Duncan Booth wrote: > Daniel Schüle wrote: > > > hi, > > > > [...] > > > >>># pseudo code > >>>i=2 > >>>lst=[i**=2 while i<1000] > >>> > >>>of course this could be easily rewritten into > >>>i=2 > >>>lst=[] > >>>while i<1000: > >>> i**=2 > >>> lst.append(i) > >>> > >> > >> > >> Neither of these loops would terminate until memory is exhausted. Do > >> you have a use case for a 'while' in a list comprehension which would > >> terminate? > > > > unless I am missing something obvious, I can not see why the loop > > should not terminate > > sure pseudo code is not executable but the other one works > > while tests the boolean expression first then decides whether to > > execute the body or not, in particular no next-iterator is > > involved(??) as it would be in > > lst=range(5) > > for i in lst: > > del lst[0] > > > Yes, I wasn't paying attention. The pseudo-code fails because you would > need to support assignment and generating a value inside the loop, but the > simple loop does work. > > The simple case in the pseudo-coded loop isn't nearly general enough, what > if you wanted a tuple, or i**2+1 as each loop value. You would need syntax > which generate more complex list values e.g: > > lst = [ (i, i**2) while i < 1000; i**=2 ] > > One current way to write this would be: > > def squaring(start, end): > i = start > while i < end: > yield i > i = i**2 > > Then you can do: > > lst = [(i, i**2) for i in squaring(1, 1000) ] > > which has the advantage of pulling all the complex logic out of the list > comprehension.
If one really wants a very messy one-liner, it is possible import operator x=[2] lst=list(((operator.setitem(x,0,x[0]**2),x[0])[1] for _ in xrange(10000000) if x[0] < 1000 or iter([]).next())) -- http://mail.python.org/mailman/listinfo/python-list
