如何获取特定子字符串后的字符串?
例如,我想要在以下字符串中获取"world"
后面的字符串:
my_string="hello python world, I'm a beginner"
在这种情况下,输出为:",我是一个初学者"
最简单的方法可能就是在目标词上进行分割
my_string="hello python world , i'm a beginner"
print(my_string.split("world",1)[1])
split函数接受一个单词(或字符)作为分隔符,并可选地指定拆分的次数上限。
在此示例中,使用“world”作为分隔符并将其限制为仅拆分一次。
我很惊讶没有人提到 partition
。
def substring_after(s, delim):
return s.partition(delim)[2]
s1="hello python world, I'm a beginner"
substring_after(s1, "world")
# ", I'm a beginner"
依我之见,这种解决方案比@arshajii的更易读。除此之外,我认为@arshajii的是最快的,因为它不会创建任何不必要的副本/子字符串。
str.split(..., 1)
更快。 - Martijn Pieterss1 = "hello python world , i'm a beginner"
s2 = "world"
print(s1[s1.index(s2) + len(s2):])
如果你想处理 s1
中没有出现 s2
的情况,那么请使用 s1.find(s2)
而不是 index
。如果该调用的返回值为 -1
,则表示 s1
中没有 s2
。print(s1[s1.index(s2) + len(s2):] is s1[s1.index(s2) + len(s2):])
- Joran Beasley您想要使用str.partition()
函数:
>>> my_string.partition("world")[2]
" , i'm a beginner "
>>> my_string.partition("Monty")[2] # delimiter missing
''
str.partition()
返回的第二个值是否为空来实现:prefix, success, result = my_string.partition(delimiter)
if not success: result = prefix
你也可以使用 str.split()
并且设置一个限制为1:
>>> my_string.split("world", 1)[-1]
" , i'm a beginner "
>>> my_string.split("Monty", 1)[-1] # delimiter missing
"hello python world , i'm a beginner "
然而,这个选项会比较慢。在最理想的情况下,str.partition()
比 str.split()
快约 15%:
missing first lower upper last
str.partition(...)[2]: [3.745 usec] [0.434 usec] [1.533 usec] <3.543 usec> [4.075 usec]
str.partition(...) and test: 3.793 usec 0.445 usec 1.597 usec 3.208 usec 4.170 usec
str.split(..., 1)[-1]: <3.817 usec> <0.518 usec> <1.632 usec> [3.191 usec] <4.173 usec>
% best vs worst: 1.9% 16.2% 6.1% 9.9% 2.3%
[...]
标记,<...>
表示最差。import random
from itertools import product
from operator import itemgetter
from pathlib import Path
from timeit import Timer
setup = "from __main__ import sentence as s, delimiter as d"
tests = {
"str.partition(...)[2]": "r = s.partition(d)[2]",
"str.partition(...) and test": (
"prefix, success, result = s.partition(d)\n"
"if not success: result = prefix"
),
"str.split(..., 1)[-1]": "r = s.split(d, 1)[-1]",
}
placement = "missing first lower upper last".split()
delimiter_count = 3
wordfile = Path("/usr/dict/words") # Linux
if not wordfile.exists():
# macos
wordfile = Path("/usr/share/dict/words")
words = [w.strip() for w in wordfile.open()]
def gen_sentence(delimiter, where="missing", l=1000):
"""Generate a random sentence of length l
The delimiter is incorporated according to the value of where:
"missing": no delimiter
"first": delimiter is the first word
"lower": delimiter is present in the first half
"upper": delimiter is present in the second half
"last": delimiter is the last word
"""
possible = [w for w in words if delimiter not in w]
sentence = random.choices(possible, k=l)
half = l // 2
if where == "first":
# best case, at the start
sentence[0] = delimiter
elif where == "lower":
# lower half
sentence[random.randrange(1, half)] = delimiter
elif where == "upper":
sentence[random.randrange(half, l)] = delimiter
elif where == "last":
sentence[-1] = delimiter
# else: worst case, no delimiter
return " ".join(sentence)
delimiters = random.choices(words, k=delimiter_count)
timings = {}
sentences = [
# where, delimiter, sentence
(w, d, gen_sentence(d, w)) for d, w in product(delimiters, placement)
]
test_mix = [
# label, test, where, delimiter sentence
(*t, *s) for t, s in product(tests.items(), sentences)
]
random.shuffle(test_mix)
for i, (label, test, where, delimiter, sentence) in enumerate(test_mix, 1):
print(f"\rRunning timed tests, {i:2d}/{len(test_mix)}", end="")
t = Timer(test, setup)
number, _ = t.autorange()
results = t.repeat(5, number)
# best time for this specific random sentence and placement
timings.setdefault(
label, {}
).setdefault(
where, []
).append(min(dt / number for dt in results))
print()
scales = [(1.0, 'sec'), (0.001, 'msec'), (1e-06, 'usec'), (1e-09, 'nsec')]
width = max(map(len, timings))
rows = []
bestrow = dict.fromkeys(placement, (float("inf"), None))
worstrow = dict.fromkeys(placement, (float("-inf"), None))
for row, label in enumerate(tests):
columns = []
worst = float("-inf")
for p in placement:
timing = min(timings[label][p])
if timing < bestrow[p][0]:
bestrow[p] = (timing, row)
if timing > worstrow[p][0]:
worstrow[p] = (timing, row)
worst = max(timing, worst)
columns.append(timing)
scale, unit = next((s, u) for s, u in scales if worst >= s)
rows.append(
[f"{label:>{width}}:", *(f" {c / scale:.3f} {unit} " for c in columns)]
)
colwidth = max(len(c) for r in rows for c in r[1:])
print(' ' * (width + 1), *(p.center(colwidth) for p in placement), sep=" ")
for r, row in enumerate(rows):
for c, p in enumerate(placement, 1):
if bestrow[p][1] == r:
row[c] = f"[{row[c][1:-1]}]"
elif worstrow[p][1] == r:
row[c] = f"<{row[c][1:-1]}>"
print(*row, sep=" ")
percentages = []
for p in placement:
best, worst = bestrow[p][0], worstrow[p][0]
ratio = ((worst - best) / worst)
percentages.append(f"{ratio:{colwidth - 1}.1%} ")
print("% best vs worst:".rjust(width + 1), *percentages, sep=" ")
result = re.search(r"(?:world)(.*)", "hello python world , i'm a beginner ").group(1)
。 - RaduSremoveprefix
的方法:>>> 'TestHook'.removeprefix('Test')
'Hook'
>>> 'BaseTestCase'.removeprefix('Test')
'BaseTestCase'
substring
的包。只需使用命令pip install substring
进行安装。您可以通过指定起始和结束字符/索引来获取子字符串。
例如:
import substring
s = substring.substringByChar("abcdefghijklmnop", startChar="d", endChar="n")
print(s)
输出:
# s = defghijklmn
这是一个老问题,但我遇到了一个非常相似的情况,我需要使用单词“low”作为分隔符来拆分字符串。对我而言,问题在于同一字符串中包含有以下单词:below和lower。
我用re模块解决了这个问题,方法如下:
import re
string = '...below...as higher prices mean lower demand to be expected. Generally, a high reading is seen as negative (or bearish), while a low reading is seen as positive (or bullish) for the Korean Won.'
# use re.split with regex to match the exact word
stringafterword = re.split('\\blow\\b',string)[-1]
print(stringafterword)
# ' reading is seen as positive (or bullish) for the Korean Won.'
# the generic code is:
re.split('\\bTHE_WORD_YOU_WANT\\b',string)[-1]
string.partition(" low ")[2]
?(注意low
两侧的空格) - Mtl Dev尝试采用以下通用方法:
import re
my_string="hello python world , i'm a beginner"
p = re.compile("world(.*)")
print(p.findall(my_string))
# [" , i'm a beginner "]
如果您更喜欢只使用 Python 正则表达式 re 库来完成此操作,您可以使用 Match.string
属性和 Match.end()
方法来操作 Match
对象:
import re
my_string="hello python world, I'm a beginner"
match = re.search("world", my_string)
if match:
print(match.string[match.end():])
# , I'm a beginner
my_string.partition("world")[-1]
(或...[2]
)更快。 - Martijn Pieters