So I have data structured in a nested list like so
data = [['A', '1'], ['B', '2'], ['C', '3'], ['A', '-2'], ['B', '4'], ['C', '1'], ['A', '2'], ['B', '1'], ['C', '-5']]
and I am trying to convert it into an output that looks like this
{'A': 1, 'C': -1, 'B': 7}
Basically sum up all of the A's, B's, and C's, put output as a dictionary.
I wrote this code which gives the correct answer
playerSum = {}
for ele in data:
if ele[0] not in playerSum:
playerSum[ele[0]] = int(ele[1])
else:
playerSum[ele[0]] += int(ele[1])
However, I am trying to convert the code block above into dictionary comprehension. I understand mostly how to do it, but I don't understand how to write the += as a dictionary comprehension. Any guidance on the structure would be great.
So far I have this
playerSum = {ele[0]: int(ele[1]) if ele[0] not in playerSum else playerSum[ele[0]] += int(ele[1]) for ele in data}
Edit: So @achampion was able to solve it. Thanks!
{key: sum(int(v) for k, v in data if k==key) for key in set(k for k, _ in data)}
It is not practical to do it as a comprehension.
Just as an exercise you can use a coroutine to do the counting for you, but you effectively create the dictionary twice:
from collections import defaultdict
def count():
cache = defaultdict(int)
k, v = yield
while True:
cache[k] += v
k, v = yield cache[k]
counter = count() # Create coroutine
next(counter) # Prime coroutine
data = [['A', '1'], ['B', '2'], ['C', '3'], ['A', '-2'], ['B', '4'],
['C', '1'], ['A', '2'], ['B', '1'], ['C', '-5']]
{k: counter.send((k, int(v))) for k, v in data} # Meets the challenge :)
Result:
{'A': 1, 'B': 7, 'C': -1}
Or a truly ugly one-liner that doesn't need a coroutine and isn't quadratic (not a comprehension):
>>> reduce(lambda d, (k,v): d.__setitem__(k, d.get(k,0)+int(v)) or d, data, {}) # Py2.7
{'A': 1, 'B': 7, 'C': -1}
And finally a very inefficient but true dict comprehension based on @Prune:
>>> {key: sum(int(v) for k, v in data if k==key) for key in set(k for k, _ in data)}
{'A': 1, 'B': 7, 'C': -1}
The best way is the most obvious way
from collections import defaultdict
playerSum = defaultdict(int)
for key, value in data:
playerSum[key] += int(value)
It is not possible to use a dict comprehension as your values would be overwritten, the dict does not get created until the comprehension is finished so even if you could there is nothing to +=. As it stands unless you have playerSum = {}
somewhere your code will error with a NameError, if you do you are simply rebinding the name to the result of your dict comp so playerSum = {}
is basically doing nothing.
The only way to do what you want is to along the lines of your own solution . For more efficient approach You can unpack the sublists and cast the second element to int, summing the value using a collections.defaultdict
:
from collections import defaultdict
d = defaultdict(int)
for a,b in data:
d[a] += int(b)
print(d)
defaultdict(<type 'int'>, {'A': 1, 'C': -1, 'B': 7})
Or using a regular dict:
d = {}
for a, b in data:
d[a] = d.get(a,0) + int(b)
print(d)
{'A': 1, 'C': -1, 'B': 7}
I did it in a single comprehension, as you asked.
dict ([(key, sum(int(elem[1]) for elem in data if elem[0]==key)) for key in [id for id in set([elem[0] for elem in data])] ])
From outer to inner:
Build a set of the ID's used in the list.
For each ID, make a list of the associated values.
Sum the list.
Emit (yield) the ID and sum as a pair.
Turn this list of tuples into a dictionary.
Test:
data = [['A', '1'], ['B', '2'], ['C', '3'],
['A', '-2'], ['B', '4'], ['C', '1'],
['A', '2'], ['B', '1'], ['C', '-5']]
playerSum = dict ([(key, sum(int(elem[1]) for elem in data if elem[0]==key))
for key in [id for id in set([elem[0] for elem in data])] ])
print data
print playerSum
Result:
[['A', '1'], ['B', '2'], ['C', '3'], ['A', '-2'], ['B', '4'], ['C', '1'], ['A', '2'], ['B', '1'], ['C', '-5']]
{'A': 1, 'C': -1, 'B': 7}
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