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# flatten a level one list

 P: n/a Is there some smart/fast way to flatten a level one list using the latest iterator/generator idioms. The problem arises in coneverting lists of (x,y) coordinates into a single list of coordinates eg f([(x0,y0),(x1,y1),....]) --> [x0,y0,x1,y1,....] or g([x0,x1,x2,......],[y0,y1,y2,....]) --> [x0,y0,x1,y1,....] clearly if f is doable then g can be done using zip. I suppose this is a special case flatten, but can flatten be done fast? The python recipes seem rather slow compared to the builtin functions. -- Robin Becker Jan 12 '06 #1
32 Replies

 P: n/a Robin Becker wrote: Is there some smart/fast way to flatten a level one list using the latest iterator/generator idioms. The problem arises in coneverting lists of (x,y) coordinates into a single list of coordinates eg f([(x0,y0),(x1,y1),....]) --> [x0,y0,x1,y1,....] or g([x0,x1,x2,......],[y0,y1,y2,....]) --> [x0,y0,x1,y1,....] clearly if f is doable then g can be done using zip. I suppose this is a special case flatten, but can flatten be done fast? The python recipes seem rather slow compared to the builtin functions. how fast is fast ? for this case, is the following good enough ? def flat(li): for x,y in li: yield x yield y Jan 12 '06 #2

 P: n/a Robin Becker schrieb: Is there some smart/fast way to flatten a level one list using the latest iterator/generator idioms. The problem arises in coneverting lists of (x,y) coordinates into a single list of coordinates eg f([(x0,y0),(x1,y1),....]) --> [x0,y0,x1,y1,....] or g([x0,x1,x2,......],[y0,y1,y2,....]) --> [x0,y0,x1,y1,....] clearly if f is doable then g can be done using zip. I suppose this is a special case flatten, but can flatten be done fast? The python recipes seem rather slow compared to the builtin functions. well then: first of all, i need to say, if speed really matters, do it in C. that being said, python can be fast, too. for this task psyco is your friend. i got this output from the script given below: without psyco: flatten1: 2.78046748059 flatten2: 2.90226239686 flatten3: 4.91070862996 goopy_flatten1: 8.22951110963 goopy_flatten2: 8.56373180172 with psyco: flatten1: 1.17390339924 flatten2: 1.7209583052 flatten3: 1.18490295558 goopy_flatten1: 1.34892236194 goopy_flatten2: 1.68568386584 the goopy function is taken from the google-functional package (but is treated a bit unfair, i must admit, being wrapped in a lambda) so, what does that show us? izip seems a bit faster than zip with these input data. you want to do your own timings with more realistic data. and all these functions are what just came to my mind, i'm sure they can be improved. hope this helps, -- David. used script: ---------------------------------------------------------------- from itertools import izip xdata = range(1000) ydata = range(1000)[::-1] def flatten1(): return [x for pair in izip(xdata, ydata) for x in pair] def flatten2(): return [x for pair in zip(xdata, ydata) for x in pair] def flatten3(): res = [] for pair in izip(xdata, ydata): for x in pair: res.append(x) return res def goopy_flatten(seq): lst = [] for x in seq: if type(x) is list or type(x) is tuple: for val in x: lst.append(val) else: lst.append(x) return lst goopy_flatten1 = lambda: goopy_flatten(izip(xdata, ydata)) goopy_flatten2 = lambda: goopy_flatten(zip(xdata, ydata)) if __name__=='__main__': from timeit import Timer functions = ['flatten1', 'flatten2', 'flatten3', 'goopy_flatten1', 'goopy_flatten2'] print 'without psyco:' print for fn in functions: t = Timer(fn+'()', 'from __main__ import '+fn) print fn+':', t.timeit(5000) try: import psyco; psyco.full() except ImportError: pass print print 'with psyco:' print for fn in functions: t = Timer(fn+'()', 'from __main__ import '+fn) print fn+':', t.timeit(5000) Jan 12 '06 #3

 P: n/a Robin Becker writes: f([(x0,y0),(x1,y1),....]) --> [x0,y0,x1,y1,....] import operator a=[(1,2),(3,4),(5,6)] reduce(operator.add,a) (1, 2, 3, 4, 5, 6) Jan 12 '06 #4

 P: n/a Paul Rubin writes: import operator a=[(1,2),(3,4),(5,6)] reduce(operator.add,a) (1, 2, 3, 4, 5, 6) (Note that the above is probably terrible if the lists are large and you're after speed.) Jan 12 '06 #5

 P: n/a > Robin Becker schrieb: Is there some smart/fast way to flatten a level one list using the latest iterator/generator idioms. .... David Murmann wrote: Some functions and timings .... Here are some more timings of David's functions, and a couple of additional contenders that time faster on my box (I don't have psyco): # From David Murman from itertools import izip xdata = range(1000) ydata = range(1000)[::-1] def flatten1(x, y): return [i for pair in izip(x, y) for i in pair] def flatten2(x, y): return [i for pair in zip(x, y) for i in pair] def flatten3(x, y): res = [] for pair in izip(x, y): for i in pair: res.append(i) return res # New attempts: from itertools import imap def flatten4(x, y): l = [] list(imap(l.extend, izip(x, y))) return l from Tkinter import _flatten def flatten5(x, y): return list(_flatten(zip(x, y))) flatten_funcs = [flatten1, flatten2, flatten3, flatten4, flatten5] def testthem(): flatten1res = flatten_funcs[0](xdata, ydata) for func in flatten_funcs: assert func(xdata, ydata) == flatten1res def timethem(): for func in flatten_funcs: print shell.timefunc(func, xdata, ydata) testthem() timethem() flatten1(...) 704 iterations, 0.71msec per call flatten2(...) 611 iterations, 0.82msec per call flatten3(...) 344 iterations, 1.46msec per call flatten4(...) 1286 iterations, 389.08usec per call flatten5(...) 1219 iterations, 410.24usec per call Michael Jan 12 '06 #6

 P: n/a Michael Spencer wrote: > Robin Becker schrieb: >> Is there some smart/fast way to flatten a level one list using the >> latest iterator/generator idioms. ... David Murmann wrote: > Some functions and timings ... Here's one more that's quite fast using Psyco, but only average without it. def flatten6(): n = min(len(xdata), len(ydata)) result = [None] * (2*n) for i in xrange(n): result[2*i] = xdata[i] result[2*i+1] = ydata[i] -tim Here are some more timings of David's functions, and a couple of additional contenders that time faster on my box (I don't have psyco): # From David Murman from itertools import izip xdata = range(1000) ydata = range(1000)[::-1] def flatten1(x, y): return [i for pair in izip(x, y) for i in pair] def flatten2(x, y): return [i for pair in zip(x, y) for i in pair] def flatten3(x, y): res = [] for pair in izip(x, y): for i in pair: res.append(i) return res # New attempts: from itertools import imap def flatten4(x, y): l = [] list(imap(l.extend, izip(x, y))) return l from Tkinter import _flatten def flatten5(x, y): return list(_flatten(zip(x, y))) flatten_funcs = [flatten1, flatten2, flatten3, flatten4, flatten5] def testthem(): flatten1res = flatten_funcs[0](xdata, ydata) for func in flatten_funcs: assert func(xdata, ydata) == flatten1res def timethem(): for func in flatten_funcs: print shell.timefunc(func, xdata, ydata) >>> testthem() >>> timethem() flatten1(...) 704 iterations, 0.71msec per call flatten2(...) 611 iterations, 0.82msec per call flatten3(...) 344 iterations, 1.46msec per call flatten4(...) 1286 iterations, 389.08usec per call flatten5(...) 1219 iterations, 410.24usec per call >>> Michael Jan 12 '06 #7

 P: n/a Tim Hochberg wrote: Michael Spencer wrote: > Robin Becker schrieb: >> Is there some smart/fast way to flatten a level one list using the >> latest iterator/generator idioms. ... David Murmann wrote: > Some functions and timings ... Here's one more that's quite fast using Psyco, but only average without it. def flatten6(): n = min(len(xdata), len(ydata)) result = [None] * (2*n) for i in xrange(n): result[2*i] = xdata[i] result[2*i+1] = ydata[i] -tim Indeed: I added yours to the list (after adding the appropriate return) testthem() timethem() flatten1(...) 702 iterations, 0.71msec per call flatten2(...) 641 iterations, 0.78msec per call flatten3(...) 346 iterations, 1.45msec per call flatten4(...) 1447 iterations, 345.66usec per call flatten5(...) 1218 iterations, 410.55usec per call flatten6(...) 531 iterations, 0.94msec per call (See earlier post for flatten1-5) Michael Jan 12 '06 #8

 P: n/a Paul Rubin wrote: Paul Rubin writes:>import operator>a=[(1,2),(3,4),(5,6)]>reduce(operator.add,a)(1, 2, 3, 4, 5, 6) (Note that the above is probably terrible if the lists are large and you're after speed.) yes, and it is all in C and so could be a contender for the speed champ. I guess what you're saying is that it's doing (1,2) (1,2)+(3,4) (1,2,3,4)+(5,6) ie we do n or n-1 tuple additions each of which requires tuple allocation etc etc A fast implementation would probably allocate the output list just once and then stream the values into place with a simple index. -- Robin Becker Jan 12 '06 #9

 P: n/a Another try: def flatten6(x, y): return list(chain(*izip(x, y))) (any case, this is shorter ;-) Cyril On 1/12/06, Michael Spencer wrote: Tim Hochberg wrote: Michael Spencer wrote: > Robin Becker schrieb: >> Is there some smart/fast way to flatten a level one list using the >> latest iterator/generator idioms. ... David Murmann wrote: > Some functions and timings ... Here's one more that's quite fast using Psyco, but only average withoutit. def flatten6(): n = min(len(xdata), len(ydata)) result = [None] * (2*n) for i in xrange(n): result[2*i] = xdata[i] result[2*i+1] = ydata[i] -tim Indeed: I added yours to the list (after adding the appropriate return) >>> testthem() >>> timethem() flatten1(...) 702 iterations, 0.71msec per call flatten2(...) 641 iterations, 0.78msec per call flatten3(...) 346 iterations, 1.45msec per call flatten4(...) 1447 iterations, 345.66usec per call flatten5(...) 1218 iterations, 410.55usec per call flatten6(...) 531 iterations, 0.94msec per call >>> (See earlier post for flatten1-5) Michael -- http://mail.python.org/mailman/listinfo/python-list Jan 12 '06 #10

 P: n/a Tim Hochberg wrote: Here's one more that's quite fast using Psyco, but only average without it. def flatten6(): n = min(len(xdata), len(ydata)) result = [None] * (2*n) for i in xrange(n): result[2*i] = xdata[i] result[2*i+1] = ydata[i] I you require len(xdata) == len(ydata) there's an easy way to move the loop into C: def flatten7(): n = len(xdata) assert len(ydata) == n result = [None] * (2*n) result[::2] = xdata result[1::2] = ydata return result \$ python -m timeit 'from flatten import flatten6 as f' 'f()' 1000 loops, best of 3: 847 usec per loop \$ python -m timeit 'from flatten import flatten7 as f' 'f()' 10000 loops, best of 3: 43.9 usec per loop Peter Jan 12 '06 #11

 P: n/a Robin Becker writes:>>reduce(operator.add,a) ... A fast implementation would probably allocate the output list just once and then stream the values into place with a simple index. That's what I hoped "sum" would do, but instead it barfs with a type error. So much for duck typing. Jan 12 '06 #12

 P: n/a Robin Becker wrote: Paul Rubin wrote: Paul Rubin writes:>>import operator>>a=[(1,2),(3,4),(5,6)]>>reduce(operator.add,a)(1, 2, 3, 4, 5, 6) (Note that the above is probably terrible if the lists are large and you're after speed.) yes, and it is all in C and so could be a contender for the speed champ. I guess what you're saying is that it's doing That is what I thought too but seems that [x for pair in li for x in pair] is the fastest on my machine and what is even stranger is that if I use psyco.full(), I got a 10x speed up for this solution(list comprehension) which is head and shoulder above all the other suggested so far. Jan 12 '06 #13

 P: n/a Well, maybe it's time to add a n-levels flatten() function to the language (or to add it to itertools). Python is open source, but I am not able to modify its C sources yet... Maybe Raymond Hettinger can find some time to do it for Py 2.5. Bye, bearophile Jan 12 '06 #14

 P: n/a [Robin Becker] Is there some smart/fast way to flatten a level one list using the latest iterator/generator idioms. The problem arises in coneverting lists of (x,y) coordinates into a single list of coordinates eg f([(x0,y0),(x1,y1),....]) --> [x0,y0,x1,y1,....] Here's one way: d = [('x0','y0'), ('x1','y1'), ('x2','y2'), ('x3', 'y3')] list(chain(*d)) ['x0', 'y0', 'x1', 'y1', 'x2', 'y2', 'x3', 'y3'] FWIW, if you're into working out puzzles, there's no end of interesting iterator algebra tricks. Here are a few identities for your entertainment: # Given s (any sequence) and n (a non-negative integer): assert zip(*izip(*tee(s,n))) == [tuple(s)]*n assert list(chain(*tee(s,n))) == list(s)*n assert map(itemgetter(0),groupby(sorted(s))) == sorted(set(s)) Raymond Jan 12 '06 #15

 P: n/a In article <7x************@ruckus.brouhaha.com>, Paul Rubin wrote:Robin Becker writes: >>>>>reduce(operator.add,a) ...That's what I hoped "sum" would do, but instead it barfs with a typeerror. So much for duck typing. sum(...) sum(sequence, start=0) -> value If you're using sum() as a 1-level flatten you need to give it start=[]. -- \S -- si***@chiark.greenend.org.uk -- http://www.chaos.org.uk/~sion/ ___ | "Frankly I have no feelings towards penguins one way or the other" \X/ | -- Arthur C. Clarke her nu becomež se bera eadward ofdun hlęddre heafdes bęce bump bump bump Jan 12 '06 #16

 P: n/a Sion Arrowsmith writes: sum(sequence, start=0) -> value If you're using sum() as a 1-level flatten you need to give it start=[]. Oh, right, I should have remembered that. Thanks. Figuring out whether it's quadratic or linear would still take an experiment or code inspection which I'm not up for at the moment. Jan 12 '06 #17

 P: n/a Peter Otten wrote: Tim Hochberg wrote:Here's one more that's quite fast using Psyco, but only average withoutit. def flatten6(): n = min(len(xdata), len(ydata)) result = [None] * (2*n) for i in xrange(n): result[2*i] = xdata[i] result[2*i+1] = ydata[i] I you require len(xdata) == len(ydata) there's an easy way to move the loop into C: def flatten7(): n = len(xdata) assert len(ydata) == n result = [None] * (2*n) result[::2] = xdata result[1::2] = ydata return result \$ python -m timeit 'from flatten import flatten6 as f' 'f()' 1000 loops, best of 3: 847 usec per loop \$ python -m timeit 'from flatten import flatten7 as f' 'f()' 10000 loops, best of 3: 43.9 usec per loop Peter That's the winner for my machine and it works in the case I need :) The numbers are microseconds/call I used 20 reps and n from 10 up to 1000. no psyco Name 10 20 100 200 500 1000 flatten1 111.383 189.298 745.709 1397.300 3499.579 6628.775 flatten2 142.923 209.496 907.182 1521.618 3565.397 7197.228 flatten3 176.224 314.342 1385.958 2733.560 6726.693 12879.067 flatten4 112.696 163.010 518.250 901.288 1979.749 3657.364 flatten5 78.334 110.768 386.949 711.794 1617.664 3255.749 flatten6 142.867 230.420 894.639 1767.012 4499.734 9017.906 flatten6a 163.093 263.330 1071.337 2084.287 5209.433 10383.610 flatten6b 180.582 275.761 1063.794 2074.705 5057.123 10043.567 flatten6c 167.898 253.664 974.202 1948.181 4821.339 9562.780 flatten6d 132.475 201.702 738.194 1406.659 3612.107 7242.038 flatten7 59.030 62.354 90.347 130.771 254.613 438.994 flatten8 88.978 173.737 1667.111 5674.297 28907.501 106330.749 flatten8a 107.388 225.951 2323.563 7088.136 34254.381 114538.384 psyco Name 10 20 100 200 500 1000 flatten1 84.424 114.596 393.374 714.728 1809.448 3197.837 flatten2 102.387 136.302 507.243 942.494 2276.770 4451.990 flatten3 85.206 111.020 379.713 715.957 1607.104 3191.188 flatten4 102.667 144.599 509.255 856.450 1839.591 3425.128 flatten5 79.898 115.490 383.904 730.484 1739.411 3515.978 flatten6 54.560 61.293 183.012 332.109 837.146 1604.366 flatten6a 79.647 108.114 405.107 752.917 1873.674 3824.620 flatten6b 111.746 132.978 473.189 907.378 2217.600 4257.357 flatten6c 98.756 110.629 376.724 730.037 1772.963 3524.247 flatten6d 59.253 69.199 172.731 295.820 717.577 1402.720 flatten7 51.291 39.754 65.707 104.902 233.214 405.694 flatten8 87.050 166.837 1665.407 5410.576 28459.567 107847.422 flatten8a 122.753 251.457 2766.944 7931.204 36353.503 120773.674 ############################### from itertools import izip import timeit _R=100 def flatten1(x, y): '''D Murman''' return [i for pair in izip(x, y) for i in pair] def flatten2(x, y): '''D Murman''' return [i for pair in zip(x, y) for i in pair] def flatten3(x, y): '''D Murman''' res = [] for pair in izip(x, y): for i in pair: res.append(i) return res # New attempts: from itertools import imap def flatten4(x, y): '''D Murman''' l = [] list(imap(l.extend, izip(x, y))) return l from Tkinter import _flatten def flatten5(x, y): '''D Murman''' return list(_flatten(zip(x, y))) def flatten6(x,y): '''Tim Hochberg''' n = min(len(x), len(y)) result = [None] * (2*n) for i in xrange(n): result[2*i] = xdata[i] result[2*i+1] = ydata[i] return result def flatten6a(x,y): '''Robin Becker variant of 6''' n = min(len(x), len(y)) result = [None] * (2*n) for i in xrange(n): result[2*i:2*i+2] = xdata[i],ydata[i] return result def flatten6b(x,y): '''Robin Becker variant of 6''' n = min(len(x), len(y)) result = [None] * (2*n) for i,pair in enumerate(zip(xdata,ydata)): result[2*i:2*i+2] = pair return result def flatten6c(x,y): '''Robin Becker variant of 6''' n = min(len(x), len(y)) result = [None] * (2*n) for i,pair in enumerate(izip(xdata,ydata)): result[2*i:2*i+2] = pair return result def flatten6d(x,y): '''Robin Becker variant of 6''' n = min(len(x), len(y)) result = [None] * (2*n) j = 0 for i in xrange(n): result[j] = xdata[i] result[j+1] = ydata[i] j+=2 return result from operator import add as operator_add def flatten8(x,y): '''Paul Rubin''' return reduce(operator_add,zip(x,y),()) def flatten8a(x,y): '''Robin Becker variant of 8''' return reduce(operator_add,(xy for xy in izip(x,y)),()) def flatten7(x,y): '''Peter Otten special case equal lengths''' n = len(x) assert len(y) == n result = [None] * (2*n) result[::2] = x result[1::2] = y return result funcs = [(n,v) for n,v in globals().items() if callable(v) and n.startswith('flatten')] funcs.sort() def testthem(): res0 = funcs[0][1](xdata, ydata) for name,func in funcs: res = list(func(xdata, ydata)) if res!=res0: print name,' fails', type(res0), type(res), res0[:5],res[:5], res0[-5:],res[-5:] def timethem(D,n): for name,func in funcs: t = timeit.Timer(name+"(xdata,ydata)",'from __main__ import xdata,ydata,'+name) D.setdefault(name,{})[n] = 1e7*t.timeit(_R)/float(_R) if __name__=='__main__': N = [10, 20, 100, 200, 500, 1000] xdata = range(N[-1]) ydata = xdata[::-1] testthem() for p in 'no psyco','psyco': D={} if p=='psyco': import psyco psyco.full() for n in N: xdata = range(n) ydata = xdata[::-1] timethem(D,n) print '\n',p fmt = '%%%ds' % max(map(len,[x[0] for x in funcs])) print (fmt + len(N)*' %9d') % (('Name',)+tuple(N)) fmt1 = fmt + len(N)*' %9.3f' for name,func in funcs: print fmt1 % tuple([name]+[D[name][n] for n in N]) print ############################### -- Robin Becker Jan 12 '06 #18

 P: n/a Robin Becker schrieb: # New attempts: from itertools import imap def flatten4(x, y): '''D Murman''' l = [] list(imap(l.extend, izip(x, y))) return l from Tkinter import _flatten def flatten5(x, y): '''D Murman''' return list(_flatten(zip(x, y))) well, i would really like to take credit for these, but they're not mine ;) (credit goes to Michael Spencer). i especially like flatten4, even if its not as fast as the phenomenally faster flatten7. -- David Murmann (NN!) ;) Jan 12 '06 #19

 P: n/a Sion Arrowsmith wrote: sum(...) sum(sequence, start=0) -> value If you're using sum() as a 1-level flatten you need to give it start=[]. Except if you are trying to sum arrays of strings... sum(["a","b","c"], "") Traceback (most recent call last): File "", line 1, in ? TypeError: sum() can't sum strings [use ''.join(seq) instead] I've no idea why this limitation is here... perhaps it is because pre python2.4 calling += on strings was very slow? -- Nick Craig-Wood -- http://www.craig-wood.com/nick Jan 12 '06 #20

 P: n/a David Murmann wrote: Robin Becker schrieb: # New attempts: from itertools import imap def flatten4(x, y): '''D Murman''' l = [] list(imap(l.extend, izip(x, y))) return l from Tkinter import _flatten def flatten5(x, y): '''D Murman''' return list(_flatten(zip(x, y))) well, i would really like to take credit for these, but they're not mine ;) (credit goes to Michael Spencer). i especially like flatten4, even if its not as fast as the phenomenally faster flatten7. Me too. And expand a bit on flatten4, I got this interesting result. bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools; a=zip(xrange(1000),xrange(1000))" "l=len(a); li=[None]*l*2;li[::2]=range(1000); li[1::2]=range(1000)" 1000 loops, best of 3: 318 usec per loop bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools,psyco; a=zip(xrange(1000),xrange(1000));li=[]" "filter(li.extend,a)" 1000 loops, best of 3: 474 usec per loop Still 50% slower but it has the advantage that it works on all kinds of sequence as they can be efficiently izip() together. Jan 13 '06 #22

 P: n/a Nick Craig-Wood wrote: Sion Arrowsmith wrote: sum(...) sum(sequence, start=0) -> value If you're using sum() as a 1-level flatten you need to give it start=[]. Except if you are trying to sum arrays of strings... >>> sum(["a","b","c"], "") Traceback (most recent call last): File "", line 1, in ? TypeError: sum() can't sum strings [use ''.join(seq) instead] >>> I've no idea why this limitation is here... perhaps it is because pre python2.4 calling += on strings was very slow? No: when I implemented sum, I originally specialcased sum on strings to map down to a ''.join -- Guido decided it was confusing and had no advantage wrt calling ''.join directly so he made me put in that check. Alex Jan 13 '06 #23

 P: n/a bo****@gmail.com wrote: David Murmann wrote: > # New attempts: > from itertools import imap > def flatten4(x, y): > '''D Murman''' > l = [] > list(imap(l.extend, izip(x, y))) > return l well, i would really like to take credit for these, but they're not mine ;) (credit goes to Michael Spencer). i especially like flatten4, even if its not as fast as the phenomenally faster flatten7. Me too. And expand a bit on flatten4, I got this interesting result. bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools; a=zip(xrange(1000),xrange(1000))" "l=len(a); li=[None]*l*2;li[::2]=range(1000); li[1::2]=range(1000)" 1000 loops, best of 3: 318 usec per loop bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools,psyco; a=zip(xrange(1000),xrange(1000));li=[]" "filter(li.extend,a)" 1000 loops, best of 3: 474 usec per loop For a fair comparison you'd have to time the zip operation. Still 50% slower but it has the advantage that it works on all kinds of sequence as they can be efficiently izip() together. Creating a list via list/map/filter just for the side effect is not only bad taste, ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];ext=lst.extend' 'for i in a: ext(i)' 1000000 loops, best of 3: 1.23 usec per loop ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];filter(lst.extend, a)' 1000000 loops, best of 3: 1.63 usec per loop it is also slower than an explicit loop. Don't do it. Peter Jan 13 '06 #24

 P: n/a Peter Otten wrote: bo****@gmail.com wrote: David Murmann wrote: > # New attempts: > from itertools import imap > def flatten4(x, y): > '''D Murman''' > l = [] > list(imap(l.extend, izip(x, y))) > return l well, i would really like to take credit for these, but they're not mine ;) (credit goes to Michael Spencer). i especially like flatten4, even if its not as fast as the phenomenally faster flatten7. Me too. And expand a bit on flatten4, I got this interesting result. bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools; a=zip(xrange(1000),xrange(1000))" "l=len(a); li=[None]*l*2;li[::2]=range(1000); li[1::2]=range(1000)" 1000 loops, best of 3: 318 usec per loop bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools,psyco; a=zip(xrange(1000),xrange(1000));li=[]" "filter(li.extend,a)" 1000 loops, best of 3: 474 usec per loop For a fair comparison you'd have to time the zip operation. Still 50% slower but it has the advantage that it works on all kinds of sequence as they can be efficiently izip() together. Creating a list via list/map/filter just for the side effect is not only bad taste, ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];ext=lst.extend' 'for i in a: ext(i)' 1000000 loops, best of 3: 1.23 usec per loop ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];filter(lst.extend, a)' 1000000 loops, best of 3: 1.63 usec per loop it is also slower than an explicit loop. Don't do it. ah, stand corrected. Jan 13 '06 #25

 P: n/a Peter Otten wrote: bo****@gmail.com wrote: David Murmann wrote: > # New attempts: > from itertools import imap > def flatten4(x, y): > '''D Murman''' > l = [] > list(imap(l.extend, izip(x, y))) > return l well, i would really like to take credit for these, but they're not mine ;) (credit goes to Michael Spencer). i especially like flatten4, even if its not as fast as the phenomenally faster flatten7. Me too. And expand a bit on flatten4, I got this interesting result. bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools; a=zip(xrange(1000),xrange(1000))" "l=len(a); li=[None]*l*2;li[::2]=range(1000); li[1::2]=range(1000)" 1000 loops, best of 3: 318 usec per loop bonono@moresing:~/bonobo/psp\$ python ~/lib/python2.4/timeit.py -s "import itertools,psyco; a=zip(xrange(1000),xrange(1000));li=[]" "filter(li.extend,a)" 1000 loops, best of 3: 474 usec per loop For a fair comparison you'd have to time the zip operation. Still 50% slower but it has the advantage that it works on all kinds of sequence as they can be efficiently izip() together. Creating a list via list/map/filter just for the side effect is not only bad taste, ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];ext=lst.extend' 'for i in a: ext(i)' 1000000 loops, best of 3: 1.23 usec per loop ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];filter(lst.extend, a)' 1000000 loops, best of 3: 1.63 usec per loop it is also slower than an explicit loop. Don't do it. Hi, but I found this result instead : bonono@moresing:~\$ python ~/lib/python2.4/timeit.py -s "a=range(10000); b=zip(*[a]*2); l=[None]*len(a)*2; e=l.extend" "l[::2]=b;l[1::2]=b" 100 loops, best of 3: 6.22 msec per loop bonono@moresing:~\$ python ~/lib/python2.4/timeit.py -s "a=range(10000); b=zip(*[a]*2); l=[]; e=l.extend" "filter(e,b)" 100 loops, best of 3: 7.25 msec per loop bonono@moresing:~\$ python ~/lib/python2.4/timeit.py -s "a=range(10000); b=zip(*[a]*2); l=[]; e=l.extend" "for x in b: e(x)" 100 loops, best of 3: 10.7 msec per loop bonono@moresing:~\$ So it seems to be faster than explicit loop. By localizing the l.extend name binding, its speed is only 20% slower than the fastest method. May be I have done something wrong in the test ? Jan 13 '06 #27

 P: n/a bo****@gmail.com wrote: Creating a list via list/map/filter just for the side effect is not only bad taste, ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];ext=lst.extend' 'for i in a: ext(i)' 1000000 loops, best of 3: 1.23 usec per loop ~ \$ python -m timeit -s'a = zip([range(1000)]*2)' 'lst=[];filter(lst.extend, a)' 1000000 loops, best of 3: 1.63 usec per loop it is also slower than an explicit loop. Don't do it. Hi, but I found this result instead : bonono@moresing:~\$ python ~/lib/python2.4/timeit.py -s "a=range(10000); b=zip(*[a]*2); l=[None]*len(a)*2; e=l.extend" "l[::2]=b;l[1::2]=b" 100 loops, best of 3: 6.22 msec per loop bonono@moresing:~\$ python ~/lib/python2.4/timeit.py -s "a=range(10000); b=zip(*[a]*2); l=[]; e=l.extend" "filter(e,b)" 100 loops, best of 3: 7.25 msec per loop bonono@moresing:~\$ python ~/lib/python2.4/timeit.py -s "a=range(10000); b=zip(*[a]*2); l=[]; e=l.extend" "for x in b: e(x)" 100 loops, best of 3: 10.7 msec per loop bonono@moresing:~\$ So it seems to be faster than explicit loop. By localizing the l.extend name binding, its speed is only 20% slower than the fastest method. May be I have done something wrong in the test ? I hate to admit it but /my/ post had a bug: zip([range(1000)]*2) should have been zip(*[range(1000)]*2). filter() may be ugly, but faster it is. Peter Jan 13 '06 #28