Python 如果给定两个常数，我如何从列表中找到平均值？

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，数字必须根据其类型进行平均。在子列表的第11个索引中可以找到要匹配的类型upond

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]}

我得到了这个

{'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]}

---

我们可以使用

itertools.groupby

；首先，我们从数据中提取关注的“列”，构成每行的第7和第11个值，并将其放入

子集

，同时按第11个值排序

然后，我们使用

groupby

将我们的子集划分为多个组，其中每个组的成员都有相同的第2个元素（原始的第11个元素）。然后，我们可以使用

dict

理解来获得每组成员的第一个元素的平均值

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)

一些样本数据：

[[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}

这在熊猫身上很容易做到

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