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
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"