如何在python中通過id拆分文本文件

Question

我有一堆包含製表符分隔的文本文件。第二列包含一個ID號，每個文件已經按該ID號排序。我想通過第2列中的id號將每個文件分隔成多個文件。這就是我所擁有的。

readpath = 'path-to-read-file' 
writepath = 'path-to-write-file' 
for filename in os.listdir(readpath): 
    with open(readpath+filename, 'r') as fh: 
      lines = fh.readlines() 
    lastid = 0 
    f = open(writepath+'checkme.txt', 'w') 
    f.write(filename) 
    for line in lines: 
      thisid = line.split("\t")[1] 
      if int(thisid) <> lastid: 
       f.close() 
       f = open(writepath+thisid+'-'+filename,'w') 
       lastid = int(thisid) 
      f.write(line) 
    f.close() 

我得到的是簡單地全部用在新的文件名前面的每個文件的第一個ID號讀取文件的副本。這就好像

thisid = line.split("\t")[1] 

只在循環中完成一次。對發生了什麼的任何線索？

編輯

使用的問題我的文件\ r而非\ r \ n至終止線。更正後的代碼（只需添加'儒的打開讀取文件和交換時爲= <>！）：

readpath = 'path-to-read-file' 
writepath = 'path-to-write-file' 
for filename in os.listdir(readpath): 
    with open(readpath+filename, 'rU') as fh: 
      lines = fh.readlines() 
    lastid = 0 
    f = open(writepath+'checkme.txt', 'w') 
    f.write(filename) 
    for line in lines: 
      thisid = line.split("\t")[1] 
      if int(thisid) != lastid: 
       f.close() 
       f = open(writepath+thisid+'-'+filename,'w') 
       lastid = int(thisid) 
      f.write(line) 
    f.close() 

Answer 1

如果你處理的製表符分隔的文件，那麼你可以使用csv模塊，並利用itertools.groupby將爲您執行之前/當前跟蹤的ID。還利用os.path.join來確保您的文件名最終正確連接。

未經測試：

import os 
import csv 
from itertools import groupby 

readpath = 'path-to-read-file' 
writepath = 'path-to-write-file' 

for filename in os.listdir(readpath): 
    with open(os.path.join(readpath, filename)) as fin: 
     tabin = csv.reader(fin, delimiter='\t') 
     for file_id, rows in groupby(tabin, lambda L: L[1]): 
      with open(os.path.join(writepath, file_id + '-' + filename), 'w') as fout: 
       tabout = csv.writer(fout, delimiter='\t') 
       tabout.writerows(rows)