Splitting a list based on a delimiter word

Question

I have a list containing various string values. I want to split the list whenever I see WORD. The result will be a list of lists (which will be the sublists of original list) containing exactly one instance of the WORD I can do this using a loop but is there a more pythonic way to do achieve this ?

Example = ['A', 'WORD', 'B' , 'C' , 'WORD' , 'D']

result = [['A'], ['WORD','B','C'],['WORD','D']]

This is what I have tried but it actually does not achieve what I want since it will put WORD in a different list that it should be in:

def split_excel_cells(delimiter, cell_data):      result = []      temp = []      for cell in cell_data:         if cell == delimiter:             temp.append(cell)             result.append(temp)             temp = []         else:             temp.append(cell)      return result 

Answer 1

import itertools  lst = ['A', 'WORD', 'B' , 'C' , 'WORD' , 'D'] w = 'WORD'  spl = [list(y) for x, y in itertools.groupby(lst, lambda z: z == w) if not x] 

this creates a splitted list without delimiters, which looks more logical to me:

[['A'], ['B', 'C'], ['D']] 

If you insist on delimiters to be included, this should do the trick:

spl = [[]] for x, y in itertools.groupby(lst, lambda z: z == w):     if x: spl.append([])     spl[-1].extend(y) 

Answer 2

I would use a generator:

def group(seq, sep):     g = []     for el in seq:         if el == sep:             yield g             g = []         g.append(el)     yield g  ex = ['A', 'WORD', 'B' , 'C' , 'WORD' , 'D'] result = list(group(ex, 'WORD')) print(result) 

This prints

[['A'], ['WORD', 'B', 'C'], ['WORD', 'D']] 

The code accepts any iterable, and produces an iterable (which you don't have to flatten into a list if you don't want to).