Python: Find X to Y in a list of strings.

Question

I have a list of maybe a 100 or so elements that is actually an email with each line as an element. The list is slightly variable because lines that have a \n in them are put in a separate element so I can't simply slice using fixed values. I essentially need a variable start and stop phrase (needs to be a partial search as well because one of my start phrases might actually be Total Cost: $13.43 so I would just use Total Cost:.) Same thing with the end phrase. I also do not wish to include the start/stop phrases in the returned list. In summary:

>>> email = ['apples','bananas','cats','dogs','elephants','fish','gee']
>>> start = 'ban'
>>> stop = 'ele'

# the magic here

>>> print new_email
['cats', 'dogs']

NOTES

• While not perfect formatting of the email, it is fairly consistent so there is a slim chance a start/stop phrase will occur more than once.
• There are also no blank elements.

SOLUTION

Just for funzies and thanks to everybody's help here is my final code:

def get_elements_positions(stringList=list(), startPhrase=None, stopPhrase=None):
    elementPositionStart, elementPositionStop = 0, -1
    if startPhrase:
        elementPositionStart = next((i for i, j in enumerate(stringList) if j.startswith(startPhrase)), 0)
    if stopPhrase:
        elementPositionStop = next((i for i, j in enumerate(stringList) if j.startswith(stopPhrase)), -1)
    if elementPositionStart + 1 == elementPositionStop - 1:
        return elementPositionStart + 1
    else:
        return [elementPositionStart, elementPositionStop]

It returns a list with the starting and ending element position and defaults to 0 and -1 if the respective value cannot be found. (0 being the first element and -1 being the last).

SOLUTION-B

I made a small change, now if the list is describing a start and stop position resulting in just 1 element between it returns that elements position as an integer instead of a list which you still get for multi-line returns.

Thanks again!

Answer 1

>>> email = ['apples','bananas','cats','dogs','elephants','fish','gee']
>>> start, stop = 'ban', 'ele'
>>> ind_s = next(i for i, j in enumerate(email) if j.startswith(start))
>>> ind_e = next(i for i, j in enumerate(email) if j.startswith(stop) and i > ind_s)
>>> email[ind_s+1:ind_e]
['cats', 'dogs']

To satisfy conditions when element might not be in the list:

>>> def get_ind(prefix, prev=-1):
    it = (i for i, j in enumerate(email) if i > prev and j.startswith(prefix))
    return next(it, None)


>>> start = get_ind('ban')
>>> start = -1 if start is None else start
>>> stop = get_ind('ele', start)
>>> email[start+1:stop]
['cats', 'dogs']

Answer 2

An itertools based approach:

import itertools
email = ['apples','bananas','cats','dogs','elephants','fish','gee']
start, stop = 'ban', 'ele'
findstart = itertools.dropwhile(lambda item: not item.startswith(start), email)
findstop = itertools.takewhile(lambda item: not item.startswith(stop), findstart)
print list(findstop)[1:]
// ['cats', 'dogs']

Answer 3

Here you go:

>>> email = ['apples','bananas','cats','dogs','elephants','fish','gee']
>>> start = 'ban'
>>> stop = 'ele'
>>> out = []
>>> appending = False
>>> for item in email:
...     if appending:
...         if stop in item:
...             out.append(item)
...             break
...         else:
...             out.append(item)
...     elif start in item:
...         out.append(item)
...         appending = True
... 
>>> out.pop(0)
'bananas'
>>> out.pop()
'elephants'
>>> print out
['cats', 'dogs']

I think my version is much more readable than the other answers and doesn't require any imports =)