How do I initialize a Counter from a list of key/initial counts pairs?

Question

If I have a sequence of (key, value) pairs, I can quickly initialize a dictionary like this:

>>> data = [ ('a', 1), ('b', 2) ]
>>> dict(data) 
{'a': 1, 'b': 2} 

I would like to do the same with a Counter dictionary; but how? Both the constructor and the update() method treat the ordered pairs as keys, not key-value pairs:

>>> from collections import Counter
>>> Counter(data)
Counter({('a', 1): 1, ('b', 2): 1})

The best I could manage was to use a temporary dictionary, which is ugly and needlessly circuitous:

>>> Counter(dict(data))
Counter({'b': 2, 'a': 1})

Is there a proper way to directly initialize a Counter from a list of (key, count) pairs? My use case involves reading lots of saved counts from files (with unique keys).

Answer 1

I would just do a loop:

for obj, cnt in [ ('a', 1), ('b', 2) ]:
    counter[obj] = cnt

You could also just call the parent dict.update method:

>>> from collections import Counter
>>> data = [ ('a', 1), ('b', 2) ]
>>> c = Counter()
>>> dict.update(c, data)
>>> c
Counter({'b': 2, 'a': 1})

Lastly, there isn't anything wrong with your original solution:

Counter(dict(list_of_pairs))

The expensive part of creating dictionaries or counters is hashing all of the keys and doing periodic resizes. Once the dictionary is made, converting it to a Counter is very cheap about as fast as a dict.copy(). The hash values are reused and the final Counter hash table is pre-sized (no need for resizing).

Answer 2

From docs:

Elements are counted from an iterable or initialized from another mapping (or counter)

So it's a No, you need to convert it to mapping and then initialize Counter. And Yes when you initialized with dict it was the right move.

UPDATE

I agree that @RaymondHettinger code looks good, and actually it's faster

from collections import Counter
from random import choice
from string import ascii_letters
a=[(choice(ascii_letters), i) for i in range(100)]

Tested with Python 3.6.1 and IPython 6

Initialization with dict:

%%timeit
c1=Counter(dict(a))

Output

12.1 µs ± 342 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Update with dict.update()

%%timeit    
c2=Counter()
dict.update(c2, a)

Output:

7.21 µs ± 236 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)