在 python - 我如何将项目转换为字典？

Question

I have following list of items (key-value pairs):

items = [('A', 1), ('B', 1), ('B', 2), ('C', 3)]

What I want to get:

{
  'A' : 1,
  'B' : [1,2]
  'C' : 3
}

My naive solution:

res = {}
for (k,v) in items:
     if k in res:
         res[k].append(v)
     else:
         res[k] = [v]

I'm looking for some optimised more pythonic solution, anyone?

Answer 1

Use could use defaultdict here.

from collections import defaultdict

res = defaultdict(list)
for (k,v) in items:
    res[k].append(v)
# Use as dict(res)

EDIT:

This is using groupby, but please note, the above is far cleaner and neater to the eyes :

>>> data = [('A', 1), ('B', 1), ('B', 2), ('C', 3)]
>>> dict([(key,list(v[1] for v in group)) for (key,group) in groupby(data, lambda x: x[0])])
{'A': [1], 'C': [3], 'B': [1, 2]}

Downside : Every element is a list. Change lists to generators as needed.

---

To convert all single item lists to individual items:

>>> res = # Array of tuples, not dict
>>> res = [(key,(value[0] if len(value) == 1 else value)) for key,value in res]
>>> res
[('A', 1), ('B', [1, 2]), ('C', 3)]

Answer 2

If you don't want to use defaultdict / groupby , the following works:

d = {}
for k,v in items:
   d.setdefault(k, []).append(v)

Answer 3

This is very easy to do with defaultdict , which can be imported from collections .

>>> from collections import defaultdict
>>> items = [('A', 1), ('B', 1), ('B', 2), ('C', 3)]
>>> d = defaultdict(list)
>>> for k, v in items:
    d[k].append(v)
>>> d
defaultdict(<class 'list'>, {'A': [1], 'C': [3], 'B': [1, 2]})

You can also use a dictionary comprehension:

>>> d = {l: [var for key, var in items if key == l] for l in {v[0] for v in items}}
>>> d
{'A': [1], 'C': [3], 'B': [1, 2]}

Answer 4

You can use the WebOb multidict implementation:

 >>> from webob import multidict
 >>> a = multidict.MultiDict(items)
 >>> a.getall('B')
 [1,2]

Answer 5

Maybe it looks ugly, but it works.

In [1]: items = [('A', 1), ('B', 1), ('B', 2), ('C', 3)]

In [2]: d = {}

In [3]: map(lambda i: d.update({i[0]: i[1] if d.get(i[0], i[1]) == i[1] else [d[i[0]], i[1]]}), items)
Out[3]: [None, None, None, None]

In [4]: print d
{'A': 1, 'C': 3, 'B': [1, 2]}

In else branch we can check if d[i[0]] returns a list.

Answer 6

To convert dictionary to list of items

dict_items = list(dict_1.items())

To convert the list of items back to the dictionary

dict2 = dict(dict_items)