2.5. Dictionaries¶

Contents

How can I get a dictionary to display its keys in a consistent order?
I want to do a complicated sort: can you do a Schwartzian Transform in Python?
How can I sort one list by values from another list?

2.5.1. How can I get a dictionary to display its keys in a consistent order?¶

You can’t. Dictionaries store their keys in an unpredictable order, so the display order of a dictionary’s elements will be similarly unpredictable.

This can be frustrating if you want to save a printable version to a file, make some changes and then compare it with some other printed dictionary. In this case, use the pprint module to pretty-print the dictionary; the items will be presented in order sorted by the key.

A more complicated solution is to subclass dict to create a SortedDict class that prints itself in a predictable order. Here’s one simpleminded implementation of such a class:

class SortedDict(dict):
    def __repr__(self):
        keys = sorted(self.keys())
        result = ("{!r}: {!r}".format(k, self[k]) for k in keys)
        return "{{{}}}".format(", ".join(result))

    __str__ = __repr__

This will work for many common situations you might encounter, though it’s far from a perfect solution. The largest flaw is that if some values in the dictionary are also dictionaries, their values won’t be presented in any particular order.

2.5.2. I want to do a complicated sort: can you do a Schwartzian Transform in Python?¶

The technique, attributed to Randal Schwartz of the Perl community, sorts the elements of a list by a metric which maps each element to its “sort value”. In Python, use the key argument for the sort() function:

Isorted = L[:]
Isorted.sort(key=lambda s: int(s[10:15]))

2.5.3. How can I sort one list by values from another list?¶

Merge them into a single list of tuples, sort the resulting list, and then pick out the element you want.

>>> list1 = ["what", "I'm", "sorting", "by"]
>>> list2 = ["something", "else", "to", "sort"]
>>> pairs = zip(list1, list2)
>>> pairs
[('what', 'something'), ("I'm", 'else'), ('sorting', 'to'), ('by', 'sort')]
>>> pairs.sort()
>>> result = [ x[1] for x in pairs ]
>>> result
['else', 'sort', 'to', 'something']

An alternative for the last step is:

>>> result = []
>>> for p in pairs: result.append(p[1])

If you find this more legible, you might prefer to use this instead of the final list comprehension. However, it is almost twice as slow for long lists. Why? First, the append() operation has to reallocate memory, and while it uses some tricks to avoid doing that each time, it still has to do it occasionally, and that costs quite a bit. Second, the expression “result.append” requires an extra attribute lookup, and third, there’s a speed reduction from having to make all those function calls.