图灵程序丛书 —《数据科学入门》— Ch1 导论

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一个tail bound定理一些基础： Markov不等式：为非负随机变量， Chebyshev不等式： Markov不等式的推论：当r为偶数时，有注记： Markov不等式要求X是非负随机变量，给出的界随着a的增长，以的速率下降 Chebyshev不等式不要求X是非负随机变量，给出的界随着a的增长，以的速率下降推论不等式在r为偶数的时候成立，给出的界随着a的增长，以的速率下降如果相较于, 等低阶矩增加的不是很多，被某种上界所限制，则推论效果会比较好，可以获得更紧的界（r越大，界越紧）；定理的Intuit. 阅读详情

Ch1 导论

此系列记录《数据科学入门》学习笔记

1.3.1 寻找关键联系人

1、定义users字典list

users = [
    {'id': 0, 'name': 'Hero'},
    {'id': 1, 'name': 'Dunn'},
    {'id': 2, 'name': 'Sue'},
    {'id': 3, 'name': 'Chi'},
    {'id': 4, 'name': 'Thor'},
    {'id': 5, 'name': 'Clive'},
    {'id': 6, 'name': 'Hicks'},
    {'id': 7, 'name': 'Devin'},
    {'id': 8, 'name': 'Kate'},
    {'id': 9, 'name': 'Klein'},
]

2、定义友邻关系list

friendships = [(0, 1), (0, 2), (1, 2), (1, 3), (2, 3), (3, 4), (4, 5), (5, 6), (5, 7), (6, 8), (7, 8), (8, 9)]

3、在users添加friends列

for user in users:
    user['friends'] = []
for i ,j in friendships:
    users[i]['friends'].append(users[j])
    users[j]['friends'].append(users[i])

4、计算友邻关系的联系数

# 计算全部联系数
def number_of_friends(user):
    """how many friends dose _user_ have?"""
    return len(user['friends'])
total_connections = sum(number_of_friends(user) for user in users)
total_connections

# 计算平均联系数
# from __future__ import division  导入整数除法,python3不需要这样操作
num_users = len(users)
avg_connections = total_connections / num_users
avg_connections

5、对users按照朋友数排序

# 按照朋友数目排序,得到（id，number）
num_friends_by_id = [(user['id'], number_of_friends(user)) for user in users]
sorted(num_friends_by_id, key=lambda x: x[1], reverse=True) #表示从大到小排序

1.3.2 你可能知道的科学家

1、对users内用户计算朋友的朋友

# 计算朋友的朋友
def friends_of_friend_ids_bad(user):
    # foaf 是朋友的朋友 英文缩写
    return [foaf['id']
        for friend in user['friends']
        for foaf in friend['friends']]
friends_of_friend_ids_bad(users[0])  #[0, 2, 3, 0, 1, 3]

# 1,2是0的朋友；0,2,3是1的朋友；0,1,3是2的朋友
print([friend['id'] for friend in users[0]['friends']])
print([friend['id'] for friend in users[1]['friends']])
print([friend['id'] for friend in users[2]['friends']])

2、找到用户与其他用户相同好友数

# 计算共同朋友
from collections import Counter
# Counter是一个简单的计数器，例如，统计字符出现的个数


def not_the_same(user, other_user):
    """two users are not the same if they have different ids"""
    return user['id'] != other_user['id']


def not_friends(user, other_user):
    """other_user is not a friend if he's not in user['friends],
    that is, if he's not_the_same as all the people in user['friends]"""
    return all(not_the_same(friend, other_user) for friend in user['friends'])


def friends_of_friend_ids(user):
    return Counter(foaf['id'] 
                   for friend in user['friends']
                   for foaf in friend['friends']
                   if not_the_same(user, foaf) and not_friends(user, foaf))

print(friends_of_friend_ids(users[3]))
# Counter({0: 2, 5: 1}) 说明3和2有两个共同好友，3和5有一个共同好友

3、寻找相同爱好的用户

interests = [(0,"Hadoop"),(0,"Big Data"),(0,"HBase"),(0,"Java"),(0,"Spark"),(0,"Storm"),
             (0,"Cassandra"),
             (1,"NoSQL"),(1,"MongoDB"),(1,"Cassandra"),(1,"HBase"),(1,"Postgres"),
             (2,"Python"),(2,"scikit-learn"),(2,"Scipy"),(2,"numpy"),(2,"satasmodels"),
             (2,"pandas"),
             (3,"R"),(3,"Python"),(3,"Statistics"),(3,"regression"),(3,"probability"),
             (4,"machine learning"),(4,"regression"),(4,"decision trees"),(4,"libsvm"),
             (5,"Python"),(5,"R"),(5,"Java"),(5,"C++"),(5,"Haskell"),
             (5,"programing languages"),
             (6,"statistics"),(6,"probability"),(6,"machine learning"),(6,"theory"),
             (7,"machine learning"),(7,"scikit-learn"),(7,"Mahout"),(7,"neural networdks"),
             (8,"neural networdks"),(8,"deep learning"),(8,"Big Data"),
             (8,"artificial intelligence"),
             (9,"Hadoop"),(9,"Java"),(9,"MapReduce"),(9,"Big Data")]

# 找出有共同爱好的用户——需要遍历整个兴趣列表
def data_scientists_who_like(target_interest):
    return [user_id 
            for user_id, user_interest in interests 
            if user_interest == target_interest]

print(data_scientists_who_like('regression'))
# [3, 4] 说明用户3和4具有共同爱好‘regeression’

from collections import defaultdict
# 使用dict时，如果引用的Key不存在，就会抛出KeyError。如果希望key不存在时，返回一个默认值，就可以用defaultdict：
# 除了在Key不存在时返回默认值，defaultdict的其他行为跟dict是完全一样的。


# 得到关于每个interest的用户id，键是interest，值是带有这个interest的user_id列表
user_ids_by_interest = defaultdict(list)


for user_id, interest in interests:
    user_ids_by_interest[interest].append(user_id)

interests_by_user_id = defaultdict(list)


# 得到关于每个用户的interest，键是user_id，值是带有这个user_id的interest列表
for user_id, interest in interests:
    interests_by_user_id[user_id].append(interest)

def most_common_interests_with(user):
    return Counter(interested_user_id
                  for interest in interests_by_user_id[user['id']]
                  for interested_user_id in user_ids_by_interest[interest]
                  if interested_user_id != user['id'])

most_common_interests_with(users[0])
# Counter({1: 2, 5: 1, 8: 1, 9: 3}) 用户1与用户0有两个兴趣相同，用户9与用户0有三个兴趣相同

1.3.3 工资与工作年限

寻找工作年限和工资的关系

salaries_and_tenures = [(83000,8.7),(88000,8.1),(48000,0.7),(76000,6),(69000,6.5),
                        (76000,7.5),(60000,2.5),(83000,10),(48000,1.9),(63000,4.2)]

# 计算每个工作年限的薪水，键是year，值是对每一个tenure的salary列表
salary_by_tenure = defaultdict(list)


for salary, tenure in salaries_and_tenures:
    salary_by_tenure[tenure].append(salary)
    
# 计算每个工作年限的平均薪水，键是year，值是tenure相对应的平均salary
    average_salary_by_tenure = {
    tenure : sum(salaries) / len(salaries)
    for tenure, salaries in salary_by_tenure.items()
}

average_salary_by_tenure
# 因为上述数据中任意两个用户都没有相同的工作年限，所以均值结果作用有限，仅表示每个用户的独立收入

# 将用户工作年限分组
def tenure_bucket(tenure):
    if tenure < 2:
        return 'less than two years'
    elif tenure < 5:
        return 'between two and five years'
    else:
        return 'more than five years'

salary_by_tenure_bucket = defaultdict(list)

for salary, tenure in salaries_and_tenures:
    bucket = tenure_bucket(tenure)
    salary_by_tenure_bucket[bucket].append(salary)
    
# 计算每个工作年限的平均薪水，键是year，值是tenure相对应的平均salary
average_salary_by_bucket = {
    tenure_bucket : sum(salaries) / len(salaries)
    for tenure_bucket, salaries in salary_by_tenure_bucket.items()
}

average_salary_by_bucket
# {'between two and five years': 61500.0,
#  'less than two years': 48000.0,
#  'more than five years': 79166.66666666667}

1.3.4 付费账户

预测用户是否付费

years_paid = [(0.7, 'paid'),
              (1.9, 'unpaid'),
              (2.5, 'paid'),
              (4.2, 'unpaid'),
              (6, 'unpaid'),
              (6.5, 'unpaid'),
              (7.5, 'unpaid'),
              (8.1, 'unpaid'),
              (8.7, 'paid'),
              (10, 'paid'),]

# 预测是否付费
def predict_paid_or_unpaid(years_experience):
    if years_experience < 3.0:
        return 'paid'
    elif years_experience < 8.5:
        return 'unpaid'
    else:
        return 'paid'

predict_paid_or_unpaid(years_paid[1][0])
# 'paid'  year == 2.5 的预测结果为paid，实际上是unpaid

1.3.5 兴趣主题

提取interests内兴趣关键词

words_and_counts = Counter(word
                           for user, interest in interests
                           for word in interest.lower().split())

for word, count in words_and_counts.most_common():
    if count > 1:
        print(word, count)

learning 4
big 3
data 3
java 3
python 
machine 3
hadoop 2
hbase 2
cassandra 2
scikit-learn 2
r 2
statistics 2
regression 2
probability 2
neural 2
networdks 2

以上是Ch1的相关内容，第一次写博客，排版不好，请见谅

2018.01.30 YR

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