查找字符串是否以列表的可变长度前缀之一开头

Question

我需要查明名称是否以列表的任何前缀开头，然后将其删除，例如：

if name[:2] in ["i_", "c_", "m_", "l_", "d_", "t_", "e_", "b_"]:
    name = name[2:]

以上仅适用于长度为2的列表前缀。我需要可变长度前缀的相同功能。

如何有效地完成（少量代码和良好的性能）？

for循环迭代每个前缀，然后检查name.startswith(prefix)以根据前缀的长度最终切片名称，但它是很多代码，可能是低效的，并且是“非Pythonic”。 / p>

有人有一个很好的解决方案吗？

Answer 1

  

str.startswith（前缀[，start [，end]]）¶

     

如果字符串以前缀开头，则返回True，否则返回   假。前缀也可以是要查找的前缀元组。同   可选的开始，从该位置开始的测试字符串。同   可选结束，停止比较该位置的字符串。

$ ipython
Python 3.5.2 (default, Nov 23 2017, 16:37:01)
Type 'copyright', 'credits' or 'license' for more information
IPython 6.4.0 -- An enhanced Interactive Python. Type '?' for help.

In [1]: prefixes = ("i_", "c_", "m_", "l_", "d_", "t_", "e_", "b_")

In [2]: 'test'.startswith(prefixes)
Out[2]: False

In [3]: 'i_'.startswith(prefixes)
Out[3]: True

In [4]: 'd_a'.startswith(prefixes)
Out[4]: True

Answer 2

有点难以阅读，但这有效：

name=name[len(filter(name.startswith,prefixes+[''])[0]):]

Answer 3

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

Answer 4

如果您将前缀定义为下划线之前的字符，则可以检查

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

Answer 5

如何使用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))

请注意，每个列表项与前缀的比较会在第一次匹配时有效停止。 any函数会在找到True值后立即返回此行为，例如：

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

或者，使用re.match代替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 6

正则表达式可能会给你最快的速度：

prefixes = ["i_", "c_", "m_", "l_", "d_", "t_", "e_", "b_", "also_longer_"]
re_prefixes = "|".join(re.escape(p) for p in prefixes)

m = re.match(re_prefixes, my_string)
if m:
    my_string = my_string[m.end()-m.start():]

Answer 7

在搜索和效率方面，总是考虑使用索引技术来改进算法。如果您有一长串前缀，则可以使用内存索引，方法是将第一个字符的前缀简单地索引到dict。

只有当你有一长串前缀并且性能成为一个问题时，这个解决方案才有价值。

pref = ["i_", "c_", "m_", "l_", "d_", "t_", "e_", "b_"]

#indexing prefixes in a dict. Do this only once.
d = dict()
for x in pref:
        if not x[0] in d:
                d[x[0]] = list()
        d[x[0]].append(x)


name = "c_abcdf"

#lookup in d to only check elements with the same first character.
result = filter(lambda x: name.startswith(x),\
                        [] if name[0] not in d else d[name[0]])
print result

Answer 8

正则表达式，测试过：

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:])

输出：

'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 9

这会动态编辑列表，删除前缀。一旦找到特定项目，就会break跳过其余的前缀。

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

输出

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

Answer 10

可以使用一个简单的正则表达式。

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

或者如果下划线之前的任何内容是有效的前缀：

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

这将删除第一个下划线之前的任意数量的字符。

Answer 11

import re

def make_multi_prefix_replacer(prefixes):
    if isinstance(prefixes,str):
        prefixes = prefixes.split()
    prefixes.sort(key = len, reverse=True)
    pat = r'\b(%s)' % "|".join(map(re.escape, prefixes))
    print 'regex patern :',repr(pat),'\n'
    def suber(x, reg = re.compile(pat)):
        return reg.sub('',x)
    return suber



pfxs = "x ya foobar yaku foo a|b z."
replacer = make_multi_prefix_replacer(pfxs)               

names = "xenon yadda yeti yakute food foob foobarre foo a|b a b z.yx zebra".split()
for name in names:
    print repr(name),'\n',repr(replacer(name)),'\n'

ss = 'the yakute xenon is a|bcdf in the barfoobaratu foobarii'
print '\n',repr(ss),'\n',repr(replacer(ss)),'\n'