如何在特定子字符串之后获取字符串？
例如，我想获取 in 之后的"world"字符串
my_string="hello python world, I'm a beginner"
...在这种情况下是", I'm a beginner"：）
问问题
699568 次
9 回答
525
最简单的方法可能只是分割你的目标词
my_string="hello python world , i'm a beginner "
print my_string.split("world",1)[1]
split 接受要拆分的单词（或字符），并且可以选择限制拆分的数量。
在此示例中，在“世界”上拆分并将其限制为仅一次拆分。
于 2012-09-24T20:27:07.063 回答
74
s1 = "hello python world , i'm a beginner "
s2 = "world"
print s1[s1.index(s2) + len(s2):]
如果你想处理不存在的情况s2，那么使用而不是. 如果该调用的返回值为 ，则不在。s1s1.find(s2)index-1s2s1
于 2012-09-24T20:27:31.080 回答
71
我很惊讶没有人提到partition。
def substring_after(s, delim):
return s.partition(delim)[2]
恕我直言，这个解决方案比@arshajii 的更具可读性。除此之外，我认为@arshajii 是最快的——它不会创建任何不必要的副本/子字符串。
于 2013-05-23T11:35:10.177 回答
42
你想使用str.partition()：
>>> my_string.partition("world")[2]
" , i'm a beginner "
因为这个选项比其他选项更快。
请注意，如果缺少分隔符，这会产生一个空字符串：
>>> my_string.partition("Monty")[2] # delimiter missing
''
如果您想要原始字符串，请测试从返回的第二个值str.partition()是否为非空：
prefix, success, result = my_string.partition(delimiter)
if not success: result = prefix
您还可以使用str.split()1 的限制：
>>> 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，与以下相比，str.partition()它很容易快 15% 左右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%
这显示了每次执行的时间，此处的输入分隔符要么丢失（最坏情况），要么放在第一位（最好的情况），要么位于下半部分、上半部分或最后一个位置。最快的时间用标记[...]和<...>最差的标记。
上表是通过对所有三个选项的综合计时试验产生的，如下所示。我在具有 2.9 GHz Intel Core i7 和 16 GB ram 的 2017 型号 15" Macbook Pro 上运行了 Python 3.7.4 测试。
这个脚本生成随机句子，有和没有随机选择的分隔符，如果存在，在生成的句子的不同位置，以随机顺序重复运行测试（产生最公平的结果，说明测试期间发生的随机操作系统事件），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=" ")
于 2019-07-16T19:24:29.967 回答
8
在 Python 3.9 中，removeprefix添加了一个新方法：
>>> 'TestHook'.removeprefix('Test')
'Hook'
>>> 'BaseTestCase'.removeprefix('Test')
'BaseTestCase'
于 2020-06-10T15:21:25.327 回答
6
这是一个老问题，但我遇到了一个非常相同的情况，我需要使用“低”这个词作为分隔符来分割一个字符串，对我来说，问题是我在同一个字符串中有下面和更低的词。
我以这种方式使用 re 模块解决了它
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 re.split 和正则表达式来匹配确切的单词
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！
于 2017-01-13T03:15:16.877 回答
6
您可以使用名为substring. 只需使用命令安装pip install substring。您只需提及开始和结束字符/索引即可获取子字符串。
例如：
import substring
s = substring.substringByChar("abcdefghijklmnop", startChar="d", endChar="n")
print(s)
输出：
# s = defghijklmn
于 2018-06-26T04:11:07.253 回答
6
试试这种通用方法：
import re
my_string="hello python world , i'm a beginner "
p = re.compile("world(.*)")
print (p.findall(my_string))
#[" , i'm a beginner "]
于 2020-02-27T23:36:56.370 回答