閱讀Python遞歸文件夾

Question

我有一個C++/Obj-C背景，我只是發現Python（已經寫了大約一個小時）。 我正在寫一個腳本遞歸讀取文件夾結構中的文本文件的內容。

我遇到的問題是我寫的代碼只能用於一個文件夾。我可以看到爲什麼在代碼中（見#hardcoded path），我只是不知道如何繼續使用Python，因爲我的經驗只是全新的。

Python代碼：

import os 
import sys 

rootdir = sys.argv[1] 

for root, subFolders, files in os.walk(rootdir): 

    for folder in subFolders: 
     outfileName = rootdir + "/" + folder + "/py-outfile.txt" # hardcoded path 
     folderOut = open(outfileName, 'w') 
     print "outfileName is " + outfileName 

     for file in files: 
      filePath = rootdir + '/' + file 
      f = open(filePath, 'r') 
      toWrite = f.read() 
      print "Writing '" + toWrite + "' to" + filePath 
      folderOut.write(toWrite) 
      f.close() 

     folderOut.close() 

Answer 1

確保您瞭解os.walk三個返回值：

for root, subdirs, files in os.walk(rootdir): 

具有以下含義：

• root：「走過」的當前路徑
• subdirs：文件中的目錄類型
• files的root：文件中的目錄相比

並請使用os.path.join，而不是用斜槓串聯的其他類型的root（不subdirs）！您的問題是filePath = rootdir + '/' + file - 您必須連接當前「走」的文件夾而不是最頂層的文件夾。所以那一定是filePath = os.path.join(root, file)。順便說一句「文件」是一個內置的，所以你通常不使用它作爲變量名稱。

的另一個問題是你的循環，這應該是這樣的，例如：

import os 
import sys 

walk_dir = sys.argv[1] 

print('walk_dir = ' + walk_dir) 

# If your current working directory may change during script execution, it's recommended to 
# immediately convert program arguments to an absolute path. Then the variable root below will 
# be an absolute path as well. Example: 
# walk_dir = os.path.abspath(walk_dir) 
print('walk_dir (absolute) = ' + os.path.abspath(walk_dir)) 

for root, subdirs, files in os.walk(walk_dir): 
    print('--\nroot = ' + root) 
    list_file_path = os.path.join(root, 'my-directory-list.txt') 
    print('list_file_path = ' + list_file_path) 

    with open(list_file_path, 'wb') as list_file: 
     for subdir in subdirs: 
      print('\t- subdirectory ' + subdir) 

     for filename in files: 
      file_path = os.path.join(root, filename) 

      print('\t- file %s (full path: %s)' % (filename, file_path)) 

      with open(file_path, 'rb') as f: 
       f_content = f.read() 
       list_file.write(('The file %s contains:\n' % filename).encode('utf-8')) 
       list_file.write(f_content) 
       list_file.write(b'\n') 

如果你不知道，with聲明的文件是一個速記：

with open('filename', 'rb') as f: 
    dosomething() 

# is effectively the same as 

f = open('filename', 'rb') 
try: 
    dosomething() 
finally: 
    f.close() 

Answer 2

我認爲這個問題是你沒有正確地處理的os.walk輸出。

首先，改變：

filePath = rootdir + '/' + file 

到：

filePath = root + '/' + file 

rootdir是你的固定起始目錄; root是由os.walk返回的目錄。

其次，你不需要縮進你的文件處理循環，因爲這對每個子目錄都是沒有意義的。你會得到root設置爲每個子目錄。你不需要手工處理子目錄，除非你想對目錄本身做些什麼。

Answer 3

使用os.path.join()來構建你的路 - 這是整潔：

import os 
import sys 
rootdir = sys.argv[1] 
for root, subFolders, files in os.walk(rootdir): 
    for folder in subFolders: 
     outfileName = os.path.join(root,folder,"py-outfile.txt") 
     folderOut = open(outfileName, 'w') 
     print "outfileName is " + outfileName 
     for file in files: 
      filePath = os.path.join(root,file) 
      toWrite = open(filePath).read() 
      print "Writing '" + toWrite + "' to" + filePath 
      folderOut.write(toWrite) 
     folderOut.close() 

Answer 4

同意與Dave Webb，os.walk將產生樹中每個目錄的項目。事實是，你只需要不在乎subFolders。

這樣的代碼應該工作：

import os 
import sys 

rootdir = sys.argv[1] 

for folder, subs, files in os.walk(rootdir): 
    with open(os.path.join(folder, 'python-outfile.txt'), 'w') as dest: 
     for filename in files: 
      with open(os.path.join(folder, filename), 'r') as src: 
       dest.write(src.read()) 

Answer 5

os.walk默認完成遞歸的步行路程。對於每一個目錄，從根開始它產生一個3元組（dirpath，dirnames中，文件名）

from os import walk 
from os.path import splitext, join 

def select_files(root, files): 
    """ 
    simple logic here to filter out interesting files 
    .py files in this example 
    """ 

    selected_files = [] 

    for file in files: 
     #do concatenation here to get full path 
     full_path = join(root, file) 
     ext = splitext(file)[1] 

     if ext == ".py": 
      selected_files.append(full_path) 

    return selected_files 

def build_recursive_dir_tree(path): 
    """ 
    path - where to begin folder scan 
    """ 
    selected_files = [] 

    for root, dirs, files in walk(path): 
     selected_files += select_files(root, files) 

    return selected_files 

Answer 6

試試這個：

import os 
import sys 

for root, subdirs, files in os.walk(path): 

    for file in os.listdir(root): 

     filePath = os.path.join(root, file) 

     if os.path.isdir(filePath): 
      pass 

     else: 
      f = open (filePath, 'r') 
      # Do Stuff 

Answer 7

如果您正在使用Python 3.5+或以上，就可以得到這在1行完成。

for filename in glob.iglob(root_dir + '**/*.txt', recursive=True): 
    print(filename) 

如前所述in documentation

If recursive is true, the pattern '**' will match any files and zero or more directories and subdirectories.

如果你想每一個文件，你可以使用

for filename in glob.iglob(root_dir + '**/*', recursive=True): 
    print(filename)