How do I create a unique value for each key using dict.fromkeys?

Question

First, I'm new to Python, so I apologize if I've overlooked something, but I would like to use dict.fromkeys (or something similar) to create a dictionary of lists, the keys of which are provided in another list. I'm performing some timing tests and I'd like for the key to be the input variable and the list to contain the times for the runs:

def benchmark(input):
    ...
    return time_taken

runs = 10
inputs = (1, 2, 3, 5, 8, 13, 21, 34, 55)
results = dict.fromkeys(inputs, [])

for run in range(0, runs):
    for i in inputs:
        results[i].append(benchmark(i))

The problem I'm having is that all the keys in the dictionary appear to share the same list, and each run simply appends to it. Is there any way to generate a unique empty list for each key using fromkeys? If not, is there another way to do this without generating the resulting dictionary by hand?

Answer 1

Check out defaultdict (requires Python 2.5 or greater).

from collections import defaultdict

def benchmark(input):
    ...
    return time_taken

runs = 10
inputs = (1, 2, 3, 5, 8, 13, 21, 34, 55)
results = defaultdict(list) # Creates a dict where the default value for any key is an empty list

for run in range(0, runs):
    for i in inputs:
        results[i].append(benchmark(i))

Answer 2

The problem is that in

results = dict.fromkeys(inputs, [])

[] is evaluated only once, right there.

I'd rewrite this code like that:

runs = 10
inputs = (1, 2, 3, 5, 8, 13, 21, 34, 55)
results = {}

for run in range(runs):
    for i in inputs:
        results.setdefault(i,[]).append(benchmark(i))

Other option is:

runs = 10
inputs = (1, 2, 3, 5, 8, 13, 21, 34, 55)
results = dict([(i,[]) for i in inputs])

for run in range(runs):
    for i in inputs:
        results[i].append(benchmark(i))

Answer 3

You can also do this if you don't want to learn anything new (although I recommend you do!) I'm curious as to which method is faster?

results = dict.fromkeys(inputs)

for run in range(0, runs):
    for i in inputs:
        if not results[i]:
            results[i] = []
        results[i].append(benchmark(i))