Answers:
最简单的方法可能只是分割目标词
my_string="hello python world , i'm a beginner "
print my_string.split("world",1)[1] split使用要拆分的单词（或字符），并且可以选择限制拆分次数。
在此示例中，对“世界”进行拆分并将其限制为仅一个拆分。
target.split('lower',1)[-1].split('low',1)[-1]
my_string.partition("world")[-1]（或...[2]）更快。
s1 = "hello python world , i'm a beginner "
s2 = "world"
print s1[s1.index(s2) + len(s2):]如果你要处理的情况下s2是不存在的s1，然后使用s1.find(s2)而不是index。如果调用的返回值-1，那么s2是不是在s1。
print( s1[s1.index(s2) + len(s2):] is s1[s1.index(s2) + len(s2):])
我很惊讶没有人提及partition。
def substring_after(s, delim):
return s.partition(delim)[2]恕我直言，此解决方案比@arshajii更具可读性。除此之外，我认为@arshajii最好的是最快的-它不会创建任何不必要的副本/子字符串。
str.split(..., 1)。
您要使用str.partition()：
>>> my_string.partition("world")[2]
" , i'm a beginner "因为此选项比其他选项要快。
请注意，如果缺少分隔符，则会生成一个空字符串：
>>> my_string.partition("Monty")[2] # delimiter missing
''如果要使用原始字符串，请测试从中返回的第二个值str.partition()是否非空：
prefix, success, result = my_string.partition(delimiter)
if not success: result = prefix您也可以使用str.split()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 "但是，此选项较慢。在最佳情况下，与相比，str.partition()轻松快约15％str.split()：
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%这显示了使用输入的每次执行的时间，此处缺少分隔符（最坏情况），放在最前面（最佳情况）或位于下半部，上半部或最后位置。最快的时间标有[...]，<...>而最坏的则标有。
上表是针对以下所有三个选项的综合时间试用得出的。我在带有2.9 GHz Intel Core i7和16 GB ram的2017年型号15“ Macbook Pro上的Python 3.7.4上运行了测试。
该脚本会生成带有或不带有随机选择的定界符的随机语句，如果存在，则在生成的语句中的不同位置，以重复的随机顺序运行测试（产生最合理的结果，说明测试期间发生的随机OS事件），然后打印结果表：
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)
这是一个古老的问题，但是我遇到了非常相似的情况，我需要使用“ low”一词作为半字形来拆分一个字符串，对我来说，问题是我在同一字符串中具有“ lower”和“ lower”这个词。
我这样用re模块解决了
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.'使用带有正则表达式的re.split来匹配确切的单词
stringafterword = re.split('\\blow\\b',string)[-1]
print(stringafterword)
' reading is seen as positive (or bullish) for the Korean Won.'通用代码是：
re.split('\\bTHE_WORD_YOU_WANT\\b',string)[-1]希望这可以帮助某人！
string.partition(" low ")[2]？（请注意low
在Python 3.9中，removeprefix添加了一个新方法：
>>> 'TestHook'.removeprefix('Test')
'Hook'
>>> 'BaseTestCase'.removeprefix('Test')
'BaseTestCase'