循环/按顺序python

时间:2012-09-06 14:56:40

标签: python loops

我有一个文件,其中我提供了步骤,然后根据要遵循的内容的步骤。 这是我读到的文本文件:

[Steps]
step1 = WebAddress
step2 = Tab
step3 = SecurityType
step4 = Criteria
step5 = Date
step6 = Click1
step7 = Results
step8 = Download
[data]
WebAddress___________________________ Destination___________ Tab_____________ SecurityType___________________________________________________ Criteria___ Date_______ Click1_ Results_ Download    
https://mbsdisclosure.fanniemae.com/  q:\\%s\\raw\\fnmapool  Advanced Search  Interim MBS: Single-Family                                      Issue Date  09/01/2012  Search  100      CSV XML
https://mbsdisclosure.fanniemae.com/  q:\\%s\\raw\\fnmapool  Advanced Search  Preliminary Mega: Fannie Mae/Ginnie Mae backed Adjustable Rate  Issue Date  09/01/2012  Search  100      CSV XML
https://mbsdisclosure.fanniemae.com/  q:\\%s\\raw\\fnmapool  Advanced Search  Preliminary Mega: Fannie Mae/Ginnie Mae backed Fixed Rate       Issue Date  09/01/2012  Search  100      CSV XML

我已经有了一个工作模型来读取文件,然后将正确的内容分配给正确的标题(例如标题WebAdress的URL)。但是,我想要做的是遵循基于步骤的循环。 处理数据的代码:

from itertools import groupby
count =0
file_name = "FNMA.tbl"
with open(file_name) as f:
      pre_data,post_data =[s.strip() for s in (f.read()).split("[data]")]
post_data_lines = post_data.splitlines()
headers = post_data_lines[0].split()
headers2 = [s.replace("_"," ").strip() for s in headers]
for line in post_data_lines[1:]:
    tmpline  = []
    pos = 0
    for itm in headers:
        tmpline.append(line[pos:pos+len(itm)])
        pos += len(itm)+1
    myDict= dict(zip(headers2,tmpline))
    count += 1
    for key, group in groupby(myDict.iteritems(), lambda x: x[0]):
        for thing in group:
            print "step: %s header: %s" % (thing[1], key)
    print "Finished processing row %s" % count

1 个答案:

答案 0 :(得分:0)

首先,创建一个将步骤名称映射到数字的字典,如下所示:

steps = dict((step.split()[2], pos) 
        for (pos, step) in enumerate(pre_data.splitlines()[1:]))

(当然,这是一个非常丑陋的Python系列,但似乎有效)

现在,您可以按照以下步骤对词典中的项目进行排序:

sorted_items = sorted(myDict.items(), 
        key=lambda item: steps[item[0]] if item[0] in steps else 999)

迭代这些项目:

for key, thing in sorted_items:
    print "step: %s header: %s" % (thing, key)