在python2.7中通过目录中的文件循环读取文件时出错

Question

我从昨天开始使用googling这个问题并无济于事;

当我遍历一个目录中的多个文件，并处理该循环中每个文件的行时，我总是关闭，但好像python打开了同一个内存空间中的所有文件，所以当我循环遍历一个文件，我从先前打开的文件中检索所有记录，就好像它们在一个指针数组中。 。 。 .wtf？

import os 
    import sys 
    import glob 
    import string 
    import cPickle 
    path2 = './' 
    columnShuffleTable = loadColumnTable('myTable') #func previously defined 
    codeScrambleTable = loadScrambleTable('theirTable') #func previously defined 
    pathToFiles2 = glob.glob(os.path.join(path2, '*.DAT')) 

    for curFile in pathToFiles2:  
     _list = ['',] 
     #this is the variable with which I'm having a problem 
     unscrambledCodes = file(curFile[-10:], 'r') 
     #this always yields the actual first line of the file at which I am currently at 
     line = unscrambledCodes.readline() 
     _list[0] = '|' + line.strip() #stripping trailing spaces 
     #the list length at this point always equates to '1', so up to here everything is great 
     print "list length:", len(_list) 
     # this always reads the 2nd line of the very first file I loaded. . .wtf? 
     line = unscrambledCodes.readline().strip() 

     while(line): 
      #for unscrambledCodes [my input file] 
      print "len list: ", len(_list), "infile", unscrambledCodes 
      nextLine = unscrambledCodes.readline().strip() 

      if not nextLine: 
       _list.append('|' + line) 
       break 
      else: 
       _list.append('|' + line[:-14] + scrambleCode(line[-12:], columnShuffleTable, codeScrambleTable)) 
      #end if 

      line = nextLine 
     unscrambledCodes.close() 
     outfile = open(curFile[-10:-4] + '.Scrambled', 'w') 
     output = '\n'.join(_list) 
     outfile.write(output) 
     outfile.close() 

的要求，这里是我的I/O样本：

输入文件1：
AB00007737106517 COSTCLASSU275
C000000010031932155750539976333693187714
C000000010031932155750539976105307608239

文件2：
AB00007736638744 COSTCLASSU275
C00000001003028490769901248060 8351468369
C000000020030284907699012480751885101503

file3的：
AB00007737148207 COSTCLASSU275
C000000010032271716759259098738354718484
C000000020032271716759259098394986919513

期望的输出文件1：
AB00007737148207 COSTCLASSU275
| C000000010031932155750539976079292077121
| C000000010031932155750539976126217711213

文件2：
AB00007736638744 COSTCLASSU275
| C000000010030284907699012480968864628712
| C000000020030284907699012480294550195814

文件3：
AB00007737106517 COSTCLASSU275
| C000000010032271716759259098216262704445
| C000000020032271716759259098085462231948

电流输出文件1：
AB00007737148207 COSTCLASSU275
| C000000010031932155750539976079292077121
| C000000010031932155750539976126217711213

文件2：
AB00007736638744 COSTCLASSU275
| C000000010031932155750539976079292077121
| C000000010031932155750539976126217711213
。
。
。
| C000000010030284907699012480968864628712
| C000000020030284907699012480294550195814
文件3：
AB00007737106517 COSTCLASSU275
| C000000010031932155750539976079292077121
| C000000010031932155750539976126217711213
。
。
。
| C000000010030284907699012480968864628712
| C000000020030284907699012480294550195814
。
。
。
| C000000010032271716759259098216262704445
| C000000020032271716759259098085462231948

Answer 1

是的，unscrambledCodes.readline（）将一次读取文件的一行，递增到下一行，直到整个文件被读入

你。可以使用类似的东西：

content = unscrambledCodes.readlines() 

这将读取每一行到数组中。然后，您可以遍历内容，并根据需要更新行。

此外，而不是文件（），我一般用

myFile = open('filename.txt','r') 
content = myFile.readlines() 
myFile.close() 

Answer 2

普遍的共识是使用开放式的，而不是文件。我会以此开始。

其次，尽量做你打开的文件发电机的理解，因为它是非常容易（下一个方法将返回一个换行符）作为new_file=[x.strip() for x in unscrambledCodes)]，那么你有任何其他操作，如new_file=["|"+line for line in new_file[:-1]]和new_file[-1]=......

另外，作为别人上文所指出的，你可能会想尝试与关键字（即使它会带来缩进的另一个层面），如

with open("....","r") as in_file, open("...","w") as out_file:

`'''.... do your stuff'''`