一大波SQL面试题+solution+转换成Python/R的写法

很多公司(e.g. Linkedin) 给data scientist的technical interview会要求interviewee用SQL完成一个data manipulation, 然后再用Python或者R重新写一遍。分享一波我整理的题目 + SQL/Python pandas/R dplyr的写法。有用的话拜托撒点大米吧~~

P.S. 都是我自己的solution， 不能保证正确/最优，如果有不对的地方或者更好的解法欢迎指出！题目出自的公司就隐去了

再附上一个有用的关于怎样translate SQL queries into Python pandas的帖子
https://medium.com/jbennetcodes/ ... d-more-149d341fc53e

Q1
Problem:  Member can make purchase via either mobile  or desktop platform. Using the following data table to determine the total number of member and revenue for mobile-only, desktop_only and mobile_desktop.
The input spending table is
member_id    date    channel   spend
1001    1/1/2018    mobile    100
1001    1/1/2018    desktop    100
1002    1/1/2018    mobile    100
1002    1/2/2018    mobile    100
1003    1/1/2018    desktop    100
1003    1/2/2018    desktop    100

The output data is
date    channel    total_spend    total_members
1/1/2018    desktop    100    1
1/1/2018    mobile    100    1
1/1/2018    both    200    1-google 1point3acres
1/2/2018    desktop    100    1
1/2/2018    mobile    100    1
1/2/2018    both    0    0


SQL solution:

1.

CREATE VIEW member_spend AS
SELECT
date,
member_id,
SUM(CASE WHEN channel == ‘mobile’ THEN spend ELSE 0 END) AS mobile_spend,
SUM(CASE WHEN channel == ‘desktop’ THEN spend ELSE 0 END) AS desktop_spend
FROM
spending
GROUP BY date, member_id;


SELECT
date,
CASE WHEN mobile_spend > 0 AND desktop_spend = 0 THEN ‘mobile’
         WHEN mobile_spend = 0 AND desktop_spend > 0 THEN ‘desktop’
         WHEN mobile_spend > 0 AND desktop_spend > 0 THEN ‘both’
         END AS channel,
SUM(mobile_spend + desktop_spend) AS total_spend,
COUNT(*) AS total_members
FROM member_spend
GROUP BY
date,
CASE WHEN mobile_spend > 0 AND desktop_spend = 0 THEN ‘mobile’
         WHEN mobile_spend = 0 AND desktop_spend > 0 THEN ‘desktop’
         WHEN mobile_spend > 0 AND desktop_spend > 0 THEN ‘both’
         END
;


Python pandas:

spending[‘mobile_spend’] = spending[spending.channel == ‘mobile’].spend
spending[‘desktop_spend’] = spending[spending.channel == ‘desktop’].spend
.google  иmember_spend = spending.group_by([‘date’, ‘member_id’]).sum([‘mobile_spend’, ‘desktop_spend’]).to_frame([‘mobile_spend’, ‘desktop_spend’].reset_index()



SQL

SELECT
date,
CASE WHEN mobile_spend > 0 AND desktop_spend = 0 THEN ‘mobile’
-baidu 1point3acres         WHEN mobile_spend = 0 AND desktop_spend > 0 THEN ‘desktop’
         WHEN mobile_spend > 0 AND desktop_spend > 0 THEN ‘both’
         END AS channel,
SUM(mobile_spend + desktop_spend) AS total_spend,
COUNT(*) AS total_members
FROM member_spend
GROUP BY
date,
CASE WHEN mobile_spend > 0 AND desktop_spend = 0 THEN ‘mobile’
         WHEN mobile_spend = 0 AND desktop_spend > 0 THEN ‘desktop’
         WHEN mobile_spend > 0 AND desktop_spend > 0 THEN ‘both’
         END
;

Python pandas

member_spend[member_spend.mobile_spend>0 & member_spend.desktop_spend==0], ‘channel’] = ‘mobile’
member_spend[member_spend.mobile_spend==0 & member_spend.desktop_spend>0], ‘channel’] = ‘desktop’
. 1point3acresmember_spend[member_spend.mobile_spend>0 & member_spend.desktop_spend>0], ‘channel’] = ‘both’


tot_members = member_spend.groupby([‘date’, ‘channel’]).size().to_frame(‘tot_members’).reset_index()
tot_spend = member_spend.groupby([‘date’, ‘channel’].agg({‘mobile_spend’:sum, ‘desktop_spend’:sum}).to_frame([‘mobile_spend’, ‘desktop_spend’])
tot_spend[‘tot_spend’] = tot_spend[‘mobile_spend’] + tot_spend[‘desktop_spend’]
output = tot_members.concat(tot_spend[‘tot_spend’])


R dplyr:

output <- spending %>%
  group_by(member_id, date)%>%
  summarise(mobile_spend = sum(spend[channel == 'mobile']),
            desktop_spend = sum(spend[channel == 'desktop'])) %>%
  mutate(channel = ifelse((mobile_spend>0 & desktop_spend>0), 'both',
                          ifelse(mobile_spend==0, 'desktop', 'mobile'))) %>%
  group_by(date, channel)%>%
  summarise(total_spend = mobile_spend + desktop_spend,
                     total_member = n())


Q2

Problem: table member_id|company_name|year_start
1): count members who ever moved from Microsoft to Google?

2):  count members who directly moved from Microsoft to Google? (Microsoft -- Linkedin -- Google doesn't count)

1) SQL solution:
(assumption: no end date. define move from company A to company B if the start date of company B is after the start date of company A, Microsoft -- Linkedin -- Google count)


SELECT
COUNT(DISTINCT member_id) AS num_member
FROM table t1
JOIN table t2
ON t1.member_id = t2.member_id
WHERE t1.year_start < t2.year_start
AND t1.company_name = ‘Microsoft’
AND t2.company_name = ‘Google’
;


Python pandas:

joined_table = table.merge(table, on =‘member_id’, how=’inner’, suffixes = (‘t1_’, ‘t2_’))
Num_member = joined_table[(joined_table.t1_company_name == ‘Microsoft’) & (joined_table.t2_company_name == ‘Google’) & (joined_table.t1_year_start < joined_table.t2_year_start)].member_id.unique()

2)
.1point3acresSQL solution:
SELECT
COUNT(DISTINCT member_id) AS num_member
FROM
(SELECT
member_id,
company_name,
LEAD(company_name, 1) OVER (PARTITION BY member_id ORDER BY year_start) AS next_company
FROM table
) t
WHERE
Company_name = “Microsoft”
AND next_company = “Google”
;

Python pandas:

table[‘next_company’] = table.sort_values(by=[‘member_id’, ‘year_start’]).groupby([‘member_id’])[‘company_name’].shift(1)

table[(table.company_name == ‘Microsoft’) & (table.next_company == ‘Google’)][‘member_id’].unique()

Q3
Problem
member_id|email_address, suppose that every member has two email address, get the table in the format of
member_id | email1 | email2

SQL solution:

SELECT
member_id,
MAX(CASE WHEN email_rank = 1 THEN email_address ELSE NULL END) AS email1,
MAX(CASE WHEN email_rank = 2 THEN email_address ELSE NULL END) AS email2
FROM
(SELECT
member_id,
email_address,
ROW_NUMBER() OVER (PARTITION BY member_id ORDER BY email_address) AS email_rank
FROM table) t
GROUP BY member_id;

Python pandas:

table[‘email_rank’] = table.groupby([‘member_id’])[‘email_address’].rank(method = ‘first’)

Use groupby rank(method=’first’) to replace row_number() partition by in SQL.
method=’min’ to replace Rank, method = ‘dense’to replace dense_rank()

Output = table.pivot(index=’member_id’, columns=’email_rank’, values=’email_address’)


补充内容 (2019-6-20 11:41):
感谢xymxym指出的几个pandas的语法错误. Q1第一部分的pandas的最后一个语句应该改为member_spend = spending.groupby([‘date’, ‘member_id’]).sum()[[‘mobile_spend’, ‘desktop_spend’]].reset_index()

补充内容 (2019-6-20 11:43):
SQL中的lead()对应的pandas应该是shift(-1)而不是shift(1); SQL中 COUNT（DISTINCT X）对应的pandas function应该是.nunique() 而不是.unique()

多谢lz分享! 觉得你的python答案可能有点小问题。.to_frame应该是把series变成dataframe, groupby之后当有两个column时应该已经是dataframe了，用.to_frame会报错; lead()对应的应该是shift(-1)而不是shift(1); .unique()应该用.nunique()。 我在学习python中，如果说的不对请lz指正。 期待lz分享更多题目！

sql转成python不一定要用panda，见过很多面经都要求无包写法

谢谢楼主分享，学习了

谢谢楼主分享！mark 了！

谢楼主分享！目前小白看不懂，先mark 了~

谢谢分享~ 还有更新吗

谢谢楼主分享，已加米，期待后续

求教怎样能验证自己写的sql、r和python的output都是正确的呢？有什么平台可以练习吗？

谢谢楼主的分享！！！努力准备中

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