Correspendence between list indices originated from dictionary

Question

I wrote the below code working with dictionary and list:

d = computeRanks() # dictionary of id : interestRank pairs
lst = list(d) # tuples (id, interestRank)
interestingIds = []
for i in range(20): # choice randomly 20 highly ranked ids
  choice = randomWeightedChoice(d.values()) # returns random index from list
  interestingIds.append(lst[choice][0])

There seems to be possible error because I'm not sure if there is a correspondence between indices in lst and d.values().

Do you know how to write this better?

Why do you need `lst`? You can randomly select a key from the `d` (using `d.keys()`) and add that to `interestingIds`. — Simeon Visser, Mar 18 '12 at 12:42
@SimeonVisser I can't do this because it's random with weights and weigths are interestRanks which are in *d.values()*. — xralf, Mar 18 '12 at 12:44
Then @Simeon is right, `list(d)` is exactly the same as `d`. — alexis, Mar 18 '12 at 12:54
But actually that's not possible: Lists don't have a `values()` method. `d` is not a list. — alexis, Mar 18 '12 at 12:55
@alexis Sorry I wrote the type of lst above. The type of `d` is dictionary of int : float pairs — xralf, Mar 18 '12 at 12:56

score 3 · Accepted Answer · answered Mar 18 '12 at 12:41

3

One of the policies of dict is that the results of dict.keys() and dict.values() will correspond so long as the contents of the dictionary are not modified.

answered Mar 18 '12 at 12:41

Ignacio Vazquez-Abrams

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I didn't say that. But this part wouldn't change a lot. – Ignacio Vazquez-Abrams Mar 18 '12 at 12:56
You showed that `dict.keys()` and `dict.values()` correspond, but in the question is `lst(dict)` instead of `dict.keys()`. Does this make a difference? – xralf Mar 18 '12 at 13:35
No. Iterating over a `dict` via `list(dict)` defers to `dict.iterkeys()`, which [is part of the same policy](http://docs.python.org/library/stdtypes.html#dict.items). – Ignacio Vazquez-Abrams Mar 18 '12 at 13:48
Oh, as for the rewrite, you should modify randomWeightedChoice(). Think of what sorted() and max() do. – Ignacio Vazquez-Abrams Mar 18 '12 at 16:17
Rewrite is not problem but I wanted to go without rewrite because I'm using this function mainly for lists and don't want to create new function for each special data structure. – xralf Mar 18 '12 at 16:27
This rewrite will generalize the function to any iterable. – Ignacio Vazquez-Abrams Mar 18 '12 at 16:31
I guess you have some design pattern on mind but have no idea how to generalize it for every data structure. You still have to specify where is a list of values in that data structure, I think. – xralf Mar 18 '12 at 16:43
Correct. Which is why I brought up those examples. – Ignacio Vazquez-Abrams Mar 18 '12 at 18:49

score 0 · Answer 2 · answered Mar 18 '12 at 13:03

0

As @Ignacio says, the index choice does correspond to the intended element of lst, so your code's logic is correct. But your code should be much simpler: d already contains IDs for the elements, so rewrite randomWeightedChoice to take a dictionary and return an ID.

Perhaps it will help you to know that you can iterate over a dictionary's key-value pairs with d.items():

for k, v in d.items():
    etc.

answered Mar 18 '12 at 13:03

alexis

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*randomWeightedChoice* takes a list and returns an index from this list. It won't be rewritten. I need to generate exactly 20 random numbers so don't need to iterate over all key-value pairs. – xralf Mar 18 '12 at 13:10
You don't have to iterate, that was a demonstration. `randomWeightedChoice` needs to examine all the values in order to do its job, so it has a loop somewhere. If you can't rewrite it, I don't think there's anything else we can help with. – alexis Mar 18 '12 at 13:20
Yes, *randomWeightedChoice* needs to examine all the values 20 times to choose 20 random values. – xralf Mar 18 '12 at 13:32
No, it doesn't. It only has to examine them once. But what does it matter, if you won't be rewriting it? – alexis Mar 18 '12 at 13:34
If I would rewrite this question would have no sense. I'd like to have unique interface instead of having the same function for various data structures (dicts, lists, sets etc.). – xralf Mar 18 '12 at 13:41

Correspendence between list indices originated from dictionary

2 Answers2