Python解析 - 一个文本文件中的多个邮件
我收到了来自多个发件人的类似邮件,并使用下面的正则表达式 m
和 n
来提取需要的内容。这部分工作得很好。
不过,正则表达式 o
让我感到困惑。我正在读取的文本文件是将9封邮件合并到一个文本文件中,并在Python中作为字符串打开。原始发件人(正则表达式 o
)出现在文件中每封新邮件的开头(总共出现9次)。
我想在找到每个 CUSIP 和 NAME 后,写上相同的原始发件人,直到匹配到不同的原始发件人为止。
我正在使用 xlwt3 和 wincom32。
这是合并邮件的文本文件中的一个示例,非常标准:
--- Original Sender: TOM MADEUPNAME, SOME BANK, N. ---
----- Original Message -----
From: TOM MADEUPNAME (SOME BANK, N.)
To: BOB THISISMYEMAIL (XYZ INVESTMENTS, INC)
At: 8/31 8:53:25
**Offerings**
Mezz ReRemics
Cusip Description Original Current Cashflow Collat Offering
05531UAB6 BCAP 2009-RR5 1A2 18,745 18,745 Snr Sup Fxd 45.000
Prime/Alt-A Fixed
Cusip Description Original Current Cashflow Collat Offering
059487AE8 BOAA 2006-6 CB5 25,940 14,350 Seq Fxd 83.000
12544XAX3 CWHL 2007-9 A13 10,190 10,190 Ssnr Nas Fxd 92.500
17312XAJ3 CMSI 2007-4 1A9 2,871 2,741 Spr Snr Fxd 86.000
--- Original Sender: JOE MADEUPNAME, EUROPEAN BANK SECURI ---
----- Original Message -----
From: JOE MADEUPNAME (EUROPEAN BANK SECURI)
To: BOB THISISMYEMAIL (XYZ INVESTMENTS, INC)
At: 8/31 8:20:16
8-31-2011
Alt-A Fixed
Bond O/F C/F Cpn FICO CAL WALB 60+ Notes Offer
CSMC 06-9 7A1 25.00 11.97 L+45 728 26 578 35.21 FLT,AS,0.0% 50-00
LXS 07-10H 2A1 68.26 34.01 L+16 744 6 125 33.98 SS,9.57% 42-00
CSMC 06-7 9A1 15.00 7.81 L+30 688 5 198 46.46 SS,0.0% 29-16
Prime Hybrid
Bond O/F C/F Cpn FICO CAL WALB 60+ Notes Offer
SARM 05-18 6A1 14.56 6.01 2.58 730 46 432 15.87 SEA,SS,5/1,12.3% 78-00
Alt-A Hybrid
Bond O/F C/F Cpn FICO CAL WALB 60+ Notes Offer
ARMT 05-12 2A1 23.78 10.71 3.07 712 48 556 35.32 SS,5/1,4.9% *SOLD
Option Arm
Bond O/F C/F Cpn FICO CAL WALB 60+ Notes Offer
DBALT 07-OA4 1A1B 10.00 7.25 L+13 716 63 562 47.17 SS,OC,42.2% 64-16
--------------------------------------------------------------------------------------
更新 - 正在工作
count_cusip = 0
count_name = 0
count_sender = 0
cur_sender = ''
for line in lines:
o = re.search(r"Original Sender:\s\b\w+\s\w+", line)
if o:
count_sender += 1
ws.write(count_sender,2,o.group(0))
ws.write(count_sender,2,cur_sender)
cur_sender = o.group(0)
m = re.search('[0-9]{3}[a-zA-Z0-9]{6}', line)
if m:
count_cusip += 1
ws.write(count_cusip,0,m.group(0))
ws.write(count_cusip,2,cur_sender)
n = re.search('[A-Z]{3,5}\s[0-9]{1,4}\D{1,3}\S{1,3}\s{1,2}\w+', line)
if n:
count_name += 1
ws.write(count_name,1,n.group(0))
ws.write(count_cusip,2,cur_sender)
o = re.search(r"Original Sender:\s\b\w+\s\w+", line)
if o:
cur_sender = o.group(0)
ws.write(count_name,2,cur_sender)
更新输出 - 如所需。
CUSIP Bond Name Original Sender
00442PAD2 ACE 2006-OP1 A2B Original Sender: Nick Madeupname
12557YAE7 ARMT 05-12 2A1 Original Sender: Bobby Madeupname
39153VAT1 CSMC 06-9 7A1 Original Sender: Bobby Madeupname
05377RAE4 LXS 07-10H 2A1 Original Sender: Jane Madeupname
02005HAF0 CSMC 06-7 9A1 Original Sender: Jane Madeupname
2 个回答
1
你能不能把格式改成标准库里mailbox
模块能识别的样子?这样的话,就可以让这个模块来处理所有的解析工作了。
1
你的问题有点不太清楚,因为你没有提供输出的示例,不过我可以做个猜测:
count_cusip = 0
count_name = 0
count_sender = 0
cur_sender = ''
for line in lines:
m = re.search('[0-9]{3}[a-zA-Z0-9]{6}', line)
if m:
count_cusip += 1
ws.write(count_cusip,0,m.group(0))
ws.write(count_cusip,2,cur_sender)
n = re.search('[A-Z]{3,5}\s[0-9]{1,4}\D{1,3}\S{1,3}\s{1,2}\w+', line)
if n:
count_name += 1
ws.write(count_name,1,n.group(0))
ws.write(count_name,2,cur_sender)
o = re.search(r"Original Sender:\s\b\w+\s\w+", line)
if o:
count_sender += 1
cur_sender = o.group(0)
当遇到原始发送者时,不是直接写出来,而是需要把它保存下来,然后对于每个cusip和名字,写上当前的值。