python：如何返回列2列/索引

Question

我有一個列表，我使用for循環生成。返回：

home1-0002_UUID 3457077784 2132011944 1307504896 62% 
home1-0003_UUID 3457077784 2088064860 1351451980 61% 
home1-0001_UUID 3457077784 2092270236 1347246604 61% 

我怎麼能只返回第三和第五列？

編輯 當我得到它說「Nonetype」對象不是可迭代

for index, elem in enumerate(my_list): 
    print (index,elem) 

我也試圖通過使用列表（枚舉（my_list））來獲得指數，但它不工作，我的錯誤獲得類型錯誤：「NoneType」對象不是可迭代

我這是怎麼填充列表：

def h1ost(): 
    p1 = subprocess.Popen("/opt/lustre-gem_s/default/bin/lfs df /lustre/home1 | sort -r -nk5",stdout=subprocess.PIPE, shell=True) 
    use = p1.communicate()[0] 
    for o in use.split('\n')[:6]: 
     if "Available" not in o and "summary" not in o: 
      print (o) 

Answer 1

據我不能發表評論，我會盡我最大的努力給ÿ對問題的解決方案。

def empty_string_filter(value): 
    return value != '' 

def h1ost(): 
    p1 = subprocess.Popen("/opt/lustre-gem_s/default/bin/lfs df /lustre/home1 | sort -r -nk5",stdout=subprocess.PIPE, shell=True) 
    use = p1.communicate()[0] 
    new_file_content_list = [] 
    # Separate by line break and filter any empty string 
    filtered_use_list = filter(empty_string_filter, use.split(os.linesep)[:6]) 
    for line in filtered_use_list : 
     # Split the line and filter the empty strings in order to keep only 
     # columns with information 
     split_line = filter(empty_string_filter, line.split(' ')) 
     # Beware! This will only work if each line has 5 or more data columns 
     # I guess the correct option is to check if it has at least 5 columns 
     # and if it hasn't do not store the information or raise an exception. 
     # It's up to you. 

     new_file_content_list.append('{0} {1}'.format(split_line[2] , split_line[4])) 

    return os.linesep.join(new_file_content_list) 

這樣的想法是分裂用空格每一行和過濾離開，以獲得第3和第5列（索引2和4分別）