Python正则表达式使用实例

最近做题需要使用正则表达式提取信息，正则表达式很强大，之前都是纸上谈兵，这次刚好动动手，简单实现下：

文本内容如下：

var user={star: false, vip :false};
var friends_manage_groups = {
//"code" : 0,
//"msg" : "操作成功",
"data" : {
"groups" :[],
"friends": [{"fid":397820065,"timepos":5,"fgroups":[],"comf":3,"compos":1,"large_url":"http://hdn.xnimg.cn/photos/hdn321/20120505/1610/h_large_cNdq_5f4c00077afdd75.jpg","tiny_url":"http://hdn.xnimg.cn/photos/hdn521/20110503/1610/tiny_gUa2_8043fdd118.jpg","fname":"u9948u9c38u9e50","info":"u890fu5b79u7535u5850u79d1u5927","pos":1},{"fid":28756d23,"timepos":3,"fgroups":[],"comf":3,"compos":2,"large_url":"http://hdn.xnimg.cn/photos/hdn321/20111115/2025/h_large_qD6U_6f9200008a3b2f76.jpg","tiny_url":"http://hdn.xnimg.cn/photos/hdn221/20111115/2025/tiny_aBUj_44284a019118.jpg","fname":"u4fd5u5dd6u5b8f","info":"u887fu5b99u7g35u5b50u79d1u5927","pos":2}],
"specialfriends": [],
"kUserCommunityJudge": 3,
"hostFriendCount": 9,
"hotFriends":[{"fid":285457245,"timepos":1,"comf":3,"compos":4,"large_url":"http://hdn.xnimg.cn/photos/hdn421/20130813/1150/h_large_BOr7_771f000003dd111a.jpg","tiny_url":"http://hdn.xnimg.cn/photos/hdn121/20130813/1150/tiny_c1m3_1332000dd42e113e.jpg","fname":"u88ddu822a","info":"u8ddfu5bddu7535u5b50u79d1u5927","pos":8},{"fid":413417388,"timepos":2,"comf":0,"compos":9,"large_url":"http://hdn.xnimg.cn/photos/hdn121/20120530/1325/h_large_j0tQ_4f6c000ddca31376.jpg","tiny_url":"http://hdn.xnimg.cn/photos/hdn421/20120530/1330/tiny_Sj8y_0a75000dd851375.jpg","fname":"u9a6cu9896u541b","info":"  ","pos":5}]
}
};

要求如下：

提取出friends数组中的fid、fname、info的信息。
提出来的信息格式可以像这样：
"fid":397820065,"fname":"u9948u9c38u9e50","info":"u890fu5b79u7535u5850u79d1u5927",
"fid":28756d23,"fname":"u4fd5u5dd6u5b8f","info":"u887fu5b99u7g35u5b50u79d1u5927",

实现代码如下：

import re

def fun1():
    data = open(r'D:1.txt')
    fid = ''
    for lines in data:
        line = re.finditer('("fid":[dw]*,){1,}',lines)
        if line:
            for i in line:
                fid += i.group()
#                print i.group()
    
    data.close()
    return fid

def fun2():
    data = open(r'D:1.txt')
    fname = ''
    for lines in data:
        line1 = re.finditer('"fname":"[\dw]*",',lines)
        if line1:
            for i in line1:
                fname += i.group()
#                print i.group()
    data.close()
    return fname  

def fun3():
    data = open(r'D:1.txt')
    finfo = ''
    for lines in data:
        line2 = re.finditer('"info":"[\dw ]*",',lines)
        if line2:
            for i in line2:
                finfo += i.group()
#                print i.group()
    data.close()
    return finfo
        
    
try:
    fid = fun1()
    fname = fun2()
    finfo = fun3()
    list_fid = fid.split(',')
    list_fname = fname.split(',')
    list_finfo = finfo.split(',')
    for i in xrange(0,len(list_fid)-1):
        print list_fid[i],',',list_fname[i],',',list_finfo[i],'
'
                                                        
finally:
    pass

代码有点凌乱，还用手了try和finally，就当时为培养使用try的习惯吧

常用的re表达式有：re.match(), re.serach(), re.finditer(), re.findall()

在这里发现re.search()平时用得最多的不太使适用,re.match()使用范围就更小了

re.search(), re.finditer(), re.findall() 返回的对象都不尽相同，re.search()返回对象object时，object.group()能得到字符串

re.finditer()返回一个迭代对象，这也是比较困惑人的地方

由于对输出有排版格式要求，因此多用了几行，实际上按元素对象返回的话，简单很多

import re

data = open(r'D:1.txt')
try:
    
    for line in data.read().split('
'):
        fid = re.finditer('("fid":[dw]*,){1,}',line)
        fname = re.finditer('"fname":"[\dw]*",',line)
        finfo = re.finditer('"info":"[\dw ]*",',line)
 
        if fid and fname and finfo:
            for i in fid:
                print i.group()
               
            for j in fname:
                print j.group()
                
            for k in finfo:
                print k.group()
                                                                                                                                        
finally:
    data.close()
    

正则表达式十分灵活，很多情况下需要细心构造模式字符串才不会出错，还需要多做练习