Can you add two lists element by element with out iterating through it as below? This works, but eh. I miss matlab.
list1 = [3, 7, 3]
list2 = [4, 5, 6]
for x in range(0, len(list1)):
list1[x] = list1[x] + list2[x]
print (list1)
Output :
[7, 12, 9]
Don't miss Matlab at all :) The numpy arrays are your typical Matlab arrays:
arr1 = np.array([3, 7, 3])
arr2 = np.array([4, 5, 6])
arr1 + arr2
gives you what you want.
If you want to stick with lists, you can do this:
list1 = [3, 7, 3]
list2 = [4, 5, 6]
[sum(x) for x in zip(list1, list2)]
After the performance comments, especially Patrick's comparison, I did a comparison myself, and verified that numpy is actually the fastest among the current solutions posted. However, there is another solution which beats numpy! Let's see:
# Numpy:
timeit arr1 + arr2 # 1.67 µs per loop
# List comprehension:
timeit [sum(x) for x in zip(list1, list2)] # 2.59 µs per loop
timeit [(x+y) for x,y in zip(list1,list2)] # 1.72 µs per loop
# Map and Reduce-based (note that reduce is really overkill, I did it just for fun):
timeit map(sum, zip(list1, list2)) # 2.58 µs per loop
timeit map(lambda a, b: a + b , list1, list2) # 2.11 µs per loop
timeit [reduce((lambda x,y: x+y), e) for e in zip(list1,list2)] # 4.05 µs per loop
# And the winner:
from operator import add
timeit map(add, list1, list2) # 1.57 µs per loop
zip the lists together, and sum the pairs
list(map(sum, zip(list1, list2)))
EDIT:
I did some basic testing, and the overwhelming winners were the list comprehension if the input data starts as lists, and numpy if you can have it as arrays before your operation
def test1(list1, list2):
l=[]
for x in range(0, len(list1)):
l.append(list1[x] + list2[x])
return l
def test2(list1, list2):
return list(map(sum, zip(list1, list2)))
def test3(list1, list2):
return [x+y for x, y in zip(list1, list2)]
import numpy as np
def test4(list1, list2):
return np.array(list1) + np.array(list2)
def test5(list1, list2):
return list1 + list2
from timeit import timeit
print(timeit('test1(list1, list2)', setup='list1 = list(range(100)); list2 = list(range(100))', globals=globals()))
print(timeit('test2(list1, list2)', setup='list1 = list(range(100)); list2 = list(range(100))', globals=globals()))
print(timeit('test3(list1, list2)', setup='list1 = list(range(100)); list2 = list(range(100))', globals=globals()))
print(timeit('test4(list1, list2)', setup='list1 = list(range(100)); list2 = list(range(100))', globals=globals()))
print(timeit('test4(list1, list2)', setup='list1 = np.array(list(range(100))); list2 = np.array(list(range(100)))', globals=globals()))
print(timeit('test5(list1, list2)', setup='list1 = np.array(list(range(100))); list2 = np.array(list(range(100)))', globals=globals()))
Gives me
15.42712744511664 # append
17.329718918073922 # my solution above
7.0252319818828255 # list comprehension
16.53089915495366 # numpy with list inputs
1.430903600063175 # numpy with array inputs that are double checked
0.6451617309357971 # numpy assuming array inputs
These are the times in seconds to execute the operation 1000000 times.
Numpy is surprisingly slow when forced to do np.array(list) every time.
Try it yourself at this repl.it
If you want matlab-like behavior, you should use numpy
In [5]: list1 = [3, 7, 3]
In [6]: list2 = [4, 5, 6]
In [7]: import numpy as np
In [8]: np.array(list1) + np.array(list2)
Out[8]: array([ 7, 12, 9]
You can also use list comprehension over both the lists
list1 = [3, 7, 3]
list2 = [4, 5, 6]
list3 = [(x+y) for x,y in zip(list1,list2)]
print(list3)
Output:
[7, 12, 9]
You can use a lambda and a map:
>>> list1 = [3, 7, 3]
>>> list2 = [4, 5, 6]
>>> list(map(lambda a, b: a + b , list1, list2))
[7, 12, 9]
