我试图使用Python从字符串中删除特定字符。这是我现在使用的代码。不幸的是,它似乎对字符串没有做任何事情。
for char in line:
if char in " ?.!/;:":
line.replace(char,'')
我如何正确地做到这一点?
我试图使用Python从字符串中删除特定字符。这是我现在使用的代码。不幸的是,它似乎对字符串没有做任何事情。
for char in line:
if char in " ?.!/;:":
line.replace(char,'')
我如何正确地做到这一点?
当前回答
试试这个:
def rm_char(original_str, need2rm):
''' Remove charecters in "need2rm" from "original_str" '''
return original_str.translate(str.maketrans('','',need2rm))
这个方法在python3中很有效
其他回答
这里有一些可能的方法来完成这个任务:
def attempt1(string):
return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])
def attempt2(string):
for v in ("a", "e", "i", "o", "u"):
string = string.replace(v, "")
return string
def attempt3(string):
import re
for v in ("a", "e", "i", "o", "u"):
string = re.sub(v, "", string)
return string
def attempt4(string):
return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")
for attempt in [attempt1, attempt2, attempt3, attempt4]:
print(attempt("murcielago"))
附注:在使用" ?.!/;:"的例子中使用元音…是的,“murcielago”在西班牙语里是蝙蝠的意思…有趣的单词,因为它包含了所有的元音:)
PS2:如果你对性能感兴趣,你可以用一个简单的代码来衡量这些尝试:
import timeit
K = 1000000
for i in range(1,5):
t = timeit.Timer(
f"attempt{i}('murcielago')",
setup=f"from __main__ import attempt{i}"
).repeat(1, K)
print(f"attempt{i}",min(t))
在我的盒子里,你会得到:
attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465
因此,对于这个特定的输入,尝试4似乎是最快的。
对于只允许字符串中某些字符的相反要求,可以使用带有集补操作符[^ABCabc]的正则表达式。例如,要删除除ascii字母、数字和连字符以外的所有字符:
>>> import string
>>> import re
>>>
>>> phrase = ' There were "nine" (9) chick-peas in my pocket!!! '
>>> allow = string.letters + string.digits + '-'
>>> re.sub('[^%s]' % allow, '', phrase)
'Therewerenine9chick-peasinmypocket'
来自python正则表达式文档:
不在范围内的字符可以通过互补来匹配 一组。如果集合的第一个字符是'^',则所有字符 不在集合中的将被匹配。例如,[^5]将匹配 除'5'以外的任何字符,[^^]将匹配除 “^”。的第一个字符没有特殊意义 集。
字符串方法replace不会修改原始字符串。它保留原始文件并返回修改后的副本。
你需要的是这样的:line = line.replace(char, ")
def replace_all(line, )for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
return line
然而,每次删除一个字符都创建一个新的字符串是非常低效的。我推荐以下方法:
def replace_all(line, baddies, *):
"""
The following is documentation on how to use the class,
without reference to the implementation details:
For implementation notes, please see comments begining with `#`
in the source file.
[*crickets chirp*]
"""
is_bad = lambda ch, baddies=baddies: return ch in baddies
filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
mahp = replace_all.map(filter_baddies, line)
return replace_all.join('', join(mahp))
# -------------------------------------------------
# WHY `baddies=baddies`?!?
# `is_bad=is_bad`
# -------------------------------------------------
# Default arguments to a lambda function are evaluated
# at the same time as when a lambda function is
# **defined**.
#
# global variables of a lambda function
# are evaluated when the lambda function is
# **called**
#
# The following prints "as yellow as snow"
#
# fleece_color = "white"
# little_lamb = lambda end: return "as " + fleece_color + end
#
# # sometime later...
#
# fleece_color = "yellow"
# print(little_lamb(" as snow"))
# --------------------------------------------------
replace_all.map = map
replace_all.join = str.join
您还可以使用函数来替换不同类型的正则表达式或使用列表的其他模式。这样,您就可以混合正则表达式、字符类和真正基本的文本模式。当您需要替换大量元素(如HTML元素)时,它非常有用。
*注意:适用于Python 3.x
import re # Regular expression library
def string_cleanup(x, notwanted):
for item in notwanted:
x = re.sub(item, '', x)
return x
line = "<title>My example: <strong>A text %very% $clean!!</strong></title>"
print("Uncleaned: ", line)
# Get rid of html elements
html_elements = ["<title>", "</title>", "<strong>", "</strong>"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)
# Get rid of special characters
special_chars = ["[!@#$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)
在函数string_cleanup中,它以字符串x和未修饰的列表作为参数。对于元素或模式列表中的每一项,如果需要替代品,就会进行替换。
输出:
Uncleaned: <title>My example: <strong>A text %very% $clean!!</strong></title>
1st clean: My example: A text %very% $clean!!
2nd clean: My example: A text very clean
line = line.translate(None, " ?.!/;:")