Finding whether a string starts with one of a list's variable-length prefixes

Question

I need to find out whether a name starts with any of a list\'s prefixes and then remove it, like:

if name[:2] in [\"i_\", \"c_\", \"m_\", \"l_\", \"d_\", \"t         


        

Answer 1

for prefix in prefixes:
    if name.startswith(prefix):
        name=name[len(prefix):]
        break

Answer 2

Regex, tested:

import re

def make_multi_prefix_matcher(prefixes):
    regex_text = "|".join(re.escape(p) for p in prefixes)
    print repr(regex_text)
    return re.compile(regex_text).match

pfxs = "x ya foobar foo a|b z.".split()
names = "xenon yadda yeti food foob foobarre foo a|b a b z.yx zebra".split()

matcher = make_multi_prefix_matcher(pfxs)
for name in names:
    m = matcher(name)
    if not m:
        print repr(name), "no match"
        continue
    n = m.end()
    print repr(name), n, repr(name[n:])

Output:

'x|ya|foobar|foo|a\\|b|z\\.'
'xenon' 1 'enon'
'yadda' 2 'dda'
'yeti' no match
'food' 3 'd'
'foob' 3 'b'
'foobarre' 6 're'
'foo' 3 ''
'a|b' 3 ''
'a' no match
'b' no match
'z.yx' 2 'yx'
'zebra' no match

Answer 3

What about using filter?

prefs = ["i_", "c_", "m_", "l_", "d_", "t_", "e_", "b_"]
name = list(filter(lambda item: not any(item.startswith(prefix) for prefix in prefs), name))

Note that the comparison of each list item against the prefixes efficiently halts on the first match. This behaviour is guaranteed by the any function that returns as soon as it finds a True value, eg:

def gen():
    print("yielding False")
    yield False
    print("yielding True")
    yield True
    print("yielding False again")
    yield False

>>> any(gen()) # last two lines of gen() are not performed
yielding False
yielding True
True

Or, using re.match instead of startswith:

import re
patt = '|'.join(["i_", "c_", "m_", "l_", "d_", "t_", "e_", "b_"])
name = list(filter(lambda item: not re.match(patt, item), name))

Answer 4

This edits the list on the fly, removing prefixes. The break skips the rest of the prefixes once one is found for a particular item.

items = ['this', 'that', 'i_blah', 'joe_cool', 'what_this']
prefixes = ['i_', 'c_', 'a_', 'joe_', 'mark_']

for i,item in enumerate(items):
    for p in prefixes:
        if item.startswith(p):
            items[i] = item[len(p):]
            break

print items

Output

['this', 'that', 'blah', 'cool', 'what_this']

Answer 5

Could use a simple regex.

import re
prefixes = ("i_", "c_", "longer_")
re.sub(r'^(%s)' % '|'.join(prefixes), '', name)

Or if anything preceding an underscore is a valid prefix:

name.split('_', 1)[-1]

This removes any number of characters before the first underscore.

Answer 6

If you define prefix to be the characters before an underscore, then you can check for

if name.partition("_")[0] in ["i", "c", "m", "l", "d", "t", "e", "b", "foo"] and name.partition("_")[1] == "_":
    name = name.partition("_")[2]