Algorithm in Python To Solve This Problem

I have a list of lists such as: [[foo,1],[baz,1],[foo,0],[bar,3],[foo,1],[bar,2],[baz,2]]. I 开发者_如何学Pythonwant to get all the different items in the inner lists and find the total number of them. I mean the result should be like: [[foo,2],[bar,5],[baz,3]]. How can I do this task?

Thanks in advance.

Create a dictionary

D = {}
for item in list:
  left,right=item
  D[left] = D.get(left, 0) + right

There may be faster ways to do this though.

As suggested in the comments by Joce, Gnibbler and Blair you coud do this to get a list again.

# To get a list of lists
pairs = map(list, D.items()) 
# To get a list of tuples
pairs = D.items()

The defaultdict makes this fairly easy:

import collections
items = [['foo',1],['baz',1],['foo',0],['bar',3],['foo',1],['bar',2],['baz',2]]

totals = collections.defaultdict(int)
for key, value in items:
    totals[key] += value

print totals

When run, this gives

defaultdict(<type 'int'>, {'bar': 5, 'foo': 2, 'baz': 3})

If you want a list output, just pull the items from the dictionary

print totals.items()

and you get

[('bar', 5), ('foo', 2), ('baz', 3)]

If you really want a list-of-lists at the end,

print [list(item) for item in totals.items()]

which gives you

[['bar', 5], ['foo', 2], ['baz', 3]]

You can use itertools.groupby.

>>> import operator
>>> import itertools
>>> data = [['foo',1],['baz',1],['foo',0],['bar',3],['foo',1],['bar',2],['baz',2
]]
>>> {i:sum(k[1] for k in j)
...  for i, j in itertools.groupby(sorted(data, key=operator.itemgetter(0)),
...                                key=operator.itemgetter(0))}
{'baz': 3, 'foo': 2, 'bar': 5}

If you are using Python 3.2 (and 2.7) then you can do:

>>> from collections import Counter
>>> items = [['foo',1],['baz',1],['foo',0],['bar',3],['foo',1],['bar',2],['baz',2]]
>>> Counter(sum(( [key]*count for key,count in items), []))
Counter({'bar': 5, 'baz': 3, 'foo': 2})
>>> Counter(sum(( [key]*count for key,count in items), [])).most_common()
[('bar', 5), ('baz', 3), ('foo', 2)]
>>>