Question

I wrote this function to filter a list to show all items after a given item is seen. Somewhat similar to the builtin string method str.rpartition(sep). I have a feeling there is a more compact way to do this, perhaps using a list comprehension. Any ideas?

def ignore_until(the_list, match):
    # Ignore all items in the_list prior to match
    found = False
    for index, item in enumerate(the_list):
        if item == match:
            found = True
            break
    if found:
        return the_list[index:]
    else:
        return []

my_list = ['red','orange','yellow','green']
assert ignore_until(my_list, 'yellow') == ['yellow','green']
assert ignore_until(my_list, 'blue') == []

EDIT:

After seeing the answers for the above question, I realized that 5 of the 6 answers focused on the index() builtin method for the list data type. Actually, I need to use a regular expression, and didn’t realize that omitting that from my question would affect people’s answers. Here’s the regex code:

import re
def ignore_until(the_list, pattern):
    # Ignore all items in the_list prior to the item containing pattern.
    found = False
    for index, item in enumerate(the_list):
        if re.search(string=item, pattern=pattern):
            found = True
            break
    if found:
        return the_list[index:]
    else:
        return []

my_list = ['red','orange','yellow','green']
assert ignore_until(my_list, 'yellow') == ['yellow','green']
assert ignore_until(my_list, 'blue') == []

Answer 1

Here’s a version that reproduces what str.partition does (i.e. returns three lists):

def partition(lst, item):
    if item in lst:
        n = lst.index(item)
        return lst[:n], [item], lst[n+1:]
    else:
        return lst, [], []

print partition(range(10), 7)

and here’s a version which works with arbitrary iterables, not necessary lists:

def partition(it, item):
    a = [[]]
    for x in it:
        if x == item and len(a) == 1:
            a.append([item])
            a.append([])
        else:
            a[-1].append(x)
    return a

print partition((x for x in range(10)), 7)

Improved version:

def partition(it, item):
    a = []
    for x in it:
        if x == item:
            return a, [item], list(it)
        a.append(x)
    return a, [], []

print partition((x for x in range(10)), 7)
print partition((x for x in range(10)), 17)