根据嵌套列表python中的类别计算用户数

Question

我有一个包含两个子列表的列表。这里看起来像这样a = [['user1', 'referral'], ['user2', 'referral'], ['user1', 'referral'], ['user1', 'affiliate'], ['user7', 'affiliate'], ['user1', 'affiliate'], ['user9', 'affiliate'], ['user4', 'cpc'], ['user4', 'referral'], ['user2', 'referral'], ['user7', 'affiliate'], ['user14', 'cpc'], ['user3', 'orgainic'], ['user2', 'orgainic'], ['user4', 'cpc'], ['user2', 'cpc'], ['user8', 'cpc'], ['user2', 'orgainic']]我想根据类别计算用户（唯一）。必需的：required = [['referral',3],['affiliate',3],['cpc',4],['orgainic',2]]我得到的输出：{'referral': 3, 'affiliate': 2, 'cpc': 4, 'orgainic': 3}算错了。这是我尝试过的代码：a = [['user1', 'referral'], ['user2', 'referral'], ['user1', 'referral'], ['user1', 'affiliate'], ['user7', 'affiliate'], ['user1', 'affiliate'], ['user9', 'affiliate'], ['user4', 'cpc'], ['user4', 'referral'], ['user2', 'referral'], ['user7', 'affiliate'], ['user14', 'cpc'], ['user3', 'orgainic'], ['user2', 'orgainic'], ['user4', 'cpc'], ['user2', 'cpc'], ['user8', 'cpc'], ['user2', 'orgainic']]required = [['referral',3],['affiliate',3],['cpc',4],['orgainic',2]]c = {}visits = []for i in a:    # print(i)    for j in i[1:]:        if j not in c and i[0] not in visits:            c[j] = 1            visits.append(i[0])        elif j in c and i[0] not in visits:            c[j] = c[j]+1print(c)帮我解决一些问题...

Answer 1

这是一种使用collections.defaultdict.

前任：

from collections import defaultdict

a = [['user1', 'referral'], ['user2', 'referral'], ['user1', 'referral'], ['user1', 'affiliate'], ['user7', 'affiliate'], ['user1', 'affiliate'], ['user9', 'affiliate'], ['user4', 'cpc'], ['user4', 'referral'], ['user2', 'referral'], ['user7', 'affiliate'], ['user14', 'cpc'], ['user3', 'orgainic'], ['user2', 'orgainic'], ['user4', 'cpc'], ['user2', 'cpc'], ['user8', 'cpc'], ['user2', 'orgainic']]

result = defaultdict(int)

seen = set()

for k, v in a:

    key = "{}_{}".format(k, v)

    if key not in seen:

        result[v] += 1

        seen.add(key)

print(list(map(list, result.items())))

输出：

[['referral', 3], ['affiliate', 3], ['cpc', 4], ['orgainic', 2]]

Answer 2

首先让我们使条目独一无二：

c = {tuple(sublist) for sublist in a}

现在我们有了一对独特的用户和类型。

对于我们不需要用户的计数，因此让我们将其设为仅包含第二个参数的列表：

c = [elem[1] for elem in c]

现在我们可以很容易地计算它：

from collections import Counter

c = Counter(c)

结果：Counter({'cpc': 4, 'affiliate': 3, 'referral': 3, 'orgainic': 2})

现在把它们放在一起：

from collections import Counter

c = Counter(elem[1] for elem in {tuple(sublist) for sublist in a})

Answer 3

defaultdict和基于循环的解决方案

这可以使用defaultdict：

d = defaultdict(set)

for user, category in a:

    d[category].add(user)

res = [[category, len(users)] for category, users in d.items()]

输出：

# [['affiliate', 3], ['cpc', 4], ['orgainic', 2], ['referral', 3]]

groupby基于解决方案

或者，这可以使用groupbyfrom来完成itertools：

from itertools import groupby

from operator import itemgetter

a = [['user1', 'referral'], ['user2', 'referral'], ['user1', 'referral'], ...]

# Sort the items according to the category so groupby will collect the pairs accordingly

res = {category: len({user for user, _ in pairs}) for category, pairs in

       groupby(sorted(a, key=itemgetter(1)), key=itemgetter(1))}

res = [list(pair) for pair in res.items()]

输出：

# [['affiliate', 3], ['cpc', 4], ['orgainic', 2], ['referral', 3]]

Answer 4

这听起来像是熊猫的案例，您的列表已经是正确的形状：

import pandas as pd

a = [['user1', 'referral'], ['user2', 'referral'], ['user1', 'referral'], ['user1', 'affiliate'], ['user7', 'affiliate'], ['user1', 'affiliate'], ['user9', 'affiliate'], ['user4', 'cpc'], ['user4', 'referral'], ['user2', 'referral'], ['user7', 'affiliate'], ['user14', 'cpc'], ['user3', 'orgainic'], ['user2', 'orgainic'], ['user4', 'cpc'], ['user2', 'cpc'], ['user8', 'cpc'], ['user2', 'orgainic']]

df = pd.DataFrame(a)

df.columns=["user", "type"]

unique_per_type = df.groupby("type")["user"].unique()

现在 unique_per_type 是：

type

affiliate            [user1, user7, user9]

cpc          [user4, user14, user2, user8]

orgainic                    [user3, user2]

referral             [user1, user2, user4]

Name: user, dtype: object

您可以执行以下操作：

# access length by key

len(unique_per_type["affiliate"]) 

# or use it like a dict

for key, val in unique_per_type.items():

    print(key, len(val)))

这个解决方案添加了 pandas，这是一个巨大的依赖。但是，一旦您将数据放入 DataFrame 中，您就可以用它做很多事情：

df["user"].unique() # shows all unique users

df.query("user=='user1'") # shows all observations involving user1