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"
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
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)))
其他回答
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)))
Strip也不局限于空白字符:
# remove all leading/trailing commas, periods and hyphens
title = title.strip(',.-')
在Pyhton中运行完成的代码或程序时,为了删除会导致大量缩进错误的“空白”。只需执行以下操作;显然,如果Python一直告诉我们错误是第1、2、3、4、5行中的缩进,等等……把那条线来回固定。
然而,如果你仍然遇到与输入错误、操作符等相关的程序问题,请确保你阅读了错误Python为什么对你大喊大叫:
首先要检查的是你是否有 缩进。如果你有,那么检查一下你是否有 在代码中混合制表符和空格。
记住:代码 将看起来很好(对您),但解释器拒绝运行它。如果 如果你怀疑这一点,一个快速的解决方法就是把你的代码带入 IDLE编辑窗口,然后选择编辑…“从 菜单系统,在选择格式之前…“Untabify地区。 如果你混合了制表符和空格,这将转换你所有的 制表符到空格一次性完成(并修复任何缩进问题)。
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。