python - Find monotonic sequences in a list?

Question

I'm new in Python but basically I want to create sub-groups of element from the list with a double loop, therefore I gonna compare the first element with the next to figure out if I can create these sublist, otherwise I will break the loop inside and I want continue with the last element but in the main loop:

Example: 5,7,8,4,11

Compare 5 with 7, is minor? yes so include in the newlist and with the inside for continue with the next 8, is minor than 5? yes, so include in newlist, but when compare with 4, I break the loop so I want continue in m with these 4 to start with the next, in this case with 11...

for m in xrange(len(path)):
    for i in xrange(m+1,len(path)):
              if (path[i] > path[m]):
                  newlist.append(path[i])

             else:
                  break

             m=m+i

Thanks for suggestions or other ideas to achieve it!

P.S. Some input will be: input: [45,78,120,47,58,50,32,34] output: [45,78,120],[47,58],50,[32,34]

The idea why i want make a double loops due to to compare sub groups of the full list,in other way is while 45 is minor than the next one just add in the new list, if not take the next to compare in this case will be 47 and start to compare with 58.

Answer 1

No loop! Well at least, no explicit looping...

import itertools

def process(lst):
    # Guard clause against empty lists
    if len(lst) < 1:
        return lst

    # use a dictionary here to work around closure limitations
    state = { 'prev': lst[0], 'n': 0 }

    def grouper(x):
        if x < state['prev']:
            state['n'] += 1

        state['prev'] = x
        return state['n']

    return [ list(g) for k, g in itertools.groupby(lst, grouper) ]

Usage (work both with Python 2 & Python 3):

>>> data = [45,78,120,47,58,50,32,34]
>>> print (list(process(data)))
[[45, 78, 120], [47, 58], [50], [32, 34]]

Joke apart, if you need to group items in a list itertools.groupby deserves a little bit of attention. Not always the easiest/best answer -- but worth to make a try...

---

EDIT: If you don't like closures -- and prefer using an object to hold the state, here is an alternative:

class process:
    def __call__(self, lst):
        if len(lst) < 1:
            return lst

        self.prev = lst[0]
        self.n = 0

        return [ list(g) for k, g in itertools.groupby(lst, self._grouper) ]

    def _grouper(self, x):
        if x < self.prev:
            self.n += 1

        self.prev = x
        return self.n



data = [45,78,120,47,58,50,32,34]
print (list(process()(data)))

---

EDIT2: Since I prefer closures ... but @torek don't like the dictionary syntax, here a third variation around the same solution:

import itertools

def process(lst):
    # Guard clause against empty lists
    if len(lst) < 1:
        return lst

    # use an object here to work around closure limitations
    state = type('State', (object,), dict(prev=lst[0], n=0))

    def grouper(x):
        if x < state.prev:
            state.n += 1

        state.prev = x
        return state.n

    return [ list(g) for k, g in itertools.groupby(lst, grouper) ]

data = [45,78,120,47,58,50,32,34]
print (list(process(data)))