Trying to think of a one-liner to achieve the following ( summing all the values of a key) :
>>> data = [('a',1),('b',3),('a',4),('c',9),('b',1),('d',3)]
>>> res = {}
>>> for tup in data:
... res[tup[0]] = res.setdefault(tup[0],0) + tup[1]
...
>>> res
{'a': 5, 'c': 9, 'b': 4, 'd': 3}
One-liner version without using any imports like itertools,collections etc.
{ tup[0] : SELF_REFERENCE.setdefault(tup[0],0) + tup[1] for tup in data }
Is it possible in Python to use a reference to the object currently being comprehended ?
If not, is there any way to achieve this in a one-liner without using any imports i.e. using
basic list/dict comprehension and inbuilt functions.
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
No, there is not. A dict comprehension produces a new item for each iteration, and your code needs to produce fewer items (consolidating values).
There is no way to access keys produced in an earlier iteration, not without using (ugly, unpythonic) side-effect tricks. The dict object that is going to be produced by the comprehension doesn’t exist yet, so there is no way to produce a self-reference either.
Just stick to your for loop, it is far more readable.
The alternative would be to use sorting and grouping, a O(NlogN) algorithm vs. the simple O(N) of your straight loop:
from itertools import groupby
from operator import itemgetter
res = {key: sum(t[1] for t in group)
for key, group in groupby(sorted(data, key=itemgetter(0)), key=itemgetter(0))}
Method 2
Use reduce and collections.Counter:
>>> from operator import add
>>> from collections import Counter
>>> reduce(add, (Counter(dict([x])) for x in data))
Counter({'c': 9, 'a': 5, 'b': 4, 'd': 3})
Method 3
Don’t use a oneliner. Instead use collections.defaultdict and a simple for loop:
>>> pairs = [('a', 1), ('b', 3), ('a', 4), ('c', 9), ('b', 1), ('d', 3)]
>>> result = defaultdict(int)
>>> for key, value in pairs:
... result[key] += value
...
>>> result
defaultdict(<class 'int'>, {'a': 5, 'c': 9, 'b': 4, 'd': 3})
It is easy to understand, pythonic and fast.
Method 4
This is almost like, what you are trying to do. But I won’t recommend this, as the readability suffers.
data = [('a',1),('b',3),('a',4),('c',9),('b',1),('d',3)]
print reduce(lambda d,i: [d.__setitem__(i[0],d.get(i[0],0)+i[1]),d][1], data, {})
Output
{'a': 5, 'c': 9, 'b': 4, 'd': 3}
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0