How would I be able to find the average from a list of lists given two constaints?

Question

So I have a list of lists, where the 7th index of each sublist contains the value I am interested in averaging, however, the numbers must be averaged according to their type. This type to be matched upond can be found at the 11th index of the sublist.

Below is some code I wrote. In this exa

# Open the csv file
opened_file = open('AppleStore.csv')
from csv import reader
read_file = reader(opened_file)
# Store the data as a list or arrays
apps_data = list(read_file)

# idx_num = index number of interest
# list_doc = the list of lists
# row_start = 1
def extract(idx_num,list_doc,row_start=1):
    a_list = []
    for row in list_doc[row_start:]:
        var = row[idx_num]
        a_list.append(var)
    return a_list

# Use the extract function to get an array
a_list = extract(11, apps_data, 0)
# Find unique elements
a_list_set = set(a_list)
# Create a dictionary with initial values at [0,0]
dic = dict.fromkeys(a_list_set,[0,0])

print(dic)
# Works as intended
#{'Weather': [0, 0], 'Sports': [0, 0], 'Productivity': [0, 0], 'Games': [0, #0], 'News': [0, 0], 'Finance': [0, 0], 'Education': [0, 0], #'Entertainment': [0, 0], 'Health & Fitness': [0, 0], 'Business': [0, 0], #'Social Networking': [0, 0], 'prime_genre': [0, 0], 'Photo & Video': [0, #0], 'Navigation': [0, 0], 'Music': [0, 0], 'Medical': [0, 0], 'Travel': #[0, 0], 'Reference': [0, 0], 'Shopping': [0, 0], 'Utilities': [0, 0], #'Food & Drink': [0, 0], 'Lifestyle': [0, 0], 'Catalogs': [0, 0], 'Book': #[0, 0]}

for row in apps_data[1:]:
    price = float(row[4])
    genre = row[11]

# Here is the issue:
# I thought that this would allow for the genre instance to be matched to the appropriate key and then I could append my values.

    if genre in dic.keys():
        dic[genre][0] += 1
        dic[genre][1] += (price)
    else:
        dic[genre][0] = 1
        dic[genre][1] = price


print(dic)

## From here I would extract the array contents of the dictionary
for genre in a_list_set:
print(str(genre) + " mean price:"  + str(round(dic[genre][1]/dic[genre][0], 2)))

I got this instead.

{'Weather': [7197, 12423.58999999945], 'Sports': [7197, 12423.58999999945], 'Productivity': [7197, 12423.58999999945], 'Games': [7197, 12423.58999999945], 'News': [7197, 12423.58999999945], 'Finance': [7197, 12423.58999999945], 'Education': [7197, 12423.58999999945], 'Entertainment': [7197, 12423.58999999945], 'Health & Fitness': [7197, 12423.58999999945], 'Business': [7197, 12423.58999999945], 'Social Networking': [7197, 12423.58999999945], 'prime_genre': [7197, 12423.58999999945], 'Photo & Video': [7197, 12423.58999999945], 'Navigation': [7197, 12423.58999999945], 'Music': [7197, 12423.58999999945], 'Medical': [7197, 12423.58999999945], 'Travel': [7197, 12423.58999999945], 'Reference': [7197, 12423.58999999945], 'Shopping': [7197, 12423.58999999945], 'Utilities': [7197, 12423.58999999945], 'Food & Drink': [7197, 12423.58999999945], 'Lifestyle': [7197, 12423.58999999945], 'Catalogs': [7197, 12423.58999999945],'Book': [7197, 12423.58999999945]}

Answer 1

We can do this with itertools.groupby; first, we extract the "columns" of concern from our data, constituting the 7th and 11th value of each row, into subset, also sorting by the 11th value.

Then, we use groupby to partition our subset into groups, where each group's members all have the same 2nd element (the original 11th element). We can then use a dict comprehension to get the mean of the 1st element of each group's members.

from itertools import groupby

from operator import itemgetter

from statistics import mean

subset = sorted(((row[6], row[10]) for row in data), key=itemgetter(1))
result = {key: mean(map(itemgetter(0), group)) for key, group in groupby(subset, itemgetter(1))}

print(result)

Some sample data:

[[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -4.926456602181107, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -4.261928508086729, 0.0, 0.0, 0.0, 'that'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 6.582427615396794, 0.0, 0.0, 0.0, 'other'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -0.08345371286375847, 0.0, 0.0, 0.0, 'other'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.6323414510835206, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -7.755177634382969, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -5.948058847184649, 0.0, 0.0, 0.0, 'that'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -5.767820549798114, 0.0, 0.0, 0.0, 'other'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 4.609131600539092, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.2106567350536854, 0.0, 0.0, 0.0, 'that'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -3.1550716372338297, 0.0, 0.0, 0.0, 'other'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -0.6037278107842077, 0.0, 0.0, 0.0, 'that'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -11.819322083983815, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 5.441817745217389, 0.0, 0.0, 0.0, 'other'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.4961079817344718, 0.0, 0.0, 0.0, 'other'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 8.269603775378254, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -0.42023137240633596, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 7.855652365179269, 0.0, 0.0, 0.0, 'this'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -8.048026683773955, 0.0, 0.0, 0.0, 'that'],
 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -4.577046681982131, 0.0, 0.0, 0.0, 'this']]

And the result:

{'other': 0.585667907075492,
 'that': -3.530217022955171,
 'this': -0.9035005758618025}