split up a list by condition?

[Reinhold Birkenfeld]

Hi,

while writing my solution for "The python way?", I came across this fragment:

vees = [c for c in wlist[::-1] if c in vocals]
cons = [c for c in wlist[::-1] if c not in vocals]

So I think: Have I overlooked a function which splits up a sequence into two,
based on a condition? Such as

vees, cons = split(wlist[::-1], lambda c: c in vocals)

Reinhold

 

[Grooooops]

Reinhold,
Thanks for your response in the previous thread.
Yours is an interesting question. I haven't come up with a solution,
but I did realize that in the previous problem, the source 'word'
doesn't really need to stay intact...
So perhaps a solution along these lines?
.... if a[1] in vowels:
.... vees.append(wlist.pop(a[0]))

I don't know if it's possible to cram a 'pop' command into the single
line solution though.

I look forward to seeing other tricks to this end...

 

[gene tani]

Sounds like itertools.groupby() is what you're looking for, just sort
your sequence on the grouping func that you will pass as 2nd arg to
groupby(), before you pass the sequence to groupby()

http://www.python.org/doc/2.4/whatsnew/node13.html
http://www.python.org/dev/doc/devel/lib/itertools-functions.html

 

[Raymond Hettinger]

while writing my solution for "The python way?", I came across this fragment:

vees = [c for c in wlist[::-1] if c in vocals]
cons = [c for c in wlist[::-1] if c not in vocals]

So I think: Have I overlooked a function which splits up a sequence
into two, based on a condition

Trying to compress too much into one line is not "the python way" ;-)


vees, cons = [], []
for c in reversed(wlist):
if c in vocals:
vees.append(c)
else:
cons.append(c)

 

[Duncan Booth]

Hi,

while writing my solution for "The python way?", I came across this
fragment:

vees = [c for c in wlist[::-1] if c in vocals]
cons = [c for c in wlist[::-1] if c not in vocals]

So I think: Have I overlooked a function which splits up a sequence
into two, based on a condition? Such as

vees, cons = split(wlist[::-1], lambda c: c in vocals)

Reinhold

If you really are being charged by the number of newline characters in your
code you could write:

wlist = list('The quick brown fox')
vowels = 'aeiuo'
cons = []
vees = [ c for c in wlist if c in vowels or cons.append(c) ]
cons ['T', 'h', ' ', 'q', 'c', 'k', ' ', 'b', 'r', 'w', 'n', ' ', 'f', 'x']
vees ['e', 'u', 'i', 'o', 'o']
cons = []
vees = [ c for c in wlist[::-1] if c in vowels or cons.append(c) ]
cons ['x', 'f', ' ', 'n', 'w', 'r', 'b', ' ', 'k', 'c', 'q', ' ', 'h', 'T']
vees

['o', 'o', 'i', 'u', 'e']

but every penny you save writing a one liner will be tuppence extra on
maintenance.

 

[Reinhold Birkenfeld]

while writing my solution for "The python way?", I came across this fragment:
vees = [c for c in wlist[::-1] if c in vocals]
cons = [c for c in wlist[::-1] if c not in vocals]

So I think: Have I overlooked a function which splits up a sequence
into two, based on a condition

Trying to compress too much into one line is not "the python way" ;-)

I know (there is a Guido quote about this, I just lost the source...)

vees, cons = [], []
for c in reversed(wlist):
if c in vocals:
vees.append(c)
else:
cons.append(c)

Well, I've found a uglier solution,

vees, cons = [], []
[(vees, cons)[ch in vocals].append(ch) for ch in wlist]

Reinhold

 

[Reinhold Birkenfeld]

Hi,

while writing my solution for "The python way?", I came across this
fragment:

vees = [c for c in wlist[::-1] if c in vocals]
cons = [c for c in wlist[::-1] if c not in vocals]

So I think: Have I overlooked a function which splits up a sequence
into two, based on a condition? Such as

vees, cons = split(wlist[::-1], lambda c: c in vocals)

Reinhold

If you really are being charged by the number of newline characters in your
code you could write:
[...]

but every penny you save writing a one liner will be tuppence extra on
maintenance.

This is clear. I actually wanted to know if there is a function which I
overlooked which does that, which wouldn't be a maintenance nightmare at all.

Reinhold

 

[Grooooops]

vees, cons = [], []

[(vees, cons)[ch in vocals].append(ch) for ch in wlist]

Wow, that's horribly twisted Reinhold...
I spent about an hour last night trying something similar, to no end...


Neat tricks people...
I like Duncan's use of "or" to solve it.
I didn't see that in the python docs on list comprehension.
Very cool.

There is a special place in my heart for obfuscated Python,
but of course, not in production code if there is a clearer solution
available.

 

[Paul McGuire]

Re-run this with the input string "The quick brown fox is named
'Aloysius'." and we discover that 'A', 'y', "'" and '.' are also
consonants. (Actually, this is bug-compatible with the OP's original
example.)

-- Paul

 

[Andrew Dalke]

So I think: Have I overlooked a function which splits up a sequence
into two, based on a condition? Such as

vees, cons = split(wlist[::-1], lambda c: c in vocals)

This is clear. I actually wanted to know if there is a function which I
overlooked which does that, which wouldn't be a maintenance nightmare at
all.

Not that I know of, but if there is one it should be named
"bifilter", or "difilter" if you prefer Greek roots.


def bifilter(test, seq):
passes = []
fails = []
for term in seq:
if test(term):
passes.append(term)
else:
fails.append(term)
return passes, fails

bifilter("aeiou".__contains__, "This is a test") (['i', 'i', 'a', 'e'], ['T', 'h', 's', ' ', 's', ' ', ' ', 't', 's', 't'])

Another implementation, though in this case I cheat because I
do the test twice, is
.... seq1, seq2 = tee(seq)
.... return ifilter(test, seq1), ifilterfalse(test, seq2)
....

bifilter("aeiou".__contains__, "This is another test")

map(list, _) [['i', 'i', 'a', 'o', 'e', 'e'], ['T', 'h', 's', ' ', 's', ' ', 'n', 't', 'h', 'r', ' ', 't', 's', 't']]

Andrew
(e-mail address removed)