如何从目录路径列表中打开文件？

Question

I have a list of folders that I need to open and read a .log file. At first, I did not have the list of folders, so I was scanning all the files in the directory. This will lead to redundancy for what I want to use this for.

path = '/home/User/Test/'
files = []

# r=root, d=directories, f = files
for r, d, f in os.walk(path):
    for file in f:
        if '.log' in file:
            files.append(os.path.join(r, file))

for f in files:
    log_file = open(f, 'r')

    lines = log_file.readlines()
    log_file.close()

Now, I have a list paths which look like ['/home/User/Test/folder_test/Process1/Task1/2019-07-31T10%3A30%3A00+00%3A00', ...., 'final_path']

How can I loop through the paths so it is opening the folder, at then extracting the .log file?

NOTE: I am getting rid of path = '/home/User/Test/' as I have the list of specific directories.

Answer 1

I think you can utilize the glob library for this problem. It allows you to pattern-match your paths.