Question

client_id <- (2260419,2260412,2260413,2260415) transaction_date <- (2016-09-03, 2016-09-04, 2016-09-06, 2016-09-07) amount <- (350.0, 250.0,431.0,220.0)

month_ARPU AS
(SELECT
  visit_month,
  Avg(revenue) AS ARPU 
  FROM
  (SELECT
    Cust_id,
    Datediff(MONTH, ‘2010-01-01’, transaction_date) AS visit_month,
    Sum(transaction_size) AS revenue 
    FROM   transactions 
    WHERE  transaction_date &gt; Dateadd(‘year’, -1, CURRENT_DATE)
    GROUP BY
    1,
    2)
  GROUP BY 1)

我理解SELECT, GROUP BY, COUNT(1), SUM()和AS（）如何单独运作，但不像上面的代码那样整体运作，主要是COUNT(1)和SUM()的运作方式。

Answer 1

好的，这是一个选项，但我做了一些假设。问题是有3种情况选择了一个未在group by函数中添加的值。因此，我假设这个值在group by子句中每个变量的每个级别都是唯一的。

library(tidyverse)

month_ARPU <- transactions %>% 
          group_by(1,2) %>%
          summarise(visit_month = length(seq(from = as.Date("2010-01-01"), to = as.Date(unique(transaction_date)), by = 'month')) - 1,
                    revenue = sum(transaction_size),
                    Cust_id = unique(Cust_id),
                    transaction_date = as.Date(unique(transaction_date)) %>% ungroup() %>%
          filter(transaction_date > as.Date(Sys.time())-365) %>%
          group_by(1,visit_month) %>% 
          summarise(ARPU = mean(revenue))

我需要将以下SQL代码应用到R中

1 个答案: