append multiple values for one key in a dictionary [duplicate]

Solution 1:

If I can rephrase your question, what you want is a dictionary with the years as keys and an array for each year containing a list of values associated with that year, right? Here's how I'd do it:

years_dict = dict()

for line in list:
    if line[0] in years_dict:
        # append the new number to the existing array at this slot
        years_dict[line[0]].append(line[1])
    else:
        # create a new array in this slot
        years_dict[line[0]] = [line[1]]

What you should end up with in years_dict is a dictionary that looks like the following:

{
    "2010": [2],
    "2009": [4,7],
    "1989": [8]
}

In general, it's poor programming practice to create "parallel arrays", where items are implicitly associated with each other by having the same index rather than being proper children of a container that encompasses them both.

Solution 2:

You would be best off using collections.defaultdict (added in Python 2.5). This allows you to specify the default object type of a missing key (such as a list).

So instead of creating a key if it doesn't exist first and then appending to the value of the key, you cut out the middle-man and just directly append to non-existing keys to get the desired result.

A quick example using your data:

>>> from collections import defaultdict
>>> data = [(2010, 2), (2009, 4), (1989, 8), (2009, 7)]
>>> d = defaultdict(list)
>>> d
defaultdict(<type 'list'>, {})
>>> for year, month in data:
...     d[year].append(month)
... 
>>> d
defaultdict(<type 'list'>, {2009: [4, 7], 2010: [2], 1989: [8]})

This way you don't have to worry about whether you've seen a digit associated with a year or not. You just append and forget, knowing that a missing key will always be a list. If a key already exists, then it will just be appended to.

Solution 3:

You can use setdefault.

for line in list:  
    d.setdefault(year, []).append(value)

This works because setdefault returns the list as well as setting it on the dictionary, and because a list is mutable, appending to the version returned by setdefault is the same as appending it to the version inside the dictionary itself. If that makes any sense.