baseN(9, 2) '1001' assert 2 = N = 10 assert isinstance(number, int) and number = 0 if number == 0: return 0 b = [] while number: b.append(str(number % N)) number /= N return ''.join(reversed(b)) Michele Simionato -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Wed, 17 Oct 2007 23:06:24 -0700, Ixiaus wrote: I know '' is shifting x over by n bits; but could you point me to some literature that would explain why it is the same as x*2**n? (Googling only returns bit shift, but doesn't necessarily explain the manner in which you are using it) In decimal, multiplying by ten shifts the digits to the left by one place: 31 * 10 = 310 310 * 10 = 3100 3100 * 10 = 31000 In binary, multiplying by two shifts the bits to the left by one place: 101 * 10 = 1010 1010 * 10 = 10100 10100 * 10 = 101000 Because the underlying numbers are the same regardless of what base we use to write it in, bit-shifts to the left are equivalent to repeated multiplication by two regardless of the display base. Hence: 7 1 14 14 1 28 28 1 56 or to do it another way: 7 3 # multiplication by 2**3 56 Similarly, a bit-shift to the right is equivalent to dividing by two and dropping any remainder. I will have to read up more on Generators, but maybe you can give me a lowdown on why sum([1 k for k, i in enumerate(reversed(val)) if int(i)]) is less efficient than using a Generator (is the generator a 'temporary' list?) sum(1 k for k, i in enumerate(reversed(val)) if int(i)) The list comprehension: [1 k for k, i in enumerate(reversed(val)) if int(i)] creates the entire list first. That potentially can require a lot of memory, leading to all sorts of complicated memory shuffles as the list is resized, paging, etc. Using a generator expression means that you avoid creating the entire list all in one go. That saves you memory, which may mean that you save time. However, for small sets of input, it is likely that the overhead of creating the generator in the first place outweighs the saving of not creating the whole list. The definition of small depends on what you are trying to do. In my tests, I have found that sum(list comprehension) is marginally faster than sum(generator expression) even for 10,000 items. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
That was a very helpful answer Steven, thank you for taking the time to write it! -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 18, 7:44 pm, MRAB [EMAIL PROTECTED] wrote: It returns '' when number == 0, so you need to test for that case: def baseN(number, N=2): baseN(9, 2) '1001' assert 2 = N = 10 assert isinstance(number, int) and number = 0 if number == 0: return 0 b = [] while number: b.append(str(number % N)) number /= N return ''.join(reversed(b)) Right, or you can just change the last line: return ''.join(reversed(b)) or '0' Michele Simionato -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On 2007-10-17, Ixiaus [EMAIL PROTECTED] wrote: val = 'string' li = list(val) print li.reverse() returns nothing, but, val = 'string' li = list(val) li.reverse() print li returns what I want. Why does Python do that? Because it does. :) Also I have been playing around with Binary math and noticed that Python treats: val = 00110 as the integer 72 instead of returning 00110, why does Python do that? In order to be compatible with the C language integer literal convensions, integer litereals staring with a '0' are base-8, so 00110 is 0 * 8^0 0 + 1 * 8^1 8 + 1 * 8^264 + 0* 8^3 0 +0 * 8^4 0 72 (and how can I get around it?) You can't. -- Grant Edwards grante Yow! Do you guys know we at just passed thru a BLACK visi.comHOLE in space? -- http://mail.python.org/mailman/listinfo/python-list
Noob questions about Python
I have recently (today) just started learning/playing with Python. So far I am excited and impressed (coming from PHP background). I have a few questions regarding Python behavior... val = 'string' li = list(val) print li.reverse() returns nothing, but, val = 'string' li = list(val) li.reverse() print li returns what I want. Why does Python do that? Also I have been playing around with Binary math and noticed that Python treats: val = 00110 as the integer 72 instead of returning 00110, why does Python do that? (and how can I get around it?) Grateful for any replies! -- http://mail.python.org/mailman/listinfo/python-list
Noob questions about Python
I have recently (today) just started learning/playing with Python. So far I am excited and impressed (coming from PHP background). I have a few questions regarding Python behavior... val = 'string' li = list(val) print li.reverse() returns nothing, but, val = 'string' li = list(val) li.reverse() print li returns what I want. Why does Python do that? Also I have been playing around with Binary math and noticed that Python treats: val = 00110 as the integer 72 instead of returning 00110, why does Python do that? (and how can I get around it?) Grateful for any replies! -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
li.reverse() does not return a list it just changes the list li. val = 00110 is evaluated in base 8 try without leading 0s On 10/17/07, Ixiaus [EMAIL PROTECTED] wrote: I have recently (today) just started learning/playing with Python. So far I am excited and impressed (coming from PHP background). I have a few questions regarding Python behavior... val = 'string' li = list(val) print li.reverse() returns nothing, but, val = 'string' li = list(val) li.reverse() print li returns what I want. Why does Python do that? Also I have been playing around with Binary math and noticed that Python treats: val = 00110 as the integer 72 instead of returning 00110, why does Python do that? (and how can I get around it?) Grateful for any replies! -- http://mail.python.org/mailman/listinfo/python-list -- Furkan Kuru -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 17, 3:37 pm, Ixiaus [EMAIL PROTECTED] wrote: I have recently (today) just started learning/playing with Python. So far I am excited and impressed (coming from PHP background). I have a few questions regarding Python behavior... val = 'string' li = list(val) print li.reverse() returns nothing, but, val = 'string' li = list(val) li.reverse() print li returns what I want. Why does Python do that? Because list.reverse() modifies a list, it doesn't create and return a new one. A common idiom for returning a reversed copy of a list is: li[::-1] You can also use the builtin reversed -- it doesn't return a list, but rather a reverse iterator. You can create a list from an iterator by passing it to the list() initializer, like list(reversed(li)). -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
Ixiaus a écrit :
I have recently (today) just started learning/playing with Python. So
far I am excited and impressed
Welcome onboard then !-)
(coming from PHP background).
I have a few questions regarding Python behavior...
val = 'string'
li = list(val)
print li.reverse()
returns nothing, but,
val = 'string'
li = list(val)
li.reverse()
print li
returns what I want. Why does Python do that?
list.reverse (like list.sort) is a destructive in-place operation. Not
returning the object is reminder of the destructive nature of the
operation. That's a design choice, whether you agree with it or not
(FWIW, I don't, but I live with it !-)
Note that there's also the reverse() function that returns a reverse
iterator over any sequence, so you could also do:
li = list('allo')
print ''.join(reverse(li))
Also I have been playing around with Binary math and noticed that
Python treats:
val = 00110
as the integer 72 instead of returning 00110, why does Python do that?
Literal integers starting with '0' (zero) are treated as octal. It's a
pretty common convention (like 0x for hexa). FWIW, PHP does just the
same thing.
(and how can I get around it?)
You can't. Python has no literal notation for binary integers so far.
Literal notation for binary ints is not a common feature anyway.
HTH
--
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Re: Noob questions about Python
On Oct 17, 8:37 pm, Ixiaus [EMAIL PROTECTED] wrote:
I have a few questions regarding Python behavior...
as the integer 72 instead of returning 00110, why does Python do that?
(and how can I get around it?)
You can do this:
def bin(x):
return int(x, 2)
val = bin('00110')
--
Paul Hankin
--
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Re: Noob questions about Python
On 2007-10-17, Bruno Desthuilliers
[EMAIL PROTECTED] wrote:
Ixiaus a écrit :
val = 00110
as the integer 72 instead of returning 00110, why does Python
do that?
Literal integers starting with '0' (zero) are treated as octal.
It's a pretty common convention (like 0x for hexa). FWIW, PHP
does just the same thing.
(and how can I get around it?)
You can't. Python has no literal notation for binary integers
so far. Literal notation for binary ints is not a common
feature anyway.
You can obtain a practical workaround using the int built-in
function.
int('110', 2)
6
--
Neil Cerutti
--
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Re: Noob questions about Python
Ixiaus wrote: val = 'string' li = list(val) print li.reverse() returns nothing, but, Yes -- li.reverse() returns None. print None prints nothing. val = 'string' li = list(val) li.reverse() print li returns what I want. I'm afraid not. li.reverse() still returns None, but this time you print li, and this shows the reversed list. As already explained, li.reverse() modifies the existing list. What to use here depends on what you want to achieve: If you really want to reverse the list (i. e. re-sort the list in memory, and perhaps work with it afterwards), use li.reverse(). If you just want to have all its members in reverse order, use the builtin reversed(li) (which is an iterator giving you li's members from back to front when iterating over it). Regards, Björn -- BOFH excuse #243: The computer fleetly, mouse and all. -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
Thank you for the quick responses. I did not know that about integer literals beginning with a '0', so thank you for the explanation. I never really use PHP except for handling basic forms and silly web stuff, this is why I picked up Python because I want to teach myself a more powerful and broad programming language. With regard to why I asked: I wanted to learn about Binary math in conjunction with Python, so I wrote a small function that would return a base 10 number from a binary number. It is nice to know about the int() function now. Just for the sake of it, this was the function I came up with: def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] res.reverse() print sum(res) Now that I look at it, I probably don't need that last reverse() because addition is commutative... def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] print sum(res) It basically does the same thing int(string, 2) does. Thank you for the responses! -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 17, 10:58 pm, Ixiaus [EMAIL PROTECTED] wrote: Thank you for the quick responses. I did not know that about integer literals beginning with a '0', so thank you for the explanation. I never really use PHP except for handling basic forms and silly web stuff, this is why I picked up Python because I want to teach myself a more powerful and broad programming language. With regard to why I asked: I wanted to learn about Binary math in conjunction with Python, so I wrote a small function that would return a base 10 number from a binary number. It is nice to know about the int() function now. Just for the sake of it, this was the function I came up with: def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] res.reverse() print sum(res) Now that I look at it, I probably don't need that last reverse() because addition is commutative... def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] print sum(res) Right idea: now to remove all those intermediate lists you construct. 1. reversed(val) creates an iterator that runs over the elements (here of a string) in reverse order. 2. enumerate() is usually better than using an explicit list index. 3. You can use a generator in your sum to avoid constructing the final list. Applying these to your function, and noting that n k is nicer than n * 2 ** k, we get a one-liner: def bin2dec(val): return sum(int(i) k for k, i in enumerate(reversed(val))) Or a slightly nicer alternative is to filter the generator using 'if': def bin2dec(val): return sum(1 k for k, i in enumerate(reversed(val)) if int(i)) -- Paul Hankin -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 17, 4:58 pm, Ixiaus [EMAIL PROTECTED] wrote: Thank you for the quick responses. I did not know that about integer literals beginning with a '0', so thank you for the explanation. I never really use PHP except for handling basic forms and silly web stuff, this is why I picked up Python because I want to teach myself a more powerful and broad programming language. With regard to why I asked: I wanted to learn about Binary math in conjunction with Python, so I wrote a small function that would return a base 10 number from a binary number. It is nice to know about the int() function now. Just for the sake of it, this was the function I came up with: def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] res.reverse() print sum(res) Now that I look at it, I probably don't need that last reverse() because addition is commutative... def bin2dec(val): li = list(val) li.reverse() res = [int(li[x])*2**x for x in range(len(li))] print sum(res) It basically does the same thing int(string, 2) does. Thank you for the responses! You could also get ahold of the gmpy module. You get conversion to binary and also some useful other binary functions as shown below: # the Collatz Conjecture in binary import gmpy n = 27 print '%4d %s' % (n,gmpy.digits(n,2).zfill(16)) sv = [] # sequence vector, blocks of contiguous LS 0's while n != 1: old_n = n n = 3*n + 1 # result always even f = gmpy.scan1(n,0) # find least significant 1 bit n = f # remove LS 0's in one fell swoop sv.append(f) # record f sequence PopC = gmpy.popcount(n) # count of 1 bits HamD = gmpy.hamdist(n,old_n) # bits changed print '%4d %s' % (n,gmpy.digits(n,2).zfill(16)), print 'PopC:%2d HamD:%2d' % (PopC,HamD) print sv ## 27 00011011 ## 41 00101001 PopC: 3 HamD: 3 ## 31 0001 PopC: 5 HamD: 4 ## 47 0010 PopC: 5 HamD: 2 ## 71 01000111 PopC: 4 HamD: 3 ## 107 01101011 PopC: 5 HamD: 3 ## 161 1011 PopC: 3 HamD: 4 ## 121 0001 PopC: 5 HamD: 4 ## 91 01011011 PopC: 5 HamD: 2 ## 137 10001001 PopC: 3 HamD: 4 ## 103 01100111 PopC: 5 HamD: 6 ## 155 10011011 PopC: 5 HamD: 6 ## 233 11101001 PopC: 5 HamD: 4 ## 175 1010 PopC: 6 HamD: 3 ## 263 00010111 PopC: 4 HamD: 4 ## 395 000110001011 PopC: 5 HamD: 3 ## 593 001001010001 PopC: 4 HamD: 7 ## 445 00011001 PopC: 7 HamD: 7 ## 167 10100111 PopC: 5 HamD: 4 ## 251 1011 PopC: 7 HamD: 4 ## 377 00010001 PopC: 6 HamD: 3 ## 283 000100011011 PopC: 5 HamD: 3 ## 425 000110101001 PopC: 5 HamD: 4 ## 319 00010011 PopC: 7 HamD: 4 ## 479 00011101 PopC: 8 HamD: 3 ## 719 00101100 PopC: 7 HamD: 3 ##1079 01110111 PopC: 6 HamD: 7 ##1619 011001010011 PopC: 6 HamD: 4 ##2429 10010101 PopC: 8 HamD: 8 ## 911 00111000 PopC: 7 HamD: 7 ##1367 010101010111 PopC: 7 HamD: 6 ##2051 1011 PopC: 3 HamD: 6 ##3077 11000101 PopC: 4 HamD: 3 ## 577 00100101 PopC: 3 HamD: 5 ## 433 000110110001 PopC: 5 HamD: 6 ## 325 000101000101 PopC: 4 HamD: 5 ## 61 0001 PopC: 5 HamD: 5 ## 23 00010111 PopC: 4 HamD: 3 ## 35 00100011 PopC: 3 HamD: 3 ## 53 00110101 PopC: 4 HamD: 3 ## 5 0101 PopC: 2 HamD: 2 ## 1 0001 PopC: 1 HamD: 1 ##[1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, ## 1, 1, 1, 2, 3, 1, 1, 2, 1, 2, 1, 1, 1, ## 1, 1, 3, 1, 1, 1, 4, 2, 2, 4, 3, 1, 1, ## 5, 4] -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
Ixiaus wrote: I have recently (today) just started learning/playing with Python. So far I am excited and impressed (coming from PHP background). I have a few questions regarding Python behavior... val = 'string' li = list(val) print li.reverse() returns nothing, but, val = 'string' li = list(val) li.reverse() print li returns what I want. Why does Python do that? Also I have been playing around with Binary math and noticed that Python treats: val = 00110 as the integer 72 instead of returning 00110, why does Python do that? (and how can I get around it?) Grateful for any replies! li.reverse does not return a list but reverses the list itself and return Noe 0xxx tells python that you work in octal. hg -- http://mail.python.org/mailman/listinfo/python-list
Re: Noob questions about Python
On Oct 17, 6:23 pm, Paul Hankin [EMAIL PROTECTED] wrote:
On Oct 17, 10:58 pm, Ixiaus [EMAIL PROTECTED] wrote:
Thank you for the quick responses.
I did not know that about integer literals beginning with a '0', so
thank you for the explanation. I never really use PHP except for
handling basic forms and silly web stuff, this is why I picked up
Python because I want to teach myself a more powerful and broad
programming language.
With regard to why I asked: I wanted to learn about Binary math in
conjunction with Python, so I wrote a small function that would return
a base 10 number from a binary number. It is nice to know about the
int() function now.
Just for the sake of it, this was the function I came up with:
def bin2dec(val):
li = list(val)
li.reverse()
res = [int(li[x])*2**x for x in range(len(li))]
res.reverse()
print sum(res)
Now that I look at it, I probably don't need that last reverse()
because addition is commutative...
def bin2dec(val):
li = list(val)
li.reverse()
res = [int(li[x])*2**x for x in range(len(li))]
print sum(res)
Right idea: now to remove all those intermediate lists you construct.
1. reversed(val) creates an iterator that runs over the elements (here
of a string) in reverse order.
2. enumerate() is usually better than using an explicit list index.
3. You can use a generator in your sum to avoid constructing the final
list.
Applying these to your function, and noting that n k is nicer than
n * 2 ** k, we get a one-liner:
def bin2dec(val):
return sum(int(i) k for k, i in enumerate(reversed(val)))
Or a slightly nicer alternative is to filter the generator using 'if':
def bin2dec(val):
return sum(1 k for k, i in enumerate(reversed(val)) if int(i))
Changing the condition to if i=='1' makes it than twice faster.
There's also a small improvement by looping forward rather than
reversed:
def bin2dec_2(val):
n = len(val)-1
return sum(1n-i for i,bit in enumerate(val) if bit=='1')
These one-liners are short and sweet and have decent performance for
most use cases. Still, for squeezing every last drop of performance
while staying in pure Python, use Psyco and a more verbose version.
The following is two orders of magnitude faster with Psyco enabled:
def bin2dec_3(val):
s = 0
p = 1 len(val)
for bit in val:
p = 1
if bit == '1':
s += p
return s
And here's the benchmark:
if __name__ == '__main__':
# uncomment for Psyco
# import psyco; psyco.full()
import timeit
setup = 'import __main__; bin=101001010001001010'
for func in bin2dec, bin2dec_2, bin2dec_3:
name = func.__name__
timer = timeit.Timer('__main__.%s(bin)' % name, setup)
print '%s: %s' % (name, timer.timeit())
### Without Psyco
bin2dec: 17.6126108206
bin2dec_2: 7.57195732977
bin2dec_3: 5.46163297291
### With Psyco
bin2dec: 17.6995679618
bin2dec_2: 8.60846224869
bin2dec_3: 0.16031255369
George
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