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'

Strip也不局限于空白字符:

# remove all leading/trailing commas, periods and hyphens
title = title.strip(',.-')

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)))

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'

好吧，作为一个初学者，看到这个帖子让我头晕目眩。于是我想到了一个简单的捷径。

虽然str.strip()可以移除开头和结尾的空格，但它对字符之间的空格没有任何作用。

words=input("Enter the word to test")
# If I have a user enter discontinous threads it becomes a problem
# input = "   he llo, ho w are y ou  "
n=words.strip()
print(n)
# output "he llo, ho w are y ou" - only leading & trailing spaces are removed 

相反，使用str.replace()更有意义，错误更少，更切题。 下面的代码可以泛化str.replace()的使用

def whitespace(words):
    r=words.replace(' ','') # removes all whitespace
    n=r.replace(',','|') # other uses of replace
    return n
def run():
    words=input("Enter the word to test") # take user input
    m=whitespace(words) #encase the def in run() to imporve usability on various functions
    o=m.count('f') # for testing
    return m,o
print(run())
output- ('hello|howareyou', 0)

在diff. functions中继承相同的函数时可以很有帮助。