Question

本周开始学习python，所以我想我会用excel来解析文件路径中的一些字段。

我有大约3000个文件都符合命名约定。 /Household/LastName.FirstName.Account.Doctype.Date.extension

例如，可以命名其中一个文件：Cosby.Bill..Profile.2006.doc 并且完整路径是/ Volumes / HD / Organized Files / Cosby，Bill / Cosby.Bill..Profile.2006.doc

在这种情况下：

考斯比，比尔将成为家庭

家庭（Cosby，Bill）是实际文件的封闭文件夹

比尔将是第一个名字

Cosby 将是姓氏

帐户字段已省略

个人资料是doctype

2006 是日期

doc 是扩展程序

所有这些文件都位于此目录/ Volumes / HD / Organized Files /我使用终端和ls将所有文件的列表放到桌面上的.txt文件中，我试图解析来自文件路径分为类似上面的示例。理想情况下，我想输出到csv，每个类别都有一列。这是我丑陋的代码：

def main():
    file = open('~/Desktop/client_docs.csv', "rb")
    output = open('~/Desktop/client_docs_parsed.txt', "wb")
    for line in file:
        i = line.find(find_nth(line, '/', 2))
        beghouse = line[i + len(find_nth(line, '/', 2)):]
        endhouse = beghouse.find('/')
        household = beghouse[:endhouse]
        lastn = (line[line.find(household):])[(line[line.find(household):]).find('/') + 1:(line[line.find(household):]).find('.')]
        firstn = line[line.find('.') + 1: line.find('.', line.find('.') + 1)]
        acct = line[line.find('{}.{}.'.format(lastn,firstn)) + len('{}.{}.'.format(lastn,firstn)):line.find('.',line.find('{}.{}.'.format(lastn,firstn)) + len('{}.{}.'.format(lastn,firstn)))]
        doctype_beg = line[line.find('{}.{}.{}.'.format(lastn, firstn, acct)) + len('{}.{}.{}.'.format(lastn, firstn, acct)):]
        doctype = doctype_beg[:doctype_beg.find('.')]
        date_beg = line[line.find('{}/{}.{}.{}.{}.'.format(household,lastn,firstn,acct,doctype)) + len('{}/{}.{}.{}.{}.'.format(household,lastn,firstn,acct,doctype)):]
        date = date_beg[:date_beg.find('.')]
        print '"',household, '"','"',lastn, '"','"',firstn, '"','"',acct, '"','"',doctype, '"','"',date,'"'

def find_nth(body, s_term, n):
    start = body[::-1].find(s_term)
    while start >= 0 and n > 1:
        start = body[::-1].find(s_term, start+len(s_term))
        n -= 1
    return ((body[::-1])[start:])[::-1]

if __name__ == "__main__": main()

它似乎工作正常，但是当有另一个封闭文件夹时我会遇到问题，然后它会转移所有我的字段...例如，而不是文件驻留在

/ Volumes / HD / Organized Files / Cosby，Bill /

at at / Volumes / HD / Organized Files / Resigned / Cosby，Bill /

我知道必须采用不那么笨重的方式来解决这个问题。

Answer 1

这是一个比您的函数find_nth()更实用的工具：
rstrip()

def find_nth(body, s_term, n):
    start = body[::-1].find(s_term)
    print '------------------------------------------------'
    print 'body[::-1]\n',body[::-1]
    print '\nstart == %s' % start
    while start >= 0 and n > 1:
        start = body[::-1].find(s_term, start+len(s_term))
        print 'n == %s    start == %s' % (n,start)
        n -= 1
    print '\n (body[::-1])[start:]\n',(body[::-1])[start:]
    print '\n((body[::-1])[start:])[::-1]\n',((body[::-1])[start:])[::-1]
    print '---------------\n'
    return ((body[::-1])[start:])[::-1]


def cool_find_nth(body, s_term, n):
    assert(len(s_term)==1)
    return body.rsplit(s_term,n)[0] + s_term


ss = 'One / Two / Three / Four / Five / Six / End'
print 'the string\n%s\n' % ss

print ('================================\n'
       "find_nth(ss, '/', 3)\n%s" % find_nth(ss, '/', 3) )

print '================================='
print "cool_find_nth(ss, '/', 3)\n%s" % cool_find_nth(ss, '/', 3)

结果

the string
One / Two / Three / Four / Five / Six / End

------------------------------------------------
body[::-1]
dnE / xiS / eviF / ruoF / eerhT / owT / enO

start == 4
n == 3    start == 10
n == 2    start == 17

 (body[::-1])[start:]
/ ruoF / eerhT / owT / enO

((body[::-1])[start:])[::-1]
One / Two / Three / Four /
---------------

================================
find_nth(ss, '/', 3)
One / Two / Three / Four /
=================================
cool_find_nth(ss, '/', 3)
One / Two / Three / Four /

编辑1

这是另一个非常实用的工具：正则表达式

import re

reg = re.compile('/'
                 '([^/.]*?)/'
                 '([^/.]*?)\.'
                 '([^/.]*?)\.'
                 '([^/.]*?)\.'
                 '([^/.]*?)\.'
                 '([^/.]*?)\.'
                 '[^/.]+\Z')

def main():
    #file = open('~/Desktop/client_docs.csv', "rb")
    #output = open('~/Desktop/client_docs_parsed.txt', "wb")
    li = ['/Household/LastName.FirstName.Account.Doctype.Date.extension',
          '- /Volumes/HD/Organized Files/Cosby, Bill/Cosby.Bill..Profile.2006.doc']
    for line in li:
        print "line == %r" % line
        household,lastn,firstn,acct,doctype,date = reg.search(line).groups('')       
        print ('household == %r\n'
               'lastn     == %r\n'
               'firstn    == %r\n'
               'acct      == %r\n'
               'doctype   == %r\n'
               'date      == %r\n'
               % (household,lastn,firstn,acct,doctype,date))

if __name__ == "__main__": main()

结果

line == '/Household/LastName.FirstName.Account.Doctype.Date.extension'
household == 'Household'
lastn     == 'LastName'
firstn    == 'FirstName'
acct      == 'Account'
doctype   == 'Doctype'
date      == 'Date'

line == '- /Volumes/HD/Organized Files/Cosby, Bill/Cosby.Bill..Profile.2006.doc'
household == 'Cosby, Bill'
lastn     == 'Cosby'
firstn    == 'Bill'
acct      == ''
doctype   == 'Profile'
date      == '2006'

编辑2

当我发布上一次编辑时，我想知道我的大脑在哪里。以下工作也是如此：

rig = re.compile('[/.]')
rig.split(line)[-7:-1]

Answer 2

从我可以收集的信息来看，我相信这将作为一种解决方案，不依赖于以前编译的文件列表

import csv
import os, os.path

# Replace this with the directory where the household directories are stored.
directory = "home"
output = open("Output.csv", "wb")
csvf = csv.writer(output)

headerRow = ["Household", "Lastname", "Firstname", "Account", "Doctype", 
              "Date", "Extension"]

csvf.writerow(headerRow)

for root, households, files in os.walk(directory):
    for household in households:
        for filename in os.listdir(os.path.join(directory, household)):
            # This will create a record for each filename within the "household"
            # Then will split the filename out, using the "." as a delimiter
            # to get the detail
            csvf.writerow([household] + filename.split("."))
output.flush()
output.close()

这使用os库来“遍历”住户列表。然后，对于每个“家庭”，它将收集文件列表。这需要这个列表，在csv文件中生成记录，使用句点作为分隔符来分隔文件的名称。

它利用csv库生成输出，看起来有点像;

"Household,LastName,Firstname,Account,Doctype,Date,Extension"

如果不需要扩展名，则可以通过更改行来省略它：

csvf.writerow([household] + filename.split("."))

到

csvf.writerow([household] + filename.split(".")[-1])

告诉它只使用文件名的最后一部分，然后从headerRow中删除“Extension”字符串。

希望这会有所帮助

Answer 3

有点不清楚问题是什么，但同时，这是让你开始的事情：

#!/usr/bin/env python

import os
import csv

with open("f1", "rb") as fin:
    reader = csv.reader(fin, delimiter='.')
    for row in reader:
        # split path
        row = list(os.path.split(row[0])) + row[1:]
        print ','.join(row)

输出：

/Household,LastName,FirstName,Account,Doctype,Date,extension

另一种解释是您希望将每个字段存储在参数中还有一条额外的道路搞砸了......

这是for循环中row的样子：

['/Household/LastName', 'FirstName', 'Account', 'Doctype', 'Date', 'extension']

然后解决方案可能是倒退。

将row[-1]分配给extension，row[-2]分配给date，依此类推。

Python按自定义字段模式解析文件名

3 个答案:

编辑1

编辑2