Michael Spencer wrote: > Peter Otten wrote: > >> If 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
> Very nice observation, Peter. Thank you :-) > Easily the "winner". It reminds me of > str.translate beating python loops in filtering applications: > http://groups.google.com/group/comp.lang.python/msg/e23cdc374144a4bd > What's more, you can generalize your approach to any number of sequences > and un-equal lengths, with only modest loss of speed: > > def interleave(*args, **kw): > """Peter Otten flatten7 (generalized by Michael Spencer) > Interleave any number of sequences, padding shorter sequences if kw > pad is supplied""" > dopad = "pad" in kw > pad = dopad and kw["pad"] > count = len(args) > lengths = map(len, args) > maxlen = max(lengths) > result = maxlen*count*[None] > for ix, input in enumerate(args): > try: > result[ix::count] = input > except ValueError: > if dopad: > result[ix::count] = input + [pad]*(maxlen-lengths[ix]) > else: > raise > return result You don't loose speed because the expensive operation is only executed if list lengths differ. Two observations: - list arithmetic like 'input + [pad]*(maxlen-lengths[ix])' should and can be avoided - You are padding twice -- once with None, and then with the real thing. def interleave2(*args, **kw): dopad = "pad" in kw pad = kw.get("pad") count = len(args) lengths = map(len, args) maxlen = max(lengths) if not dopad and min(lengths) != maxlen: raise ValueError result = maxlen*count*[pad] for ix, input in enumerate(args): result[ix:len(input)*count:count] = input return result $ python -m timeit -s 'from interleave_spencer import xdata, ydata, interleave as i;xdata=xdata[:-1]' 'i(xdata, ydata, pad=None)' 10000 loops, best of 3: 69.7 usec per loop $ python -m timeit -s 'from interleave_spencer import xdata, ydata, interleave2 as i;xdata=xdata[:-1]' 'i(xdata, ydata, pad=None)' 10000 loops, best of 3: 46.4 usec per loop Not overwhelming, but I expect the difference to grow when the arguments occupy a significant portion of the available memory. Peter -- http://mail.python.org/mailman/listinfo/python-list
