Is there any nice way to unpack a list of unknown size??

[Fredrik Lundh]

1. first, second, third, *rest = foo

update to Python 3.0 (as others have pointed out), or just do

first, second, third = foo[:3]
rest = foo[3:]

for item in list_of_lists:
a, b, c = item[:3]
rest = item[3:]
...

and move on to more interesting parts of your program.

</F>

[Tim Chase]

1. first, second, third, *rest = foo

Python 3.0 has exactly this feature. No current Python 2.x version has it.

I asked something similar[1] on c.l.p a while back and Diez
Roggisch gave this nice workaround/hack[2]

It's a bit ugly in the implementation (sniffing the stack), but
elegant in the use, doing exactly what I needed.

It doesn't accommodate the "rest" portion that you reference, but
allows for arbitrary unpacking into a fixed-length tuple.

-tkc


[1]
http://mail.python.org/pipermail/python-list/2006-May/381386.html

[2]
http://mail.python.org/pipermail/python-list/2006-May/381399.html

[Gary Herron]

1. first, second, third, *rest = foo

Python 3.0 has exactly this feature. No current Python 2.x version has it.

Gary Herron

Please suggest.
Thanks,
Srini
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[Grant Edwards]

Post by Fredrik Lundh

1. first, second, third, *rest = foo

update to Python 3.0 (as others have pointed out), or just do
first, second, third = foo[:3]
rest = foo[3:]

Of course you can do that in one line if you want it to look a
bit more like the original pseudocode:

(a,b,c),rest = foo[:3],foo[3:]

That still requires you to manually count the number of
"non-rest" destination elements on the LHS and type that number
twice on the RHS. If you wanted to elminate a tiny bit of the
redundancy you could define a split() function:

def split(seq,n):
return seq[:n],seq[n:]

(a,b,c),rest = split(foo,3)

[srinivasan srinivas]

I want to do something like below:

1. first, second, third, *rest = foo

?2. for (a,b,c,*rest) in list_of_lists:

Please suggest.

Thanks,
Srini


Bring your gang together. Do your thing. Find your favourite Yahoo! group at http://in.promos.yahoo.com/groups/

[Arnaud Delobelle]

1. first, second, third, *rest = foo

Python 3.0 has exactly this feature. ?No current Python 2.x version has it.
Gary Herron

Please suggest.
Thanks,
Srini
? ? ? Bring your gang together. Do your thing. Find your favourite Yahoo! group athttp://in.promos.yahoo.com/groups/
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In python >= 2.4, you can define a function like this:

def truncate(iterable, n=1):
iterator = iter(iterable)
for i in iterator:
if n == 0:
yield iterator
return
yield i
n -= 1

a, b, c, tail = truncate([1,2,3,4,5,6], 3)
a

1

b

2

c

3

tail

<listiterator object at 0x78990>

list(tail)

[5, 6]

--
Arnaud