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
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