在字典中存储一个键的多个值

Question

I have a list of data, which has 2 values:

a    12
a    11
a    5
a    12
a    11

I would like to use a dictionary, so I can end up with a list of values for each of the key. Column 1 may have a different entry, like 'b' , so I can arrange data based on column 1 as key, while column 2 is the data for each key

[a:12,11,5]

How do I achieve this? From what I read, if 2 values has the same key, the last one override the previous one, so only one key is in the dictionary.

d={}
for line in results:
    templist=line.split(' ')
    thekey=templist[0]  
    thevalue=templist[1]

    if thevalue in d:
        d[thekey].append(thevalue)
    else:
        d[thekey]=[thevalue]

Am I approaching the problem using the wrong way?

Answer 1

Python dicts can have only one value for a key, so you cannot assign multiple values in the fashion you are trying to.

Instead, store the mutiple values in a list corresponding to the key so that the list becomes the one value corresponding to the key :

d = {}
d["a"] = []
d["a"].append(1)
d["a"].append(2)

>>> print d
{'a': [1, 2]}

You can use a defaultdict to simplify this, which will initialise the key if it doesn't exist with an empty list as below:

from collections import defaultdict
d = defaultdict(list)
d["a"].append(1)
d["a"].append(2)

>>> print d
defaultdict(<type 'list'>, {'a': [1, 2]})

If you don't want repeat values for a key, you can use a set instead of list. But do note that a set is unordered.

from collections import defaultdict
d = defaultdict(set)
d["a"].add(1)
d["a"].add(2)
d["a"].add(1)

>>> print d
defaultdict(<type 'set'>, {'a': set([1, 2])})

If you need to maintain order, either use sorted at runtime, or use a list with an if clause to check for values

from collections import defaultdict
d = defaultdict(list)
for item in (1, 2, 1, 2, 3, 4, 1, 2):
    if item not in d["a"]:
        d["a"].append(item)

>>> print d
defaultdict(<type 'list'>, {'a': [1, 2, 3, 4]})