How do I initialize a counter from a list of key-value pairs?
If I have a sequence of pairs (key, value)
, I can quickly initialize the dictionary like this:
>>> data = [ ('a', 1), ('b', 2) ]
>>> dict(data)
{'a': 1, 'b': 2}
I would like to do the same with a dictionary Counter
; but how? Both the constructor and the method update()
treat 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 do is use a temporary dictionary, which is an ugly and unnecessary workaround:
>>> Counter(dict(data))
Counter({'b': 2, 'a': 1})
Is there a way to properly initialize Counter
from a list of pairs (key, count)
? My use case involves reading a large number of stored samples from files (with unique keys).
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I would just do a loop:
for obj, cnt in [ ('a', 1), ('b', 2) ]:
counter[obj] = cnt
You can also just call the parent method dict.update
:
>>> from collections import Counter
>>> data = [ ('a', 1), ('b', 2) ]
>>> c = Counter()
>>> dict.update(c, data)
>>> c
Counter({'b': 2, 'a': 1})
Finally, there is nothing wrong with your original solution:
Counter(dict(list_of_pairs))
An expensive part of creating dictionaries or counters is hashing all keys and resizing them periodically. Once the dictionary is done, converting it to a counter is very cheap, about as fast as dict.copy (). The hash values are reused and the final hash hash table is predefined (no need to resize).
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From the docs :
Items are counted from iteration or initialized from another mapping (or counter)
So it is not, you need to convert it to a mapping and then initialize it Counter
. And yes, when you initialized with dict
, it was the right move.
UPDATE
I agree that @RaymondHettinger's code looks good and is actually 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)
Updating 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)
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If your list of keys in pairs is (key, value)
already unique - no duplicates - you can use Raymond Hettinger's excellent solution .
Beware if you only get the last value for any given key, if there are duplicate keys:
>>> data=[ ('a', 1), ('b', 2), ('a', 3), ('b', 4) ]
>>> c=Counter()
>>> dict.update(c, data)
>>> c
Counter({'b': 4, 'a': 3}) # note 'a' and 'b' are only the last value...
Same with dict
:
>>> Counter(dict(data))
Counter({'b': 4, 'a': 3})
But Counters are most often used to count totals, including duplicates. If you want the sum of the records "a" and "b", you need to iterate over all the pairs:
>>> c=Counter()
>>> for k, v in data:
... c[k]+=v
...
>>> c
Counter({'b': 6, 'a': 4}) # the sum of the 'k' entries given 'v'
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