Make Python Sublists from a list using a Separator

Question

I have for example the following list:

[\'|\', u\'MOM\', u\'DAD\', \'|\', u\'GRAND\', \'|\', u\'MOM\', u\'MAX\', u\'JULES\', \'|\']

and wan

Answer 1

>>> [list(x[1]) for x in itertools.groupby(['|', u'MOM', u'DAD', '|', u'GRAND', '|', u'MOM', u'MAX', u'JULES', '|'], lambda x: x=='|') if not x[0]]
[[u'MOM', u'DAD'], [u'GRAND'], [u'MOM', u'MAX', u'JULES']]

Answer 2

Simple solution using plain old for-loop (was beaten to it for the groupby solution, which BTW is better!)

seq = ['|', u'MOM', u'DAD', '|', u'GRAND', '|', u'MOM', u'MAX', u'JULES', '|']

S=[]
tmp=[]

for i in seq:
    if i == '|':
        S.append(tmp)
        tmp = []
    else:
        tmp.append(i)

# Remove empty lists
while True:
    try:
        S.remove([])
    except ValueError:
        break

print S

Gives

[[u'MOM', u'DAD'], [u'GRAND'], [u'MOM', u'MAX', u'JULES']]

Answer 3

>>> reduce(
        lambda acc,x: acc+[[]] if x=='|' else acc[:-1]+[acc[-1]+[x]], 
        myList,
        [[]]
    )
[[], ['MOM', 'DAD'], ['GRAND'], ['MOM', 'MAX', 'JULES'], []]

Of course you'd want to use itertools.groupby, though you may wish to note that my approach "correctly" puts empty lists on the ends. =)

Answer 4

itertools.groupby() does this very nicely...

>>> import itertools
>>> l = ['|', u'MOM', u'DAD', '|', u'GRAND', '|', u'MOM', u'MAX', u'JULES', '|']
>>> key = lambda sep: sep == '|'
>>> [list(group) for is_key, group in itertools.groupby(l, key) if not is_key]
[[u'MOM', u'DAD'], [u'GRAND'], [u'MOM', u'MAX', u'JULES']]