I have a list of tuples similar to this:
l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]
I want to create a simple one-liner that will give me
Use zip()
and sum()
:
In [1]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]
In [2]: [sum(x) for x in zip(*l)]
Out[2]: [25, 20]
or:
In [4]: map(sum, zip(*l))
Out[4]: [25, 20]
timeit
results:
In [16]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]*1000
In [17]: %timeit [sum(x) for x in zip(*l)]
1000 loops, best of 3: 1.46 ms per loop
In [18]: %timeit [sum(x) for x in izip(*l)] #prefer itertools.izip
1000 loops, best of 3: 1.28 ms per loop
In [19]: %timeit map(sum, zip(*l))
100 loops, best of 3: 1.48 ms per loop
In [20]: %timeit map(sum, izip(*l)) #prefer itertools.izip
1000 loops, best of 3: 1.29 ms per loop
Without using zip
sum(e[0] for e in l), sum(e[1] for e in l)
I want to add something to the given answer:
If I have an array of dict e.g.
l = [{'quantity': 10, 'price': 5},{'quantity': 6, 'price': 15},{'quantity': 2, 'price': 3},{'quantity': 100, 'price': 2}]
and i want to obtain two (or more) sums of calculated quantity over the values e.g. sum of quantities and of price*quantity
I can do:
(total_quantity, total_price) = (
sum(x) for x in zip(*((item['quantity'],
item['price'] * item['quantity'])
for item in l)))
Instead of:
total_quantity = 0
total_price = 0
for item in l:
total_quantity += item['quantity']
total_price += item['price'] * item['quantity']
Maybe the first solution is less readable, but is more "pythonesque" :)