Differences between revisions 2 and 13 (spanning 11 versions)
Revision 2 as of 2003-11-21 19:32:38
Size: 579
Editor: dsl254-010-130
Comment: Question: This a good way?
Revision 13 as of 2010-03-30 19:08:00
Size: 2180
Editor: wl-ol-s246-130
Comment:
Deletions are marked like this. Additions are marked like this.
Line 11: Line 11:
a = { "key1":5, "key2":8, "key3":2 }
b = { "key1":7, "key2":4, "key3":9 }
c = { "key1":6, "key2":1, "key3":1 }
l = [ a,b,c ] # how do you sort this list?
a = {"key1": 5 , "key2": 8, "key3": 2}
b = {"key1": 7 , "key2": 4, "key3": 9}
c = {"key1": 6 , "key2": 1, "key3": 1}
undecorated = [a, b, c] # how do you sort this list?
Line 17: Line 17:
There are many ways to do this; Here's one way I've been doing it: There are many ways to do this. Here's the fastest way to do it, as it avoids using a custom comparison function, instead using builtin comparisons. This is the ''decorate-sort-undecorate'' pattern, or the ''Schwartzian transform'' if you're coming from Perl.
Line 21: Line 21:
sort_key = "key2"
l = [(d[sort_key],d) for d in l]
l.sort()
l = [d for (k,d) in l]
sort_on = "key2"
decorated = [(dict_[sort_on], dict_) for dict_ in undecorated]
decorated.sort()
result = [dict_ for (key, dict_) in decorated]
Line 27: Line 27:
== Questions == (The variable was named {{{dict_}}} because {{{dict}}} is already a builtin.)
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This a good way? -- LionKimbro Starting with Py2.4 the {{{list.sort()}}} method provides a {{{key=}}} argument for doing the transform in a single step. The new {{{sorted()}}} built-in function goes a step further and encapsulates making a new sorted list while leaving the original intact. Also, the new {{{operator.itemgetter()}}} function helps by constructing functions for key access:

{{{
>>> from operator import itemgetter
>>> result = sorted(undecorated, key=itemgetter('key2'))
}}}

This will sort on arbitrary multiple columns of the dictionary.
{{{
#!python
def multikeysort(items, columns):
    from operator import itemgetter
    comparers = [ ((itemgetter(col[1:].strip()), -1) if col.startswith('-') else (itemgetter(col.strip()), 1)) for col in columns]
    def comparer(left, right):
        for fn, mult in comparers:
            result = cmp(fn(left), fn(right))
            if result:
                return mult * result
        else:
            return 0
    return sorted(items, cmp=comparer)
}}}
You can call it like this:
{{{
>>> result = multikeysort(undecorated, ['key1', 'key2', 'key3'])
}}}
Column names preceded by '-' are sorted in descending order:
{{{
>>> result = multikeysort(undecorated, ['-key1', '-key2', '-key3'])
}}}

== See Also ==

 * [[HowTo/Sorting]]

Sorting Lists of Dictionaries

Frequently you want to sort a list of dictionaries, based on some particular key.

For example:

   1 a = {"key1": 5 , "key2": 8, "key3": 2}
   2 b = {"key1": 7 , "key2": 4, "key3": 9}
   3 c = {"key1": 6 , "key2": 1, "key3": 1}
   4 undecorated = [a, b, c] # how do you sort this list?

There are many ways to do this. Here's the fastest way to do it, as it avoids using a custom comparison function, instead using builtin comparisons. This is the decorate-sort-undecorate pattern, or the Schwartzian transform if you're coming from Perl.

   1 sort_on = "key2"
   2 decorated = [(dict_[sort_on], dict_) for dict_ in undecorated]
   3 decorated.sort()
   4 result = [dict_ for (key, dict_) in decorated]

(The variable was named dict_ because dict is already a builtin.)

Starting with Py2.4 the list.sort() method provides a key= argument for doing the transform in a single step. The new sorted() built-in function goes a step further and encapsulates making a new sorted list while leaving the original intact. Also, the new operator.itemgetter() function helps by constructing functions for key access:

>>> from operator import itemgetter
>>> result = sorted(undecorated, key=itemgetter('key2'))

This will sort on arbitrary multiple columns of the dictionary.

   1 def multikeysort(items, columns):
   2     from operator import itemgetter
   3     comparers = [ ((itemgetter(col[1:].strip()), -1) if col.startswith('-') else (itemgetter(col.strip()), 1)) for col in columns]
   4     def comparer(left, right):
   5         for fn, mult in comparers:
   6             result = cmp(fn(left), fn(right))
   7             if result:
   8                 return mult * result
   9         else:
  10             return 0
  11     return sorted(items, cmp=comparer)

You can call it like this:

>>> result = multikeysort(undecorated, ['key1', 'key2', 'key3'])

Column names preceded by '-' are sorted in descending order:

>>> result = multikeysort(undecorated, ['-key1', '-key2', '-key3'])

See Also

SortingListsOfDictionaries (last edited 2010-03-30 19:08:00 by wl-ol-s246-130)

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