Python 解析 - 一个文本文件中的多封电子邮件

Question

我收到来自多个发件人的类似电子邮件，并使用下面的正则表达式 m 和 n 提取所需的字符串。那部分工作正常。

正则表达式 o 然而让我感到困惑。我正在阅读的文本文件是 9 封电子邮件的组合，保存到一个文本文件中，并在 Python 中作为字符串打开。原始发件人(正则表达式 o)出现在文件中每条新消息的开头(9 次)

我想在找到的每个 CUSIP 和名称之后写入相同的原始发件人，直到匹配到不同的原始发件人。

我正在使用 xlwt3 和 wincom32。

来自文本文件的示例以及组合电子邮件，这是非常标准的:

--- 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
--------------------------------------------------------------------------------------

已更新 - 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 - 根据需要。

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

Answer 1

您的问题并不完全清楚，因为您没有显示输出示例，但这是一个有根据的猜测:

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)

您需要保存它并写入每个 cusip 和名称的当前值，而不是在遇到时写入原始发件人。