Question

我有一个包含这样的文件路径的列表：

my_paths = ['/home/mark/results/chilo/15381_chilo_140618_099_X/15381_chilo.csv','/home/mark/results/chilo/15382_chilo_140610_099_X/15382_chilo.csv','/home/mark/results/chilo/15383_chilo_140616_099_X/15383_chilo.csv','/home/mark/results/chilo/15384_chilo_140620_099_X/15384_chilo.csv']

我喜欢根据第二级日期对列表进行排序，例如140616 in 15383_chilo_140616_099_X。所以输出应该是：

['/home/mark/results/chilo/15382_chilo_140610_099_X/15382_chilo.csv', '/home/mark/results/chilo/15383_chilo_140616_099_X/15383_chilo.csv', '/home/mark/results/chilo/15381_chilo_140618_099_X/15381_chilo.csv', '/home/mark/results/chilo/15384_chilo_140620_099_X/15384_chilo.csv']

这样做的最佳方法是什么。我无法理解我是否应首先遍历路径，采取如下的第二级：

for my_path in my_paths:    
    (SeqDir,seqFileName) = os.path.split(my_path)
    (SeqDir_remaining,second_level) = os.path.split(SeqDir)

....然后拆分下划线，取日期然后对其进行排序并采取该日期的路径，或使用字典并将日期作为键，将路径作为值（但随后出现问题排序）。

感谢您的帮助。

谢谢！

Answer 1

在下划线上拆分三次并将第三个元素转换为int，路径分隔符无关紧要，您只需要第二个和第三个下划线之间的数字：

my_paths = ['/home/mark/results/chilo/15381_chilo_140618_099_X/15381_chilo.csv','/home/mark/results/chilo/15382_chilo_140610_099_X/15382_chilo.csv','/home/mark/results/chilo/15383_chilo_140616_099_X/15383_chilo.csv','/home/mark/results/chilo/15384_chilo_140620_099_X/15384_chilo.csv']

my_list.sort(key=lambda x: int(x.split("_", 3)[2])))

输出：

['/home/mark/results/chilo/15382_chilo_140610_099_X/15382_chilo.csv', 
'/home/mark/results/chilo/15383_chilo_140616_099_X/15383_chilo.csv', 
'/home/mark/results/chilo/15381_chilo_140618_099_X/15381_chilo.csv', 
'/home/mark/results/chilo/15384_chilo_140620_099_X/15384_chilo.csv']

如果它们实际上是年/月/日日期，则不需要使用int。

Answer 2

编写一个函数来提取要排序的东西：

def getdate(item):
    ...

然后

my_paths.sort(key=getdate)

您的getdate功能可能需要比这更好，但您明白了这一点：

>>> import pprint
>>> pprint.pprint(my_paths)
['/home/mark/results/chilo/15381_chilo_140618_099_X/15381_chilo.csv',
 '/home/mark/results/chilo/15382_chilo_140610_099_X/15382_chilo.csv',
 '/home/mark/results/chilo/15383_chilo_140616_099_X/15383_chilo.csv',
 '/home/mark/results/chilo/15384_chilo_140620_099_X/15384_chilo.csv']
>>> def getdate(item):
...     start = len('/home/mark/results/chilo/15381_chilo_')
...     end = start + 6
...     return item[start:end]
...
>>> getdate(my_paths[0])
'140618'
>>> my_paths.sort(key=getdate)
>>> pprint.pprint(my_paths)
['/home/mark/results/chilo/15382_chilo_140610_099_X/15382_chilo.csv',
 '/home/mark/results/chilo/15383_chilo_140616_099_X/15383_chilo.csv',
 '/home/mark/results/chilo/15381_chilo_140618_099_X/15381_chilo.csv',
 '/home/mark/results/chilo/15384_chilo_140620_099_X/15384_chilo.csv']
>>>

Answer 3

def sort_links(my_paths, pattern):
# to sort by chilo_xxxxxx
# pattern = r'(chilo_\d+)'
  import re
  my_paths = sorted(my_paths,key=lambda x : re.search(pattern,x).groups(1)[0])
  return my_paths


my_paths = sorted(my_paths,key=f)
return my_paths

    print(sort_links(my_paths,r'(chilo_\d+)'))

['/home/mark/results/chilo/15382_chilo_140610_099_X/15382_chilo.csv', '/home/mark/results/chilo/15383_chilo_140616_099_X/15383_chilo.csv', '/home/mark/results/chilo/15381_chilo_140618_099_X/15381_chilo.csv', '/home/mark/results/chilo/15384_chilo_140620_099_X/15384_chilo.csv']

python如何基于正则表达式拆分列表并对其进行排序

3 个答案: