Re: lambda strangeness??
On Sun, 27 Feb 2005 09:39:49 GMT, Roel Schroeven [EMAIL PROTECTED] wrote: adds = [lambda y: (y + n) for n in range(10)] You're picking it up not as y but as n, since n in the lambda is evaluated when you call the lambde, not when you define it. for n in range(10): def f(y, n=n): return y+n adds.append(f) adds = [lambda x, y=n: (x + y) for n in range(10)] Works perfectly, thanks Roel. I knew I'd got this to work in the past with a for loop so I knew it should be possible in a comprehension. I just forgot the default argument trick! And then confused myself by coincidentally using n both inside and outside the LC, thus apparently getting inconsistent results! Thanks again, Alan G. Author of the Learn to Program website http://www.freenetpages.co.uk/hp/alan.gauld -- http://mail.python.org/mailman/listinfo/python-list
Re: lambda strangeness??
On Sun, 27 Feb 2005 09:07:28 + (UTC), Alan Gauld [EMAIL PROTECTED] wrote: adds = [lambda y: (y + n) for n in range(10)] adds[0](0) 9 for n in range(5): print adds[n](42) ... 42 43 the for loop... It seems to somehow be related to the last value in the range(), am I somehow picking that up as y? Further exploration suggests I'm picking it up as n not y, if that indeed is what's happening... Alan G. Author of the Learn to Program website http://www.freenetpages.co.uk/hp/alan.gauld -- http://mail.python.org/mailman/listinfo/python-list
Re: lambda strangeness??
Alan Gauld wrote: I was playing with lambdas and list compregensions and came across this unexpected behaviour: adds = [lambda y: (y + n) for n in range(10)] adds[0](0) 9 for n in range(5): print adds[n](42) ... 42 43 44 45 46 adds[0](0) 4 Can anyone explain the different answers I'm getting? FWIW the behaviour I expected was what seems to happen inside the for loop... It seems to somehow be related to the last value in the range(), am I somehow picking that up as y? If so why? You're picking it up not as y but as n, since n in the lambda is evaluated when you call the lambde, not when you define it. Or is that just a coincidence? And why did it work inside the for loop? In the loop you are giving n exactly the values you intended it to have inside the lambda. Check what happens when you use a different loop variable: for i in range(5): print adds[i](0) 9 9 9 9 9 I guess you could something like this instead: adds=[] for n in range(10): def f(y, n=n): return y+n adds.append(f) adds[0](0) 0 adds[0](5) 5 adds[9](5) 14 -- Codito ergo sum Roel Schroeven -- http://mail.python.org/mailman/listinfo/python-list
Re: lambda strangeness??
Alan Gauld wrote: On Sun, 27 Feb 2005 09:07:28 + (UTC), Alan Gauld [EMAIL PROTECTED] wrote: adds = [lambda y: (y + n) for n in range(10)] adds[0](0) 9 for n in range(5): print adds[n](42) ... 42 43 the for loop... It seems to somehow be related to the last value in the range(), am I somehow picking that up as y? Further exploration suggests I'm picking it up as n not y, if that indeed is what's happening... Your intent is to create lambda's that are equivalent to adds[0] = lambda y: y + 0 adds[1] = lambda y: y + 1 etc. but what actually happens is that you get lambda's that are equivalent to adds[0] = lambda y: y + n adds[1] = lambda y: y + n etc. which obviously depend on the value of n at the moment you call the lambda. -- Codito ergo sum Roel Schroeven -- http://mail.python.org/mailman/listinfo/python-list
Re: lambda strangeness??
Alan Gauld [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] adds = [lambda y: (y + n) for n in range(10)] To bind the definition-time value of n to the function, [lambda y,n=n:(y+n) for n in range(10)] Terry J. Reedy -- http://mail.python.org/mailman/listinfo/python-list
