appropriate data structure for sorting a multi dimensional array in python?

Question

I have three numeric values (weight, count, contribution) for various strings (words) that i would like to organise into one multidimensional array, and then sort. To do this, I made lists within a dictionary, where the numeric values are in the list and the string is the key:

print_dictionary[word] = [weight,count,contribution]

How can I sort, first in ascending order and then in descending order, by 'contribution' (the third value in the list), and show the first 10 items of the sorted list. How can I do this?

For example, for the following print_dictionary:

print_dictionary[sam] = [2,7,1]
print_dictionary[sun] = [4,1,3]
print_dictionary[dog] = [1,3,2]

I want to them be able to sort contribution in ascending order:

Word:   Weight:   Count:    Contribution:
sam     2         7         1
dog     1         3         2
sun     4         1         3

I don't see how itemegetter can be used for this:

sorted(print_dictionary, key=itemgetter(2))

Lists are ordered in python. I think you meant "dicts are unordered". — Joel Cornett
– Joel Cornett, Commented May 27, 2012 at 2:48
Also, you can't use itemgetter on a dictionary like that. You have to call print_dictionary.items(). — Joel Cornett
– Joel Cornett, Commented May 27, 2012 at 2:53
Do you just want the keys in order so that you can access it in contribution order? — Michael
– Michael, Commented May 27, 2012 at 3:09
I want to sort between sets, not within sets. I added an example in the question. — Zach
– Zach, Commented May 27, 2012 at 3:16

dank.game · Accepted Answer · 2012-05-27 03:34:29Z

4

You can pass an anonymous function as the key to sorted. This uses the third member of the multi-dimensional dict as the key:

>>> d = {'a': [1, 4, 7], 'b': [2, 3, 9], 'c': [3, 2, 8]}
>>> for key in sorted(d, key=lambda x: d[x][2]):
...    print key, d[key]
a [1, 4, 7]
c [3, 2, 8]
b [2, 3, 9]

For descending order, use reverse=True. To limit the results, add [:N]:

sorted(d, key=lambda x: d[x][2], reverse=True)[:2]

# b [2, 3, 9]
# c [3, 2, 8]

More about sorted and sorting in Python.

edited May 27, 2012 at 3:34

answered May 27, 2012 at 3:17

dank.game

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dank.game Over a year ago

Close, just put the slice statement on sorted: sorted(d, key=lambda x: d[x][2])[:10]

Mark Reed · Accepted Answer · 2012-05-27 03:28:38Z

1

You can't really sort a dictionary; when you try, you are really just sorting the list of keys from the dictionary. You can do that using a custom sort comparison that looks at the third item in the value.

sorted(print_dictionary, key=lambda word: print_dictionary[word][2])

So to generate your report, something like this would work:

sorted_keys = sorted(print_dictionary, key=lambda word: print_dictionary[word][2])

print "Word:\tWeight:\tCount:\tContribution"      
for i in range(10): # or however many you want
    word = sorted_keys[i]
    values = print_dictionary[word]
    print "\t".join([word]+[str(n) for n in values])

edited May 27, 2012 at 3:28

answered May 27, 2012 at 3:17

Mark Reed

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