我正在寻找一种方法来获取字母表中的所有字母:但我不知道从哪里开始。我会使用正则表达式吗?如果是这样的话?
string = "Username: How are you today?"
有人能告诉我一个关于我能做什么的例子吗?
答案 0 :(得分:82)
只需使用split
功能即可。它返回一个列表,因此您可以保留第一个元素:
>>> s1.split(':')
['Username', ' How are you today?']
>>> s1.split(':')[0]
'Username'
答案 1 :(得分:24)
使用index
:
>>> string = "Username: How are you today?"
>>> string[:string.index(":")]
'Username'
索引会在字符串中给出“:”的位置,然后你可以将其分割
如果你想使用正则表达式:
>>> import re
>>> re.match("(.*?):",string).group()
'Username'
match
匹配字符串的开头,
答案 2 :(得分:14)
您不需要regex
>>> s = "Username: How are you today?"
您可以使用split
方法在':'
字符
>>> s.split(':')
['Username', ' How are you today?']
切出元素[0]
以获取字符串的第一部分
>>> s.split(':')[0]
'Username'
答案 3 :(得分:2)
我已经在Python 3.7.0(IPython)下对这些各种技术进行了基准测试。
c
时):预编译的正则表达式。s.partition(c)[0]
。c
可能不在s
中):分区,拆分。import string, random, re
SYMBOLS = string.ascii_uppercase + string.digits
SIZE = 100
def create_test_set(string_length):
for _ in range(SIZE):
random_string = ''.join(random.choices(SYMBOLS, k=string_length))
yield (random.choice(random_string), random_string)
for string_length in (2**4, 2**8, 2**16, 2**32):
print("\nString length:", string_length)
print(" regex (compiled):", end=" ")
test_set_for_regex = ((re.compile("(.*?)" + c).match, s) for (c, s) in test_set)
%timeit [re_match(s).group() for (re_match, s) in test_set_for_regex]
test_set = list(create_test_set(16))
print(" partition: ", end=" ")
%timeit [s.partition(c)[0] for (c, s) in test_set]
print(" index: ", end=" ")
%timeit [s[:s.index(c)] for (c, s) in test_set]
print(" split (limited): ", end=" ")
%timeit [s.split(c, 1)[0] for (c, s) in test_set]
print(" split: ", end=" ")
%timeit [s.split(c)[0] for (c, s) in test_set]
print(" regex: ", end=" ")
%timeit [re.match("(.*?)" + c, s).group() for (c, s) in test_set]
String length: 16
regex (compiled): 156 ns ± 4.41 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
partition: 19.3 µs ± 430 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
index: 26.1 µs ± 341 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split (limited): 26.8 µs ± 1.26 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split: 26.3 µs ± 835 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
regex: 128 µs ± 4.02 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
String length: 256
regex (compiled): 167 ns ± 2.7 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
partition: 20.9 µs ± 694 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
index: 28.6 µs ± 2.73 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split (limited): 27.4 µs ± 979 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split: 31.5 µs ± 4.86 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
regex: 148 µs ± 7.05 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
String length: 65536
regex (compiled): 173 ns ± 3.95 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
partition: 20.9 µs ± 613 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
index: 27.7 µs ± 515 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split (limited): 27.2 µs ± 796 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split: 26.5 µs ± 377 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
regex: 128 µs ± 1.5 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
String length: 4294967296
regex (compiled): 165 ns ± 1.2 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
partition: 19.9 µs ± 144 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
index: 27.7 µs ± 571 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split (limited): 26.1 µs ± 472 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
split: 28.1 µs ± 1.69 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
regex: 137 µs ± 6.53 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
答案 4 :(得分:2)
partition()可能比split()更好,因为它在没有分隔符或更多分隔符的情况下具有更好的可预测结果。