python词典列表排序多个键

Question

I have a list of dictionaries that have keys of path and type (type 0 => file and type 1 => directory) initial looks like this

everything = [{u'path': u'/1/', u'type': 1},
     {u'path': u'/2/', u'type': 1},
     {u'path': u'/3/', u'type': 1},
     {u'path': u'/1/something/a.py', u'type': 0},
     {u'path': u'/1/something/b.py', u'type': 0},
     {u'path': u'/1/something/c.py', u'type': 0},
     {u'path': u'/1/foobar/', u'type': 1},
     {u'path': u'/2/baz/', u'type': 1},
     {u'path': u'/1/something/baz/snap/pop/a.py', u'type': 0},
     {u'path': u'/1/something/baz', u'type': 1}]

I want to first sort on path then off of type so I get something like this

everything = [{u'path': u'/1/', u'type': 1},
     {u'path': u'/1/foobar/', u'type': 1},
     {u'path': u'/1/something/baz', u'type': 1}, # < --- I want this here
     {u'path': u'/1/something/a.py', u'type': 0},
     {u'path': u'/1/something/b.py', u'type': 0},
     {u'path': u'/1/something/baz/snap/pop/a.py', u'type': 0},
     {u'path': u'/1/something/c.py', u'type': 0},
     {u'path': u'/2/', u'type': 1},
     {u'path': u'/2/baz/', u'type': 1},
     {u'path': u'/3/', u'type': 1}]

where the "something" section starts with type 1.

I thought when I do

everything.sort(key=lambda x: (x['path'], x['type']))

I would get what I want, but I get

everything = [{u'path': u'/1/', u'type': 1},
     {u'path': u'/1/foobar/', u'type': 1},
     {u'path': u'/1/something/a.py', u'type': 0},
     {u'path': u'/1/something/b.py', u'type': 0},
     {u'path': u'/1/something/baz', u'type': 1}, # < --- I don't want this here
     {u'path': u'/1/something/baz/snap/pop/a.py', u'type': 0},
     {u'path': u'/1/something/c.py', u'type': 0},
     {u'path': u'/2/', u'type': 1},
     {u'path': u'/2/baz/', u'type': 1},
     {u'path': u'/3/', u'type': 1}]

Is there an easy way to sort everything in this fashion or do I have to write my own sort?

EDIT:

Perhaps this will help explain what I want.

In linux when you do a ls -lR you get a nice list of dirs and files in the dirs listed under it.

Thanks

Answer 1

You need to only include in the key what you actually want to be sorted. Since you appear to want to reduce the priority of everything after the last slash-delimited segment in the path, you should remove that or move it later in the sort order:

everything.sort(key=lambda x: (
    '/'.join(x['path'].split('/')[:-1]),
    x['type'],
    ''.join(x['path'].split('/')[-1:]),
  )
)

Answer 2

你的排序标准是如此深奥，我认为你只能通过编写比较函数来做到这一点。