How to efficiently get the mean of the elements in two list of lists in Python

You can use the Pandas library to avoid writing any sort of loops yourself.

Your code would be really concise and clean.

Install Pandas like: pip install pandas.

Then try this:

In [132]: import pandas as pd

In [109]: df1 = pd.DataFrame(mylist1)

In [110]: df2 = pd.DataFrame(mylist2)

In [117]: res = pd.merge(df1, df2, on=0)

In [121]: res['mean'] = res.mean(axis=1)

In [125]: res.drop(['1_x', '1_y'], 1, inplace=True)

In [131]: res.values.tolist()
Out[131]: [['egg', 0.45], ['chocolate', 0.5]]

Edit

Pandas is crazy fast because it uses numpy under the hood. Numpy implements highly efficient array operations.

Please check the post : Why is Pandas so madly fast? for more details on calculating mean through pure Python vs Pandas.

You can do it in O(n) (single pass over each list) by converting 1 to a dict, then per item in the 2nd list access that dict (in O(1)), like this:

mylist1 = [["lemon", 0.1], ["egg", 0.1], ["muffin", 0.3], ["chocolate", 0.5]]
mylist2 = [["chocolate", 0.5], ["milk", 0.2], ["carrot", 0.8], ["egg", 0.8]]

l1_as_dict = dict(mylist1)

myoutput = []
for item,price2 in mylist2:
    if item in l1_as_dict:
        price1 = l1_as_dict[item]
        myoutput.append([item, (price1+price2)/2])

print(myoutput)

Output:

[['chocolate', 0.5], ['egg', 0.45]]

Here is one solution that uses collections.defaultdict to group the items and calculates the averages with statistics.mean:

from collections import defaultdict
from statistics import mean

mylist1 = [["lemon", 0.1], ["egg", 0.1], ["muffin", 0.3], ["chocolate", 0.5]]
mylist2 = [["chocolate", 0.5], ["milk", 0.2], ["carrot", 0.8], ["egg", 0.8]]

d = defaultdict(list)
for lst in (mylist1, mylist2):
    for k, v in lst:
        d[k].append(v)

result = [[k, mean(v)] for k, v in d.items()]

print(result)
# [['lemon', 0.1], ['egg', 0.45], ['muffin', 0.3], ['chocolate', 0.5], ['milk', 0.2], ['carrot', 0.8]]

If we only want common keys, just check if the values are more than 1:

result = [[k, mean(v)] for k, v in d.items() if len(v) > 1]

print(result)
# [['egg', 0.45], ['chocolate', 0.5]]

We could also just build the result from set intersection:

mylist1 = [["lemon", 0.1], ["egg", 0.1], ["muffin", 0.3], ["chocolate", 0.5]]
mylist2 = [["chocolate", 0.5], ["milk", 0.2], ["carrot", 0.8], ["egg", 0.8]]

d1, d2 = dict(mylist1), dict(mylist2)

result = [[k, (d1[k] + d2[k]) / 2] for k in d1.keys() & d2.keys()]

print(result)
# [['egg', 0.45], ['chocolate', 0.5]]

An O(n) solution that will average all items.
Construct a dictionary with a list of the values and then average that dictionary afterwards:

In []:
d = {}
for lst in (mylist1, mylist2):
    for i, v in lst:
        d.setdefault(i, []).append(v)   # alternative use collections.defaultdict

[(k, sum(v)/len(v)) for k, v in d.items()]

Out[]:
[('lemon', 0.1), ('egg', 0.45), ('muffin', 0.3), ('chocolate', 0.5), ('milk', 0.2), ('carrot', 0.8)]

Then if you just want the common ones you can add a guard:

In []:
[(k, sum(v)/len(v)) for k, v in d.items() if len(v) > 1]

Out[]:
[('egg', 0.45), ('chocolate', 0.5)]

This extends to any number of lists and makes no assumption around the number of common elements.