Question

我有一个使用for循环生成的列表。它返回：

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

如何只返回第三列和第五列？

EDIT 当我收到错误时，它会说'Nonetype＆＃39;对象不可迭代

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

我也尝试使用list（enumerate（my_list））获取索引，但它没有工作我得到TypeError：＆＃39; NoneType＆＃39;对象不可迭代

这就是我填充列表的方式：

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

就我无法发表评论而言，我会尽力为您提供问题的解决方案。

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)

所以这个想法是用空格分隔每一行并过滤掉任何空字符串以获得第3和第5列（分别为索引2和4）

python：如何在列表中返回2列/索引

1 个答案: