我怎样才能从给定两个常数的列表中找到平均值？

Question

所以我有一个列表列表，其中每个子列表的第7个索引包含我感兴趣的平均值，但是，这些数字必须根据它们的类型进行平均。在子列表的第11个索引中可以找到要匹配的类型。你知道吗

下面是我写的一些代码。在这个例子中

# 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)))

我得到了这个。你知道吗

{'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

我们可以用itertools.groupby来实现这一点；首先，我们从我们的数据中提取相关的“列”，构成每行的第7和第11个值，放入subset，也按第11个值排序。你知道吗

然后，我们使用groupby将子集划分为多个组，每个组的成员都有相同的第2个元素（原来的第11个元素）。然后我们可以使用dict理解来获得每个组成员的第一个元素的平均值。你知道吗

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)

一些示例数据：

[[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']]

结果是：

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