f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
Packing tail recursion into one line is bad for both understanding and
refactoring. Use better names and a docstring gives
def group(seq, n):
'Yield from seq successive disjoint slices of length n plus the
remainder'
On 6/6/2011 9:42 AM, jyoun...@kc.rr.com wrote:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
Packing tail recursion into one line is bad for both understanding and
refactoring. Use better names and a docstring gives
def group(seq, n):
'Yield from seq successive
On Jun 5, 11:33 pm, Terry Reedy tjre...@udel.edu wrote:
On 6/5/2011 5:31 AM, Alain Ketterlin wrote:
jyoun...@kc.rr.com writes:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
f=lambda ... statements are inferior for practical purposes to the
equivalent def f statements
On Mon, 06 Jun 2011 12:52:31 -0400, Terry Reedy wrote:
Let me add something not said much here about designing functions: start
with both a clear and succinct definition *and* test cases. (I only
started writing tests first a year ago or so.)
For any non-trivial function, I usually start by
Steven D'Aprano wrote:
For any non-trivial function, I usually start by writing the
documentation (a docstring and doctests) first. How else do you know what
the function is supposed to do if you don't have it documented?
Yes. In my early years I was no different than any other hacker in terms
On 6/6/2011 1:29 PM, rusi wrote:
On Jun 5, 11:33 pm, Terry Reedytjre...@udel.edu wrote:
Let me add something not said much here about designing functions: start
with both a clear and succinct definition *and* test cases. (I only
started writing tests first a year ago or so.)
I am still one
On 6/6/2011 12:52 PM, Terry Reedy wrote:
def group(seq, n):
'Yield from seq successive disjoint slices of length n the remainder'
if n=0: raise ValueError('group size must be positive')
for i in range(0,len(seq), n):
yield seq[i:i+n]
for inn,out in (
(('',1), []), # no input, no output
jyoun...@kc.rr.com writes:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
f(Hallo Welt, 3)
['Hal', 'lo ', 'Wel', 't']
http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-s
ized-chunks-in-python/312644
It doesn't work with a huge list,
On 2:59 PM, Ian Kelly wrote:
On Sat, Jun 4, 2011 at 12:09 PM, Chris Angelico ros...@gmail.com wrote:
Python doesn't seem to have an inbuilt function to divide strings in
this way. At least, I can't find it (except the special case where n
is 1, which is simply 'list(string)'). Pike allows you
On 6/5/2011 5:31 AM, Alain Ketterlin wrote:
jyoun...@kc.rr.com writes:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
f=lambda ... statements are inferior for practical purposes to the
equivalent def f statements because the resulting object is missing a
useful name
On Sun, Jun 5, 2011 at 3:46 AM, jyoun...@kc.rr.com wrote:
It doesn't work with a huge list, but looks like it could be handy in certain
circumstances. I'm trying to understand this code, but am totally lost. I
know a little bit about lambda, as well as the ternary operator, but how
does
jyoun...@kc.rr.com wrote:
I was surfing around looking for a way to split a list into equal
sections. I came upon this algorithm:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
f(Hallo Welt, 3)
['Hal', 'lo ', 'Wel', 't']
On Sat, Jun 4, 2011 at 12:09 PM, Chris Angelico ros...@gmail.com wrote:
Python doesn't seem to have an inbuilt function to divide strings in
this way. At least, I can't find it (except the special case where n
is 1, which is simply 'list(string)'). Pike allows you to use the
division operator:
jyoun...@kc.rr.com wrote:
I was surfing around looking for a way to split a list into equal
sections.
non-recursive, same-unreadeable (worse?) one liner alternative:
def chunks(s, j):
return [''.join(filter(None,c))for c in map(None,*(s[i::j]for i in
range(j)))]
--
By ZeD
--
On Fri, Jun 11, 2010 at 10:11 PM, Ian Kelly ian.g.ke...@gmail.com wrote:
On Fri, Jun 11, 2010 at 9:31 PM, Vincent Davis vinc...@vincentdavis.net
wrote:
Starting with an example.
In [23]: x = [1,2,3,4,4,4,5,5,3,2,2,]
In [24]: y = set(x)
In [25]: y
Out[25]: set([1, 2, 3, 4, 5])
In [26]: y2
On Fri, Jun 11, 2010 at 9:31 PM, Vincent Davis vinc...@vincentdavis.net wrote:
Starting with an example.
In [23]: x = [1,2,3,4,4,4,5,5,3,2,2,]
In [24]: y = set(x)
In [25]: y
Out[25]: set([1, 2, 3, 4, 5])
In [26]: y2 = len(set(x))
In [27]: y2
Out[27]: 5
How would I do the above y2 =
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