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list comprehension without if but with else

My question aims to use the else condition of a for-loop in a list comprehension.
example:
empty_list = []
def example_func(text):
for a in text.split():
for b in a.split(","):
empty_list.append(b)
else:
empty_list.append(" ")
I would like to make it cleaner by using a list comprehension with both for-loops.
But how can I do this by including an escape-clause for one of the loops (in this case the 2nd).
I know I can use if with and without else in a list comprehension. But how about using else without an if statement.
Is there a way, so the interpreter will understand it as escape-clause of a for loop?
Any help is much appreciated!
EDIT:
Thanks for the answers! In fact im trying to translate morse code.
The input is a string, containing morse codes.
Each word is separated by 3 spaces. Each letter of each word is separated by 1 space.
def decoder(code):
str_list = []
for i in code.split(" "):
for e in i.split():
str_list.append(morse_code_dic[e])
else:
str_list.append(" ")
return "".join(str_list[:-1]).capitalize()
print(decoder(".. - .-- .- ... .- --. --- --- -.. -.. .- -.--"))
I want to break down the whole sentence into words, then translate each word.
After the inner loop (translation of one word) is finished, it will launch its escape-clause else, adding a space, so that the structure of the whole sentence will be preserved. That way, the 3 Spaces will be translated to one space.

As noted in comments, that else does not really make all that much sense, since the purpose of an else after a for loop is actually to hold code for conditional execution if the loop terminates normally (i.e. not via break), which your loop always does, thus it is always executed.
So this is not really an answer to the question how to do that in a list comprehension, but more of an alternative. Instead of adding spaces after all words, then removing the last space and joining everything together, you could just use two nested join generator expressions, one for the sentence and one for the words:
def decoder(code):
return " ".join("".join(morse_code_dic[e] for e in i.split())
for i in code.split(" ")).capitalize()

As mentioned in the comments, the else clause in your particular example is pointless because it always runs. Let's contrive an example that would let us investigate the possibility of simulating a break and else.
Take the following string:
s = 'a,b,c b,c,d c,d,e, d,e,f'
Let's say you wanted to split the string by spaces and commas as before, but you only wanted to preserve the elements of the inner split up to the first occurrence of c:
out = []
for i in s.split():
for e in i.split(','):
if e == 'c':
break
out.append(e)
else:
out.append('-')
The break can be simulated using the arcane two-arg form of iter, which accepts a callable and a termination value:
>>> x = list('abcd')
>>> list(iter(iter(x).__next__, 'c'))
['a', 'b']
You can implement the else by chaining the inner iterable with ['-'].
>>> from itertools import chain
>>> x = list('abcd')
>>> list(iter(chain(x, ['-'])
.__next__, 'c'))
['a', 'b']
>>> y = list('def')
>>> list(iter(chain(y, ['-'])
.__next__, 'c'))
['d', 'e', 'f', '-']
Notice that the placement of chain is crucial here. If you were to chain the dash to the outer iterator, it would always be appended, not only when c is not encountered:
>>> list(chain(iter(iter(x).__next__, 'c'), ['-']))
['a', 'b', '-']
You can now simulate the entire nested loop with a single expression:
from itertools import chain
out = [e for i in s.split() for e in iter(chain(i.split(','), ['-']).__next__, 'c')]

Finding element in a string and smarter manipulation

I have a list of characters
a = ["s", "a"]
I have some words.
b = "asp"
c= "lat"
d = "kasst"
I know that the characters in the list can appear only once or in linear order(or at most on small set can appear in the bigger one).
I would like to split my words by putting the elements in a in the middle, an the rest on the left or on the right (and put a "=" if there is nothing)
so b = ["*", "as", "p"]
If a bigger set of characters which contains
d = ["k", "ass", "t"]
I know that the combinations can be at most of length 4.
So I have divided the possible combinations depending on the length:
import itertools
c4 = [''.join(i) for i in itertools.product(a, repeat = 4)]
c3 = [''.join(i) for i in itertools.product(a, repeat = 3)]
c2 = [''.join(i) for i in itertools.product(a, repeat = 2)]
c1 = [''.join(i) for i in itertools.product(a, repeat = 1)]
For each c, starting with the greater
For simplicity, let's say I start with c3 in this case and not with length 4.
I have to do this with a lot of data.
Is there a way to simplify the code ?

You can do something similar using a regular expression:
>>> import re
>>> p = re.compile(r'([sa]{1,4})')
p matches the characters 's' or 'a' repeated between 1 and 4 times.
To split a given string at this pattern, use p.split. The use of capturing parentheses in the pattern leads to the pattern itself being included in the result.
>>> p.split('asp')
['', 'as', 'p']
>>> p.split('lat')
['l', 'a', 't']
>>> p.split('kasst')
['k', 'ass', 't']

Use regex ?
import re
a = ["s", "a"]
text = "kasst"
pattern = re.compile("[" + "".join(a) + "]{1,4}")
match = pattern.search(text)
parts = [text[:match.start()], text[match.start():match.end()], text[match.end():]]
parts = [part if part else "*" for part in parts]
However, note that this won't handle the case when there is no match on the elements in a

I would do a regular expression to simplify the matching.
import re
splitters = ''.join(a)
pattern = re.compile("([^%s]*)([%s]+)([^%s]*)" % (splitters, splitters, splitters))
words = [v if v else '=' for v in pattern.match(s).groups() ]
This doesn't allow the characters in the first or last group, so not all string will match correctly (and throw an exception). You can allow them if you want. Feel free to modify the regular expression to better match what you want it to do.
Also you only need to run the re.compile once, not for every string you are trying to match.

Python find element in list that ends with number

I have a list of strings, and I want to all the strings that end with _1234 where 1234 can be any 4-digit number. It's ideal to find all the elements, and what the digits actually are, or at least return the 1st matching element, and what the 4 digit is.
For example, I have
['A', 'BB_1024', 'CQ_2', 'x_0510', 'y_98765']
I want to get
['1024', '0510']
Okay so far I got, _\d{4}$ will match _1234 and return a match object, and the match_object.group(0) is the actual matched string. But is there a better way to look for _\d{4}$ but only return \d{4} without the _?

Use re.search():
import re
lst = ['A', 'BB_1024', 'CQ_2', 'x_0510']
newlst = []
for item in lst:
match = re.search(r'_(\d{4})\Z', item)
if match:
newlst.append(match.group(1))
print(newlst) # ['1024', '0510']
As for the regex, the pattern matches an underscore and exactly 4 digits at the end of the string, capturing only the digits (note the parens). The captured group is then accessible via match.group(1) (remember that group(0) is the entire match).

import re
src = ['A', 'BB_1024', 'CQ_2', 'x_0510', 'y_98765', 'AB2421', 'D3&1345']
res = []
p = re.compile('.*\D(\d{4})$')
for s in src:
m = p.match(s)
if m:
res.append(m.group(1))
print(res)
Works fine, \D means not a number, so it will match 'AB2421', 'D3&1345' and so on.

Please show some code next time you ask a question here, even if it doesn't work at all. It makes it easier for people to help you.
If you're interested in a solution without any regex, here's a way with list comprehensions:
>>> data = ['A', 'BB_1024', 'CQ_2', 'x_0510', 'y_98765']
>>> endings = [text.split('_')[-1] for text in data]
>>> endings
['A', '1024', '2', '0510', '98765']
>>> [x for x in endings if x.isdigit() and len(x)==4]
['1024', '0510']

Try this:
[s[-4:] for s in lst if s[-4:].isdigit() and len(s) > 4]
Just check the last four characters if it's a number or not.
added the len(s) > 4 to correct the mistake Joran pointed out.

Try this code:
r = re.compile(".*?([0-9]+)$")
newlist = filter(r.match, mylist)
print newlist

How to return the regex that matches some text?

The answer to Javascript regex question Return the part of the regex that matched is "No, because compilation destroys the relationship between the regex text and the matching logic."
But Python preserves Match Objects, and re.groups() returns the specific group(s) that triggered a match. It should be simple to preserve the regex text of each group as part of a Match Object and return it, but there doesn't appear to be a call to do so.
import re
pat = "(^\d+$)|(^\w+$)|(^\W+$)"
test = ['a', 'c3', '36d', '51', '29.5', '#$%&']
for t in test:
m = re.search(pat, t)
s = (m.lastindex, m.groups()) if m else ''
print(str(bool(m)), s)
This returns:
True (2, (None, 'a', None))
True (2, (None, 'c3', None))
True (1, ('51', None, None))
False
True (3, (None, None, '#$%&'))
The compiler obviously knows that there are three groups in this pattern. Is there a way to extract the subpattern in each group in a regex, with something like:
>>> print(m.regex_group_text)
('^\d+$', '^\w+$', '^\W+$')
Yes, it would be possible to write a custom pattern parser, for example to split on '|' for this particular pattern. But it would be far easier and more reliable to use the re compiler's understanding of the text in each group.

If the indices are not sufficient and you absolutely need to know the exact part of the regex, there is probably no other possibility but to parse the expression's groups on your own.
All in all, this is no big deal, since you can simply count opening and closing brackets and log their indices:
def locateBraces(inp):
bracePositions = []
braceStack = []
depth = 0
for i in range(len(inp)):
if inp[i] == '(':
braceStack.append(i)
depth += 1
if inp[i] == ')':
bracePositions.append((braceStack.pop(), i))
depth -= 1
if depth < 0:
raise SyntaxError('Too many closing braces.')
if depth != 0:
raise SyntaxError('Too many opening braces.')
return bracePositions
Edited: This dumb implementation only counts opening and closing braces. However, regexes may contain escaped braces, e.g. \(, which are
counted as regular group-defining braces using this method. You may
want to adapt it to omit braces that have an uneven number of
backslashes right before them. I leave this issue as a task for you ;)
With this function, your example becomes:
pat = "(^\d+$)|(^\w+$)|(^\W+$)"
bloc = locateBraces(pat)
test = ['a', 'c3', '36d', '51', '29.5', '#$%&']
for t in test:
m = re.search(pat, t)
print(str(bool(m)), end='')
if m:
h = bloc[m.lastindex - 1]
print(' %s' % (pat[h[0]:h[1] + 1]))
else:
print()
Which returns:
True (^\w+$)
True (^\w+$)
True (^\w+$)
True (^\d+$)
False
True (^\W+$)
Edited: To get the list of your groups, of course a simple comprehension would do:
gtxt = [pat[b[0]:b[1] + 1] for b in bloc]

This may or may not be helpful depending on the problem that you are actually trying to solve ... But python lets you name the groups:
r = re.compile('(?P<int>^\d+$)|(?P<word>^\w+$)')
From there, when you have a match, you can inspect the groupdict to see which groups are present:
r.match('foo').groupdict() # {'int': None, 'word': 'foo'}
r.match('10').groupdict() # {'int': '10', 'word': None}
Of course, this doesn't tell you the exact regular expression associated with the match -- You'd need to keep track of that yourself based on the group name.
If you really want to go beyond this, you probably want something a little more sophisticated than simple regular expression parsing. In that case, I might suggest something like pyparsing. Don't let the old-school styling on the website fool you (or the lack of a PEP-8 compliant API) -- the library is actually quite powerful once you get used to it.

It will remain up to you to track what regular expressions you are feeding into re.search. Something like:
import re
patts = {
'a': '\d+',
'b': '^\w+',
'c': '\W+'
}
pat = '^' + '|'.join('({})'.format(x) for x in patts.values()) + '$'
test = ['a', 'c3', '36d', '51', '29.5', '#$%&']
for t in test:
m = re.search(pat, t)
if m:
for g in m.groups():
for key, regex in patts.iteritems():
if g and re.search(regex, g):
print "t={} matched regex={} ({})".format(t, key, regex)
break

Most efficient way to split strings in Python

My current Python project will require a lot of string splitting to process incoming packages. Since I will be running it on a pretty slow system, I was wondering what the most efficient way to go about this would be. The strings would be formatted something like this:
Item 1 | Item 2 | Item 3 <> Item 4 <> Item 5
Explanation: This particular example would come from a list where the first two items are a title and a date, while item 3 to item 5 would be invited people (the number of those can be anything from zero to n, where n is the number of registered users on the server).
From what I see, I have the following options:
repeatedly use split()
Use a regular expression and regular expression functions
Some other Python functions I have not thought about yet (there are probably some)
Solution 1 would include splitting at | and then splitting the last element of the resulting list at <> for this example, while solution 2 would probably result in a regular expression like:
((.+)|)+((.+)(<>)?)+
Okay, this regular expression is horrible, I can see that myself. It is also untested. But you get the idea.
Now, I am looking for the way that a) takes the least amount of time and b) ideally uses the least amount of memory. If only one of the two is possible, I would prefer less time. The ideal solution would also work for strings that have more items separated with | and strings that completely lack the <>. At least the regular expression-based solution would do that.
My understanding would be that split() would use more memory (since you basically get two resulting lists, one that splits at | and the second one that splits at <>), but I don't know enough about Python's implementation of regular expressions to judge how the regular expression would perform. split() is also less dynamic than a regular expression if it comes to different numbers of items and the absence of the second separator. Still, I can't shake the impression that Python can do this better without regular expressions, and that's why I am asking.
Some notes:
Yes, I could just benchmark both solutions, but I'm trying to learn something about Python in general and how it works here, and if I just benchmark these two, I still don't know what Python functions I have missed.
Yes, optimizing at this level is only really required for high-performance stuff, but as I said, I am trying to learn things about Python.
Addition: in the original question, I completely forgot to mention that I need to be able to distinguish the parts that were separated by | from the parts with the separator <>, so a simple flat list as generated by re.split(\||<>,input) (as proposed by obmarg) will not work too well. Solutions fitting this criterium are much appreciated.
To sum the question up: Which solution would be the most efficient one, for what reasons?
Due to multiple requests, I have run some timeit on the split()-solution and the first proposed regular expression by obmarg, as well as the solutions by mgibsonbr and duncan:
import timeit
import re
def splitit(input):
res0 = input.split("|")
res = []
for element in res0:
t = element.split("<>")
if t != [element]:
res0.remove(element)
res.append(t)
return (res0, res)
def regexit(input):
return re.split( "\||<>", input )
def mgibsonbr(input): # Solution by mgibsonbr
items = re.split(r'\||<>', input) # Split input in items
offset = 0
result = [] # The result: strings for regular items, lists for <> separated ones
acc = None
for i in items:
delimiter = '|' if offset+len(i) < len(input) and input[offset+len(i)] == '|' else '<>'
offset += len(i) + len(delimiter)
if delimiter == '<>': # Will always put the item in a list
if acc is None:
acc = [i] # Create one if doesn't exist
result.append(acc)
else:
acc.append(i)
else:
if acc is not None: # If there was a list, put the last item in it
acc.append(i)
else:
result.append(i) # Add the regular items
acc = None # Clear the list, since what will come next is a regular item or a new list
return result
def split2(input): # Solution by duncan
res0 = input.split("|")
res1, res2 = [], []
for r in res0:
if "<>" in r:
res2.append(r.split("<>"))
else:
res1.append(r)
return res1, res2
print "mgibs:", timeit.Timer("mgibsonbr('a|b|c|de|f<>ge<>ah')","from __main__ import mgibsonbr").timeit()
print "split:", timeit.Timer("splitit('a|b|c|de|f<>ge<>ah')","from __main__ import splitit").timeit()
print "split2:", timeit.Timer("split2('a|b|c|de|f<>ge<>ah')","from __main__ import split2").timeit()
print "regex:", timeit.Timer("regexit('a|b|c|de|f<>ge<>ah')","from __main__ import regexit").timeit()
print "mgibs:", timeit.Timer("mgibsonbr('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import mgibsonbr").timeit()
print "split:", timeit.Timer("splitit('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import splitit").timeit()
print "split:", timeit.Timer("split2('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import split2").timeit()
print "regex:", timeit.Timer("regexit('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import regexit").timeit()
The results:
mgibs: 14.7349407408
split: 6.403942732
split2: 3.68306812233
regex: 5.28414318792
mgibs: 107.046683735
split: 46.0844590775
split2: 26.5595985591
regex: 28.6513302646
At the moment, it looks like split2 by duncan beats all other algorithms, regardless of length (with this limited dataset at least), and it also looks like mgibsonbr's solution has some performance issues (sorry about that, but thanks for the solution regardless).

I was slightly surprised that split() performed so badly in your code, so I looked at it a bit more closely and noticed that you're calling list.remove() in the inner loop. Also you're calling split() an extra time on each string. Get rid of those and a solution using split() beats the regex hands down on shorter strings and comes a pretty close second on the longer one.
import timeit
import re
def splitit(input):
res0 = input.split("|")
res = []
for element in res0:
t = element.split("<>")
if t != [element]:
res0.remove(element)
res.append(t)
return (res0, res)
def split2(input):
res0 = input.split("|")
res1, res2 = [], []
for r in res0:
if "<>" in r:
res2.append(r.split("<>"))
else:
res1.append(r)
return res1, res2
def regexit(input):
return re.split( "\||<>", input )
rSplitter = re.compile("\||<>")
def regexit2(input):
return rSplitter.split(input)
print("split: ", timeit.Timer("splitit( 'a|b|c|de|f<>ge<>ah')","from __main__ import splitit").timeit())
print("split2:", timeit.Timer("split2( 'a|b|c|de|f<>ge<>ah')","from __main__ import split2").timeit())
print("regex: ", timeit.Timer("regexit( 'a|b|c|de|f<>ge<>ah')","from __main__ import regexit").timeit())
print("regex2:", timeit.Timer("regexit2('a|b|c|de|f<>ge<>ah')","from __main__ import regexit2").timeit())
print("split: ", timeit.Timer("splitit( 'a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import splitit").timeit())
print("split2:", timeit.Timer("split2( 'a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import split2").timeit())
print("regex: ", timeit.Timer("regexit( 'a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import regexit").timeit())
print("regex2:", timeit.Timer("regexit2('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import regexit2").timeit())
Which gives the following result:
split: 1.8427431439631619
split2: 1.0897291360306554
regex: 1.6694280610536225
regex2: 1.2277749050408602
split: 14.356198082969058
split2: 8.009285948995966
regex: 9.526430513011292
regex2: 9.083608677960001
And of course split2() gives the nested lists that you wanted whereas the regex solution doesn't.
Compiling the regex will improve performance. It does make a slight difference, but Python caches compiled regular expressions so the saving is not as much as you might expect. I think usually it isn't worth doing it for speed (though it can be in some cases), but it is often worthwhile to make the code clearer.

I'm not sure if it's the most efficient, but certainly the easiest to code seems to be something like this:
>>> input = "Item 1 | Item 2 | Item 3 <> Item 4 <> Item 5"
>>> re.split( "\||<>", input )
>>> ['Item 1 ', ' Item 2 ', ' Item 3 ', ' Item 4 ', ' Item 5']
I would think there's a fair chance of it being more efficient than a plain old split as well (depending on the input data) since you'd need to perform the second split operation on every string output from the first split, which doesn't seem likely to be efficient for either memory or time.
Though having said that I could easily be wrong, and the only way to be sure would be to time it.

Calling split multiple times is likely to be inneficient, because it might create unneeded intermediary strings. Using a regex like you proposed won't work, since the capturing group will only get the last item, not every of them. Splitting using a regex, like obmarg suggested, seems to be the best route, assuming a "flattened" list is what you're looking for.
If you don't want a flattened list, you can split using a regex first and then iterate over the results, checking the original input to see which delimiter was used:
items = re.split(r'\||<>', input)
offset = 0
for i in items:
delimiter = '|' if input[offset+len(i)] == '|' else '<>'
offset += len(i) + len(delimiter)
# Do something with i, depending on whether | or <> was the delimiter
At last, if you don't want the substrings created at all (using only the start and end indices to save space, for instance), re.finditer might do the job. Iterate over the delimiters, and do something to the text between them depending on which delimiter (| or <>) was found. It's a more complex operation, since you'll have to handle many corner cases, but might be worth it depending on your needs.
Update: for your particular case, where the input format is uniform, obmarg's solutions is the best one. If you must, post-process the result to have a nested list:
split_result = re.split( "\||<>", input )
result = [split_result[0], split_result[1], [i for i in split_result[2:] if i]]
(that last list comprehension is to ensure you'll get [] instead of [''] if there are no items after the last |)
Update 2: After reading the updated question, I finally understood what you're trying to achieve. Here's the full example, using the framework suggested earlier:
items = re.split(r'\||<>', input) # Split input in items
offset = 0
result = [] # The result: strings for regular itens, lists for <> separated ones
acc = None
for i in items:
delimiter = '|' if offset+len(i) < len(input) and input[offset+len(i)] == '|' else '<>'
offset += len(i) + len(delimiter)
if delimiter == '<>': # Will always put the item in a list
if acc is None:
acc = [i] # Create one if doesn't exist
result.append(acc)
else:
acc.append(i)
else:
if acc is not None: # If there was a list, put the last item in it
acc.append(i)
else:
result.append(i) # Add the regular items
acc = None # Clear the list, since what will come next is a regular item or a new list
print result
Tested it with your example, the result was:
['a', 'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c','de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'aha'],
'b', 'c', 'de', ['f', 'ge', 'ah']]

If you know that <> is not going to appear elsewhere in the string then you could replace '<>' with '|' followed by a single split:
>>> input = "Item 1 | Item 2 | Item 3 <> Item 4 <> Item 5"
>>> input.replace("<>", "|").split("|")
['Item 1 ', ' Item 2 ', ' Item 3 ', ' Item 4 ', ' Item 5']
This will almost certainly be faster than doing multiple splits. It may or may not be faster than using re.split - timeit is your friend.
On my system with the sample string you supplied, my version is more than three times faster than re.split:
>>> timeit input.replace("<>", "|").split("|")
1000000 loops, best of 3: 980 ns per loop
>>> import re
>>> timeit re.split(r"\||<>", input)
100000 loops, best of 3: 3.07 us per loop
(N.B.: This is using IPython, which has timeit as a built-in command).

You can make use of replace. First replace <> with |, and then split by |.
def replace_way(input):
return input.replace('<>','|').split('|')
Time performance:
import timeit
import re
def splitit(input):
res0 = input.split("|")
res = []
for element in res0:
t = element.split("<>")
if t != [element]:
res0.remove(element)
res.append(t)
return (res0, res)
def regexit(input):
return re.split( "\||<>", input )
def replace_way(input):
return input.replace('<>','|').split('|')
print "split:", timeit.Timer("splitit('a|b|c|de|f<>ge<>ah')","from __main__ import splitit").timeit()
print "regex:", timeit.Timer("regexit('a|b|c|de|f<>ge<>ah')","from __main__ import regexit").timeit()
print "replace:",timeit.Timer("replace_way('a|b|c|de|f<>ge<>ah')","from __main__ import replace_way").timeit()
print "split:", timeit.Timer("splitit('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import splitit").timeit()
print "regex:", timeit.Timer("regexit('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import regexit").timeit()
print "replace:",timeit.Timer("replace_way('a|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>aha|b|c|de|f<>ge<>ah')","from __main__ import replace_way").timeit()
Results on my machine:
split: 11.8682055461
regex: 12.7430856814
replace: 2.54299265006
split: 79.2124379066
regex: 68.6917008003
replace: 10.944842347