Python basis | string manipulation

Attributes

Code	Features	Remark
len(s)	String length
type(s).__name__	type name	If the string is outputstr
s[0],s[1],s[-1]	Take the string 1, 2, and the last character
s.find('x')	Determine the location of the development of character in the string
s.count('x')	Specified character (string) appeared several times in s
s.index('1')	Specify the location of the character (string) for the first time in s (default)

Judge

Code	Features	Remark
s.isalpha()	Are all letters, and at least one character
s.isdigit()	It is all digital, and there is at least one character
s.isspace()	Whether all whitespace characters, and there is at least one character
s.islower()	Whether all lowercase letters
s.isupper()	Whether the letter is upper case
s.istitle()	Whether it is capitalized
s.startswith('a')	The first letter of verification
s.endswith('a')	Verify the last letter
'xyz' in s	If they contain characters developed (string)

format

operating	Code	Remark
All capital letters	s.upper()
All lowercase letters	s.lower()
Capitalized	s.capitalize()
Invert case	s.swapcase()	The original uppercase letters to lowercase, and vice versa

Formatted output

# 整数
s = 1
print('整数示例 %d'%s)

Output:
整数示例 1

# 小数
s = 1.2345
print('小数示例 %5.2f'%s) # 整数保留5位，小数保留2位

Output:
小数示例 1.23

# 百分比
s = 0.058
print('百分比示例 %5.2f%%'%(s*100))

Output:
百分比示例 5.80%

deformation

Overturn

s = '12345'
s[::-1]

Output:
'54321'

splice

s1 = 'abc'
s2 = '123'
s1+s2

Output:
'abc123'

'.'.join(s1)
'.'.join([s1,s2])
s1.join(s2)

Output:
'a.b.c'
'abc.123'
'1abc2abc3'

Split

s = 'I am Groot'
s.split(' ') #此处以空格作为分隔符
s.partition(' ')

Output:
['I', 'am', 'Groot']
('I', ' ', 'am Groot')

repeat

s = 'abc'
s*3

Output:
'abcabcabc'

filling

s = '12'
s.zfill(5)

Output:
'00012'

s = 'begin'
s.center(30,'*')

Output:
'************begin*************'

Wash

White spaces

s.strip() # 去除两端的空格
s.rstrip() # 去除右侧的空格
s.lstrip() # 去除左侧的空格

Letter substitutions

One replacement

s = 'this is a test'
s.replace('t','T')

Output:
'This is a TesT'

Many-to-replace

# 多个字符串替换成单个字符串
import re
re.sub('t|T|s','H',s)

tab characters are replaced by eight spaces:s.expandtabs

Character matches

Whether it contains a character (string)

'x' in s
'xyz' in s

If it contains a few characters (string) of

items_list = re.findall('(字符串1|字符串2|字符串3)',s)

# 如果匹配到了，那么item_list长度>0
if len(items_list)>0:
    print('匹配到相关字符')

Extraction according to the rules of
usage of regular expressions refer
Details can be found the most complete encyclopedia of common regular expressions

coding

Reference to this article