Question

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In what order does python display dictionary keys?

There was this very weird Python question which I was trying to answer (on SO but looks like the author deleted it). This has gotten me head-over-heels and I cannot seem to plausibly explain myself what could be the cause behind this flummoxing phenomenon.
I have a Python dictionary (associative array) which I initialize as below.

d = {"word1": 1, "word2": 2, "word3": 3}

And loop through it using a for..in construct in which i print the keys of “d” separated by a space.
Expected output: word1 word2 word3
Actual output: word1 word3 word2

I then tried to print “d” as

 print d 

Output

{'word1': 1, 'word3': 3, 'word2': 2}

I was scanning the online Python docs, existing questionnaire on SO, blogs but fully failed to figure out the cause. Am I missing something here? Furthermore, there’s a kind request to all those who down-vote this question to please elaborate on the reason for the same.

Cheers. 🙂

Answer 1

From the documentation:

It is best to think of a dictionary as an unordered set of key: value pairs, with the requirement that the keys are unique (within one dictionary). A pair of braces creates an empty dictionary: {}. Placing a comma-separated list of key:value pairs within the braces adds initial key:value pairs to the dictionary; this is also the way dictionaries are written on output.

The main operations on a dictionary are storing a value with some key and extracting the value given the key. It is also possible to delete a key:value pair with del. If you store using a key that is already in use, the old value associated with that key is forgotten. It is an error to extract a value using a non-existent key.

The keys() method of a dictionary object returns a list of all the keys used in the dictionary, in arbitrary order (if you want it sorted, just apply the sorted() function to it). To check whether a single key is in the dictionary, use the in keyword.

http://docs.python.org/tutorial/datastructures.html

The last paragraph says it all.