使用python同时从两个目录中读取同名文件

我有目录A

A

包含

A.txt

，

b.txt

，

c.txt

我有目录B，

B

包含

a.csv

，

B.csv

，

c.csv

我希望能够分别读取和处理文件，如下所示：

a.py

src = '/home/A'
file_paths = [os.path.join(src, f) for f in os.listdir(src)]
for file in file_paths:

    for line in open(file, 'rb').readlines():
    #do some some
    # print some stuff  

src = '/home/B'
file_paths = [os.path.join(src, f) for f in os.listdir(src)]
for file in file_paths:

    with open(filename, 'r') as ff:
    for line in ff:
        _, end, label = line.strip().split('\t')

    #do some some
    # print some stuff  

b.py

src = '/home/A'
file_paths = [os.path.join(src, f) for f in os.listdir(src)]
for file in file_paths:

    for line in open(file, 'rb').readlines():
    #do some some
    # print some stuff  

src = '/home/B'
file_paths = [os.path.join(src, f) for f in os.listdir(src)]
for file in file_paths:

    with open(filename, 'r') as ff:
    for line in ff:
        _, end, label = line.strip().split('\t')

    #do some some
    # print some stuff  

因此，

a.py

的结果如下所示：

file_processed: a.txt   
output    
file processed: b.txt  
output  
file processed: c.txt  
output

同样，从

b.py

file_processed: a.csv   
output    
file processed: b.csv  
output  
file processed: c.csv  
output

我不想执行此过程，因为很难直观地比较.csv和.txt的结果，因为它们位于两个不同的终端或文件上等等

我想做的是：立即从dir读取同名文件，然后执行计算，然后打印结果

示例：我应该从/a中读取a.txt并对其进行处理，然后从/B中读取a.csv并对其进行处理。然后打印两个结果。然后从/A转到b.txt，从/Band转到b.csv，依此类推

---

请告知。如果您想让我更新

的逻辑，请告诉我做一些事情
代码部分。
只需形成两个文件路径列表，然后同时循环它们

a_root = '/home/A'
b_root = '/home/B'
a_paths = [os.path.join(a_root, f) for f in os.listdir(a_root)]
b_paths = [os.path.join(b_root, f) for f in os.listdir(b_root)]
for a_path, b_path in zip(a_paths, b_paths):
    with open(a_path) as a_file, open(b_path) as b_file:
        a_data = a_file.read()
        b_data = b_file.read()
    # do stuff with a and b

只需形成两个文件路径列表，然后同时循环它们

a_root = '/home/A'
b_root = '/home/B'
a_paths = [os.path.join(a_root, f) for f in os.listdir(a_root)]
b_paths = [os.path.join(b_root, f) for f in os.listdir(b_root)]
for a_path, b_path in zip(a_paths, b_paths):
    with open(a_path) as a_file, open(b_path) as b_file:
        a_data = a_file.read()
        b_data = b_file.read()
    # do stuff with a and b

示例：我应该从/a中读取a.txt并对其进行处理，然后读取
a、 csv from/B并对其进行处理。然后打印两个结果。然后转到
b、 txt来自/A，b.csv来自/Band，依此类推

我觉得你已经回答了你自己的问题。这甚至可以用作伪代码。这里有一个不雅但有效的例子

filenames = [f for f in however_you_scrape_you_filenames]
directory_extension_map = {'/path/to/a': 'txt',
                           '/path/to/b': 'csv'
                          }
for fname in filenames:
    for directory in directory_extension_map:
        extension = directory_extension_map[directory]
        file_path = os.path.join(directory, f'{fname}.{extension}')
        # Read and parse file_path

我觉得你可以从这里开始

示例：我应该从/a中读取a.txt并对其进行处理，然后读取
a、 csv from/B并对其进行处理。然后打印两个结果。然后转到
b、 txt来自/A，b.csv来自/Band，依此类推

我觉得你已经回答了你自己的问题。这甚至可以用作伪代码。这里有一个不雅但有效的例子

filenames = [f for f in however_you_scrape_you_filenames]
directory_extension_map = {'/path/to/a': 'txt',
                           '/path/to/b': 'csv'
                          }
for fname in filenames:
    for directory in directory_extension_map:
        extension = directory_extension_map[directory]
        file_path = os.path.join(directory, f'{fname}.{extension}')
        # Read and parse file_path

我觉得你可以从这里接手。
请提供一份报告。另外，为什么要在Python代码中使用C风格的注释？使用python注释。很抱歉，我将更改该注释。请提供一个。另外，为什么要在Python代码中使用C风格的注释？使用python注释。很抱歉，我会更改它。进展如何？这回答了你的问题吗？进展如何？这回答了你的问题吗？