(编辑:jimmy 日期: 2024/11/13 浏览:2)
学习python爬虫时遇到了一个问题,书上有示例如下:
import re line='Cats are smarter than dogs' matchObj=re.match(r'(.*)are(.*"htmlcode">matchObj=re.match(r'(.*)are(.*"htmlcode">import re line='Cats are smarter than dogs' matchObj=re.match(r'(.*)are(.*"matchObj.group():",matchObj.group()) print("matchObj.group(1):", matchObj.group(1)) print("matchObj.group(2):", matchObj.group(2)) print("matchObj.group(3):", matchObj.group(3)) else: print('No match!\n')得到的结果是:
matchObj.group(): Cats are smarter than dogs
matchObj.group(1): Cats
matchObj.group(2):
matchObj.group(3): smarter than dogs
可见第二个括号里的内容被默认为空了,然后删去那个?,可以看到结果变成:
matchObj.group(): Cats are smarter than dogs
matchObj.group(1): Cats
matchObj.group(2): smarter than dogs
matchObj.group(3):
那么这是否就意味着?的默认值很可能是0次,那?这个符号到底有什么用呢
仔细想来这个说法并不是很严谨。尝试使用单独的."htmlcode">
import re line='Cats are smarter than dogs' matchObj=re.match(r'(.*) are(.*)"matchObj.group():",matchObj.group()) print("matchObj.group(1):", matchObj.group(1)) print("matchObj.group(2):", matchObj.group(2))也能在组别2中正常提取到are之后的字符内容,但稍微改动一下将?放到第二个括号内,
就什么也提取不到,同时导致group(0)中匹配的字符到Cats are就截止了(也就是第二个括号匹配失败)。
令人感到奇怪的是,如果将上面的代码改成
import re line='Cats are smarter than dogs' matchObj=re.match(r'(.*) are (.*)+',line) if matchObj: print("matchObj.group():",matchObj.group()) print("matchObj.group(1):", matchObj.group(1)) print("matchObj.group(2):", matchObj.group(2))也就是仅仅将?改为+,虽然能成功匹配整个line但group(2)中没有内容,
如果把+放到第二个括号中就会产生报错,匹配失败。
那么是否可以认为.*"htmlcode">
import re line='Cats are smarter than dogs' matchObj=re.match(r'(.*) are (.*r).*',line) if matchObj: print("matchObj.group():",matchObj.group()) print("matchObj.group(1):", matchObj.group(1)) print("matchObj.group(2):", matchObj.group(2)) #print("matchObj.group(3):", matchObj.group(3)) else: print('No match!\n')为了泛用性尝试了一下把r改成‘ '但是得到的结果是‘smarter than '。于是尝试把.换成表示任意字母的
[a-zA-Z],成功提取出了单个smarter,代码如下:
import re line='Cats are smarter than dogs' matchObj=re.match(r'(.*) are ([a-zA-Z]* ).*',line) if matchObj: print("matchObj.group():",matchObj.group()) print("matchObj.group(1):", matchObj.group(1)) print("matchObj.group(2):", matchObj.group(2)) #print("matchObj.group(3):", matchObj.group(3)) else: print('No match!\n')