FB DS 电面

rochellehx 发表于 2018-1-6 05:35
2.         SQL部分
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Follow up: The conversion rate for the experiment is ...

For the close friend notification question, I think we can break down the answer to three steps.

First, if the feature were successful, would it be good for Facebook? That is, would it move the key metric to the right direction? If we use engagement (time spent on Facebook, number of actions) as the metric, then the answer is yes. Sending text notification will potentially make people spend more time on Facebook and perform more actions (react to friends' posts).

Second, also the key to this question, is that we need to find a proxy for demand for this feature based on what users are doing today. The safest way for a data scientist to drive new features is to look at the data. If you find a demand for that feature (maybe users are going to friend's profile page to check for updates), then you can incentivize this behavior by simplifying the process. Sending text messages is a good way to do this.

Third, once you establish that the feature is good for Facebook's target metric and there's a demand for that today, we can run an A/B test to decide if we want to add it.

很有用信息

starfoe 发表于 2018-5-30 07:49
不知道对不对啊求指正！

假设检验

2/sqrt(4.5/1000)

这里的4.5是不是错了？应该是3？
variance=1.5

补充内容 (2019-4-1 07:40):
上面的写错了，应该是2/sqrt(1.5/2000)

补充内容 (2019-4-1 07:45):
不好意思刚才又写错了。。。最后的更正： 2/sqrt(1.5/500)

maohappy2013 发表于 2018-3-1 23:47
select adv.advertiser_id,
     cost = sum(spend),
     rev = sum(isnull(ad.price,0)),

同意，应该先aggregate再join啊。
ROI：
select sum(revenue)/sum(spend) as ROI
from
        (select ads_id, sum(spend) as spend
        from adv_info  a
        where date= current_date()
        group by 1
        ) s
left join
        (select ads_id, sum(user_spend) as revenue
        from ad_info b
        where date = current_date()
        group by 1
        ) r
on s.ads_id = r.ads_id

感觉做hypothesis testing的时候应该是H0: diff in time spent control = diff in time spent treatment vs. HA:diff in time spent control = diff in time spent treatment.

而不是用mean？不然给的before and after就没什么意义了吧

fiibb 发表于 2019-9-16 07:05
感觉做hypothesis testing的时候应该是H0: diff in time spent control = diff in time spent treatment vs ...

我觉得没错诶，h0是launch feature (treatment group) 不会导致significant difference in mean time spent (compared with control group)

starfoe 发表于 2018-5-30 06:13
第一问：
select count(distinct adv.advertiser_id)/(select count(distinct advertiser_id from adv_inf ...

. 1point3acreshi! 想问一下你的
select count(distinct adv.advertiser_id)/(select count(distinct advertiser_id from adv_info)
from adv_info adv join ad_info ad on adv.ad_id = ad.ad_id

里面加粗的部分用法叫什么呀？我之前从没见过，但是google又搜不到，所以想请教一下。谢谢！

biubiula 发表于 2019-11-8 10:32
hi! 想问一下你的
select count(distinct adv.advertiser_id)/(select count(distinct advertiser_id fr ...

嗯。。可能那个括号的位置需要调整


(select count(distinct advertiser_id）from adv_info)
这样看会不会容易理解一些？

计算adv_info中不重复的advertiser_id，是外层表达式的分母

starfoe 发表于 2019-11-18 03:13
嗯。。可能那个括号的位置需要调整

谢谢回复！我能看懂那一段，但是我以前没用过这种可以直接把select from another table套在calculation里的用法，我试了一下好像也有syntax error，可能是不同的database用法不一样把～

starfoe 发表于 2019-11-18 03:13
嗯。。可能那个括号的位置需要调整

啊楼主 我今天试了一下 发现是可以用那种用法的 哈哈 还是谢谢啦