在python2.7中通過目錄中的文件循環讀取文件時出錯

Question

我從昨天開始使用googling這個問題並無濟於事;

當我遍歷一個目錄中的多個文件，並處理該循環中每個文件的行時，我總是關閉，但好像python打開了同一個內存空間中的所有文件，所以當我循環遍歷一個文件，我從先前打開的文件中檢索所有記錄，就好像它們在一個指針數組中。 。 。 .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，這裏是我的I/O樣本：

輸入文件1：
AB00007737106517 COSTCLASSU275
C000000010031932155750539976333693187714
C000000010031932155750539976105307608239

文件2：
AB00007736638744 COSTCLASSU275
C00000001003028490769901248060 8351468369
C000000020030284907699012480751885101503

file3的：
AB00007737148207 COSTCLASSU275
C000000010032271716759259098738354718484
C000000020032271716759259098394986919513

期望的輸出文件1：
AB00007737148207 COSTCLASSU275
| C000000010031932155750539976079292077121
| C000000010031932155750539976126217711213

文件2：
AB00007736638744 COSTCLASSU275
| C000000010030284907699012480968864628712
| C000000020030284907699012480294550195814

文件3：
AB00007737106517 COSTCLASSU275
| C000000010032271716759259098216262704445
| C000000020032271716759259098085462231948

電流輸出文件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（）將一次讀取文件的一行，遞增到下一行，直到整個文件被讀入

你。可以使用類似的東西：

content = unscrambledCodes.readlines() 

這將讀取每一行到數組中。然後，您可以遍歷內容，並根據需要更新行。

此外，而不是文件（），我一般用

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

Answer 2

普遍的共識是使用開放式的，而不是文件。我會以此開始。

其次，儘量做你打開的文件發電機的理解，因爲它是非常容易（下一個方法將返回一個換行符）作爲new_file=[x.strip() for x in unscrambledCodes)]，那麼你有任何其他操作，如new_file=["|"+line for line in new_file[:-1]]和new_file[-1]=......

另外，作爲別人上文所指出的，你可能會想嘗試與關鍵字（即使它會帶來縮進的另一個層面），如

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

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