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。

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)))

你需要strip():

myphrases = [" Hello ", " Hello", "Hello ", "Bob has a cat"]

for phrase in myphrases:
    print(phrase.strip())

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"

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'

正如上面的答案所指出的

my_string.strip()

将删除所有前导和后面的空白字符，如\n， \r， \t， \f，空格。

为了获得更大的灵活性，请使用以下方法

只删除前导空白字符: 只移除尾随的空格字符: 删除特定的空白字符:my_string.strip('\n')或my_string.lstrip('\n\r')或my_string.rstrip('\n\t')等。

更多细节可以在文档中找到。