数据科学(简单案例)
2018-11-20 本文已影响97人
GHope
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数据产生
网站:用户每次点击
手机:位置和速度
智能手表、手环:心率、行动、饮食、睡眠
智能汽车:驾驶习惯
智能家居:生活习惯
数据科学家
从混乱数据中理出价值的人
案例:寻找关键联系人
根据用户网络关系数据识别关键联系人
用户列表
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"},
{"id": 10, "name": "Jen"}]
用户好友关系
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),(8,10),(10,3),(4,8)]
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为每个用户创建朋友列表
for user in users:
user["friends"] = []
填充好友数据
for i, j in friendships:
users[i]["friends"].append(users[j])
users[j]["friends"].append(users[i])
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问题: 平均朋友联系数是多少?
答: 全部联系数除以用户个数
def number_of_friends(user):
return len(user["friends"])
total_connections = sum(number_of_friends(user)for user in users)
num_users = len(users)
avg_connections = total_connections / num_users
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按朋友数多少排序
# 取(user_id, number_of_friends)
num_friends_by_id = [(user['id'], number_of_friends(user)) for user in users]
sorted(num_friends_by_id, key=lambda item: item[1], reverse=True)
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案例: 你可能知道的人
找朋友的朋友
def friends_of_friend_ids_bad(user):
# 双层循环
return [foaf["id"]for friend in user["friends"]for foaf in friend["friends"]]
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查找共同的朋友
from collections import Counter
# 判断是否是自己
def not_the_same(user, other_user):
return user["id"] != other_user["id"]
# 判断是否是自己的朋友
def not_friends(user, other_user):
return all(not_the_same(friend, other_user)for friend in user["friends"])
# 在之前的基础上进行判断即可
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))
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找共同兴趣的人
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, "statsmodels"), (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, "programming languages"),
(6, "statistics"),(6, "probability"), (6, "mathematics"), (6, "theory"),
(7, "machine learning"), (7, "scikit-learn"), (7, "Mahout"),(7, "neural networks"),
(8, "neural networks"), (8, "deep learning"),(8, "Big Data"), (8, "artificial intelligence"),
(9, "Hadoop"), (9, "Java"), (9, "MapReduce"), (9, "Big Data") ]
def data_scientists_who_like(target_interest):
return [user_idfor user_id, user_interest in interestsif user_interest == target_interest]
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每次搜索都要遍历列表,性能差,建立一个字典
from collections import defaultdict
user_ids_by_interest = defaultdict(list)
for user_id, interest in interests:
user_ids_by_interest[interest].append(user_id)
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interests_by_user_id = defaultdict(list)
for user_id, interest in interests:
interests_by_user_id[user_id].append(interest)
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此处所生成的两个字典采用以内存换时间的计算机基本原理,极大程度上提高了查找的速度。
找与指定用户爱好最多相似的用户
def most_common_interests_with(user_id):
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)
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案例:工资与工作年限
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)]
绘图
def make_chart_salaries_by_tenure():
tenures = [tenure for salary, tenure in salaries_and_tenures]
salaries = [salary for salary, tenure in salaries_and_tenures]
plt.scatter(tenures, salaries)
plt.xlabel("Years Experience")
plt.ylabel("Salary")
plt.show()
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按工作年线算平均收入
salary_by_tenure = defaultdict(list)
for salary, tenure in salaries_and_tenures:
salary_by_tenure[tenure].append(salary)
average_salary_by_tenure = {
tenure : sum(salaries) / len(salaries)
for tenure, salaries in salary_by_tenure.items() }
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分组后计算
def tenure_bucket(tenure):
if tenure < 2: return "less than two"
elif tenure < 5: return "between two and five"
else: return "more than five"
salary_by_tenure_bucket = defaultdict(list)
for salary, tenure in salaries_and_tenures:
bucket = tenure_bucket(tenure)
salary_by_tenure_bucket[bucket].append(salary)
average_salary_by_bucket = {
tenure_bucket : sum(salaries) / len(salaries)
for tenure_bucket, salaries in salary_by_tenure_bucket.items() }
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案例:兴趣主题
简单方法:数兴趣词汇个数
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)
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