如何在Python中删除字符串的前导和尾随空白?

" Hello world " --> "Hello world"
" Hello world"  --> "Hello world"
"Hello world "  --> "Hello world"
"Hello world"   --> "Hello world"

当前回答

这也可以用正则表达式来实现

import re

input  = " Hello "
output = re.sub(r'^\s+|\s+$', '', input)
# output = 'Hello'

其他回答

这将删除myString中所有前导和尾部的空格:

myString.strip()

一种方法是使用.strip()方法(删除所有周围的空白)

str = "  Hello World  "
str = str.strip()
**result: str = "Hello World"**

请注意,.strip()返回字符串的副本,并且不会更改下划线对象(因为字符串是不可变的)。

如果您希望删除所有空白(不仅仅是修剪边缘):

str = ' abcd efgh ijk  '
str = str.replace(' ', '')
**result: str = 'abcdefghijk'

要移除字符串周围的所有空白,请使用.strip()。例子:

>>> ' Hello '.strip()
'Hello'
>>> ' Hello'.strip()
'Hello'
>>> 'Bob has a cat'.strip()
'Bob has a cat'
>>> '   Hello   '.strip()  # ALL consecutive spaces at both ends removed
'Hello'

注意str.strip()删除所有空白字符,包括制表符和换行符。若要仅删除空格,请指定要删除的特定字符作为strip的参数:

>>> "  Hello\n  ".strip(" ")
'Hello\n'

最多只删除一个空格:

def strip_one_space(s):
    if s.endswith(" "): s = s[:-1]
    if s.startswith(" "): s = s[1:]
    return s

>>> strip_one_space("   Hello ")
'  Hello'

我无法找到我正在寻找的解决方案,所以我创建了一些自定义函数。你可以试试。

def cleansed(s: str):
    """:param s: String to be cleansed"""
    assert s is not (None or "")
    # return trimmed(s.replace('"', '').replace("'", ""))
    return trimmed(s)


def trimmed(s: str):
    """:param s: String to be cleansed"""
    assert s is not (None or "")
    ss = trim_start_and_end(s).replace('  ', ' ')
    while '  ' in ss:
        ss = ss.replace('  ', ' ')
    return ss


def trim_start_and_end(s: str):
    """:param s: String to be cleansed"""
    assert s is not (None or "")
    return trim_start(trim_end(s))


def trim_start(s: str):
    """:param s: String to be cleansed"""
    assert s is not (None or "")
    chars = []
    for c in s:
        if c is not ' ' or len(chars) > 0:
            chars.append(c)
    return "".join(chars).lower()


def trim_end(s: str):
    """:param s: String to be cleansed"""
    assert s is not (None or "")
    chars = []
    for c in reversed(s):
        if c is not ' ' or len(chars) > 0:
            chars.append(c)
    return "".join(reversed(chars)).lower()


s1 = '  b Beer '
s2 = 'Beer  b    '
s3 = '      Beer  b    '
s4 = '  bread butter    Beer  b    '

cdd = trim_start(s1)
cddd = trim_end(s2)
clean1 = cleansed(s3)
clean2 = cleansed(s4)

print("\nStr: {0} Len: {1} Cleansed: {2} Len: {3}".format(s1, len(s1), cdd, len(cdd)))
print("\nStr: {0} Len: {1} Cleansed: {2} Len: {3}".format(s2, len(s2), cddd, len(cddd)))
print("\nStr: {0} Len: {1} Cleansed: {2} Len: {3}".format(s3, len(s3), clean1, len(clean1)))
print("\nStr: {0} Len: {1} Cleansed: {2} Len: {3}".format(s4, len(s4), clean2, len(clean2)))

如果你想从左边和右边修剪指定数量的空格,你可以这样做:

def remove_outer_spaces(text, num_of_leading, num_of_trailing):
    text = list(text)
    for i in range(num_of_leading):
        if text[i] == " ":
            text[i] = ""
        else:
            break

    for i in range(1, num_of_trailing+1):
        if text[-i] == " ":
            text[-i] = ""
        else:
            break
    return ''.join(text)

txt1 = "   MY name is     "
print(remove_outer_spaces(txt1, 1, 1))  # result is: "  MY name is    "
print(remove_outer_spaces(txt1, 2, 3))  # result is: " MY name is  "
print(remove_outer_spaces(txt1, 6, 8))  # result is: "MY name is"