import re
text = '这个单词是由-EabcdxyxG共11个字符组成的吗？'
# 单个字符
pattern = r'\w'
print(re.findall(pattern, text))['这', '个', '单', '词', '是', '由', 'E', 'a', 'b', 'c', 'd', 'x', 'y', 'x', 'G', '共', '1', '1', '个', '字', '符', '组', '成', '的', '吗']9 正则
9.1 正则表达式
9.1.1 字母数字字符(Word character)
不包含标点符号
# 多个字符
pattern = r'\w+'
print(re.findall(pattern, text))['这个单词是由', 'EabcdxyxG共11个字符组成的吗']9.1.2 非单词字符
# 非字符，包括标点符号
pattern = r'\W'
print(re.findall(pattern, text))['-', '？']9.2 分割
# 要分割的文本
text = "Hello,world!How are you?"
# 按照逗号、句点、感叹号和问号来分割文本
result = re.split(r'[,.!?]', text)
# 输出分割后的结果
print(result)
# `maxsplit`设置分割的次数，
result = re.split(r'[,.!?]', text, maxsplit=1)
print(result)
# 如果为0，则按最大可能次数分割，它是默认值
result = re.split(r'[,.!?]', text, maxsplit=0)
print(result)['Hello', 'world', 'How are you', '']
['Hello', 'world!How are you?']
['Hello', 'world', 'How are you', '']9.2.1 保留分割符
text = "Hello|World@How|Are|You"
# 分割符为|或者@
result = re.split(fr"(?<=[@|])", text)
print(result)['Hello|', 'World@', 'How|', 'Are|', 'You']?re.split9.3 替换
import re
# 定义一个字符串
text = "apple banana cherry"
# 使用 re.sub() 将 "banana" 替换为 "orange"
new_text = re.sub(r"banana", "orange", text)
print(new_text) # 输出："apple orange cherry"apple orange cherry9.3.1 支持替换函数
# 定义一个替换函数，将匹配到的字符串转换为大写
def to_uppercase(match):
return match.group(0).upper()
# 定义一个字符串
text = "apple banana cherry"
# 使用 re.sub() 将字符串中的小写单词转换为大写
new_text = re.sub(r"\b[a-z]+\b", to_uppercase, text)
print(new_text) # 输出："APPLE BANANA CHERRY"APPLE BANANA CHERRY9.4 待分类入库
9.5 quiz 1
Check if a string contains the word word in it (case insensitive). If you have no idea, I guess you could try /word/.
import re
ptn = r'\bword\b'
txt = "I have no word to say, because i always have words with him using Word."
re.findall(ptn, txt) # 不忽略大小写
re.findall(re.compile(ptn, re.I), txt) # 忽略大小写, I代表incasesensitivity['word', 'Word']9.6 quiz 2
Use substitution to replace every occurrence of the word i with the word I (uppercase, I as in me). E.g.: i’‘m replacing it. am i not? -> I’’m replacing it. am I not?.
txt = '''i''ll use it on input to fix my lazy spelling! i mean... i just wanted to check if you understood how it worked.'''
ptn = r'\bi\b'
re.sub(ptn, "I", txt)"I''ll use it on input to fix my lazy spelling! I mean... I just wanted to check if you understood how it worked."10 字母
10.1 quiz 3
With regex you can count the number of matches. Can you make it return the number of uppercase consonants (B,C,D,F,..,X,Y,Z) in a given string? E.g.: it should return 3 with the text ABcDeFO!. Note: Only ASCII. We consider Y to be a consonant!
txt = 'You should use the WHO guide to treat COVID-19.'
ptn = r'[B-DF-HJ-NP-TV-Z]'
re.findall(ptn, txt)['Y', 'W', 'H', 'C', 'V', 'D']10.2 quiz 4
Oh no! It seems my friends spilled beer all over my keyboard last night and my keys are super sticky now. Some of the time whennn I press a key, I get two duplicates.
Can you ppplease help me fix thhhis?
txt = '''Oh no! It seems my friends spilled beer all over my keyboard last night and my keys are super sticky now. Some of the time whennn I press a key, I get two duplicates.
Can you ppplease help me fix thhhis?'''
ptn = r'([nph]){3}'
re.sub(ptn, r"\1", txt, 0, re.M) # r"\1"指匹配到的group 1，即n,p,h。有些语言中用$1表示；默认 0 表示替换所有的匹配'Oh no! It seems my friends spilled beer all over my keyboard last night and my keys are super sticky now. Some of the time when I press a key, I get two duplicates.\n\nCan you please help me fix this?'10.3 quiz 5
Match positive integers less than or equal to 255
import re
ptn = r"\b1[0-9][0-9]\b|\b2[0-5][0-5]\b|\b\d{1,2}\b" # \b设定好边界
txt = "11 123 501 9 10 31 30 2530 0 255 -1 256 -125 100 199"
txt = re.sub(r'-\d+', '', txt) # 删除负整数
re.findall(ptn, txt)['11', '123', '9', '10', '31', '30', '0', '255', '100', '199']11 quiz 6
Validate IPv4 address
import re
ptn1 = r'^((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.){3}(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])$' # 前导0不正确
ptn2 = r'\b^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)(\.)){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$\b' # 前导0也正确
txt1 = "1.1.23.269"
txt2 = "1.1.23.169"
txt3 = "01.1.023.169"
for txt in [txt1, txt2, txt3]:
if re.findall(ptn1, txt):
print("{} : Valid IPv4 address using ptn1.".format(txt))
else:
print("{} : Invalid IPv4 address using ptn1.".format(txt))
if re.findall(ptn2, txt):
print("{} : Valid IPv4 address using ptn2.".format(txt))
else:
print("{} : Invalid IPv4 address using ptn2.".format(txt))
if not re.findall(ptn1, txt) and not re.findall(ptn2, txt):
print("{} : Invalid IPv4 address using ptn1 and ptn2.".format(txt))1.1.23.269 : Invalid IPv4 address using ptn1.
1.1.23.269 : Invalid IPv4 address using ptn2.
1.1.23.269 : Invalid IPv4 address using ptn1 and ptn2.
1.1.23.169 : Valid IPv4 address using ptn1.
1.1.23.169 : Valid IPv4 address using ptn2.
01.1.023.169 : Invalid IPv4 address using ptn1.
01.1.023.169 : Valid IPv4 address using ptn2.12 quiz 7
Find IPv4 address
import re
ptn1 = r'((25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.){3}(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])' # 前导0不正确
ptn2 = r'((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)(\.)){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)' # 前导0也正确
txt1 = "1.1.23.269是IP地址"
txt2 = "判断10.1.23.169是IP地址，203.245.25.67也是IP地址"
txt3 = "01.1.023.169"
re.search(ptn2, txt2) # 只能找到第1个
# 找到所有，search返回的是整体的match，而不是分组内容
def regexMatchAny(pattern:str, rawText):
ptn = re.compile(pattern)
res = ptn.search(rawText)
lst = []
while res:
start, end = res.span()
lst.append(res.group(0))
res = ptn.search(rawText, end+1)
return lst
regexMatchAny(r'((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)(\.)){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)', txt2)['10.1.23.169', '203.245.25.67']