優雅的方式來檢查Python 2.7中有效的文件目錄路徑

Question

我想讀取特定目錄下的所有文件的內容。我發現如果路徑名不是以/結尾，那麼我的代碼將不起作用（將會產生I/O異常，因爲pathName+f無效 - 在中間缺少/）。下面是一個代碼示例，以顯示它工作時，當它不工作，

實際上，我可以檢查如果pathname採用的endsWith與/結束，如果更優雅的解決方案，只是不知道什麼時候一個全名串連路徑和文件名？

我的要求是，我想給輸入pathName更靈活，以\結尾，而不是以\結束。

使用Python 2.7。

from os import listdir 
from os.path import isfile, join 

#pathName = '/Users/foo/Downloads/test/' # working 
pathName = '/Users/foo/Downloads/test' # not working, since not ends with/ 
onlyfiles = [f for f in listdir(pathName) if isfile(join(pathName, f))] 
for f in onlyfiles: 
    with open(pathName+f, 'r') as content_file: 
     content = content_file.read() 
     print content 

Answer 1

你只想用再次加入：

pathName = '/Users/foo/Downloads/test' # not working, since not ends with/ 
onlyfiles = [f for f in listdir(pathName) if isfile(join(pathName, f))] 
for f in onlyfiles: 
    with open(join(pathName, f), 'r') as content_file: 
     content = content_file.read() 
     print content 

或者你可以使用水珠忘記加入：

from glob import glob 

pathName = '/Users/foo/Downloads/test' # not working, since not ends with/ 

onlyfiles = (f for f in glob(join(pathName,"*")) if isfile(f)) 

for f in onlyfiles: 
    with open(f, 'r') as content_file: 

或過濾器結合起來的一個更簡潔的解決方案：

onlyfiles = filter(isfile, glob(join(pathName,"*")))