question about list extension
in [Python]
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From: J on 16 Apr 2010 09:41 Ok... I know pretty much how .extend works on a list... basically it just tacks the second list to the first list... like so: >>> lista=[1] >>> listb=[2,3] >>> lista.extend(listb) >>> print lista; [1, 2, 3] what I'm confused on is why this returns None: >>> lista=[1] >>> listb=[2,3] >>> print lista.extend(listb) None >>> print lista [1, 2, 3] So why the None? Is this because what's really happening is that extend() and append() are directly manipulating the lista object and thus not actuall returning a new object? Even if that assumption of mine is correct, I would have expected something like this to work: >>> lista=[1] >>> listb=[2,3] >>> print (lista.extend(listb)) None The reason this is bugging me right now is that I'm working on some code. My program has a debugger class that takes a list as it's only (for now) argument. That list, is comprised of messages I want to spit out during debug, and full output from system commands being run with Popen... So the class looks something like this: class debugger(): def printout(self,info): for line in info: print "DEBUG: %s" % line and later on, it get's called like this meminfo = Popen('cat /proc/meminfo',shell=True,stdout=PIPE).communicate[0].splitlines() if debug: debugger.printout(meminfo) easy enough.... BUT, what if I have several lists I want to pass on multiple lists... So changing debugger.printout() to: def printout(self,*info): lets me pass in multiple lists... and I can do this: for tlist in info: for item in tlist: print item which works well... So, what I'm curious about, is there a list comprehension or other means to reduce that to a single line? I tried this, which didn't work because I'm messing up the syntax, somehow: def func(*info) print [ i for tlist in info for i in tlist ] so can someone help me understand a list comprehension that will turn those three lines into on? It's more of a curiosity thing at this point... and not a huge difference in code... I was just curious about how to make that work. Cheers, Jeff
From: Bruno Desthuilliers on 16 Apr 2010 10:23 J a �crit : > Ok... I know pretty much how .extend works on a list... basically it > just tacks the second list to the first list... like so: > >>>> lista=[1] >>>> listb=[2,3] >>>> lista.extend(listb) >>>> print lista; > [1, 2, 3] > > what I'm confused on is why this returns None: > So why the None? Is this because what's really happening is that > extend() and append() are directly manipulating the lista object and > thus not actuall returning a new object? Exactly. > Even if that assumption of mine is correct, I would have expected > something like this to work: > >>>> lista=[1] >>>> listb=[2,3] >>>> print (lista.extend(listb)) > None So what ? It JustWork(tm). list.extend returns None, so "None" is printed !-) (snip) > So changing debugger.printout() to: > > def printout(self,*info): > > lets me pass in multiple lists... and I can do this: > > for tlist in info: > for item in tlist: > print item > > which works well... > > So, what I'm curious about, is there a list comprehension or other > means to reduce that to a single line? Why do you think the above code needs to be "reduced to a single line" ? What's wrong with this snippet ??? > It's more of a curiosity thing at this point... and not a huge > difference in code... I was just curious about how to make that work. Uh, ok. What about: from itertools import chain def printout(*infos): print "\n".join("%s" % item for item in chain(*infos)) HTH
From: Lie Ryan on 16 Apr 2010 10:37 On 04/16/10 23:41, J wrote: > Ok... I know pretty much how .extend works on a list... basically it > just tacks the second list to the first list... like so: > >>>> lista=[1] >>>> listb=[2,3] >>>> lista.extend(listb) >>>> print lista; > [1, 2, 3] > > what I'm confused on is why this returns None: > >>>> lista=[1] >>>> listb=[2,3] >>>> print lista.extend(listb) > None >>>> print lista > [1, 2, 3] > > So why the None? Is this because what's really happening is that > extend() and append() are directly manipulating the lista object and > thus not actuall returning a new object? In python every function that does not explicitly returns a value or use a bare return returns None. So: def foo(): pass def bar(): return print foo() print bar() you can say that returning None is python's equivalent to void return type in other languages. > Even if that assumption of mine is correct, I would have expected > something like this to work: > >>>> lista=[1] >>>> listb=[2,3] >>>> print (lista.extend(listb)) > None Why would these has to be different? print None print (None) <snip> > So, what I'm curious about, is there a list comprehension or other > means to reduce that to a single line? from itertools import chain def printout(*info): print '\n'.join(map(str, chain(*info))) or using generator comprehension from itertools import chain def printout(*info): print '\n'.join(str(x) for x in chain(*info))
From: J. Cliff Dyer on 16 Apr 2010 10:59 On Sat, 2010-04-17 at 00:37 +1000, Lie Ryan wrote: > On 04/16/10 23:41, J wrote: > > So, what I'm curious about, is there a list comprehension or other > > means to reduce that to a single line? > > from itertools import chain > def printout(*info): > print '\n'.join(map(str, chain(*info))) > > or using generator comprehension > > from itertools import chain > def printout(*info): > print '\n'.join(str(x) for x in chain(*info)) It's even easier if you don't need to modify lista. print lista + listb
From: Terry Reedy on 16 Apr 2010 15:16
On 4/16/2010 9:41 AM, J wrote: > Ok... I know pretty much how .extend works on a list... basically it > just tacks the second list to the first list... like so: > >>>> lista=[1] >>>> listb=[2,3] >>>> lista.extend(listb) >>>> print lista; > [1, 2, 3] This shows right here that lista is extended in place. If you are not convinced, print(id(lista)) before and after. > what I'm confused on is why this returns None: > >>>> lista=[1] >>>> listb=[2,3] >>>> print lista.extend(listb) > None It is conventional in Python (at least the stdlib) that methods that mutate mutable objects 'in-place' return None to clearly differentiate them from methods that return new objects. There are pluses and minuses but such it is. Terry Jan Reedy |