Question

我对R和sqldf很新，似乎无法解决一个基本问题。我有一个包含交易的文件，其中每行代表购买的产品。

该文件如下所示：

customer_id,order_number,order_date, amount, product_name
1, 202, 21/04/2015, 58, "xlfd"
1, 275, 16//08/2015, 74, "ghb"
1, 275, 16//08/2015, 36, "fjk"
2, 987, 12/03/2015, 27, "xlgm"
3, 376, 16/05/2015, 98, "fgt"
3, 368, 30/07/2015, 46, "ade"

我需要找到每个order_number在单个交易中花费的最大金额（相同customer_id）。例如，如果是customer_id "1"，则为(74+36)=110。

Answer 1

假设数据框名为orders，以下将完成工作：

sqldf("select customer_id, order_number, sum(amount) 
       from orders 
       group by customer_id, order_number")

使用嵌套查询

更新：，以下内容将提供所需的输出：

sqldf("select customer_id, max(total) 
       from (select customer_id, order_number, sum(amount) as total 
             from orders 
             group by customer_id, order_number) 
       group by customer_id")

输出：

    customer_id max(total)
1           1        110
2           2         27
3           3         98

Answer 2

如果sqldf不是严格的要求。

将您的输入视为dft，您可以尝试：

require(dplyr)
require(magrittr)
dft %>% 
    group_by(customer_id, order_number) %>% 
    summarise(amt = sum(amount)) %>% 
    group_by(customer_id) %>% 
    summarise(max_amt = max(amt))

给出：

Source: local data frame [3 x 2]
Groups: customer_id [3]

  customer_id max_amt
        <int>   <int>
1           1     110
2           2      27
3           3      98

Answer 3

我们也可以使用data.table。将'data.frame'转换为'data.table'（setDT(df1)），按'customer_id'，'order_number'分组，我们得到'{1}}'amount'，再做第二组'customer_id'并获得'Sumamount'的sum

max

或者使其更加紧凑，在按“customer_id”进行分组后，我们{'1}}'ordernumber'的'amount'，循环遍历library(data.table) setDT(df1)[, .(Sumamount = sum(amount)) , .(customer_id, order_number) ][,.(MaxAmount = max(Sumamount)) , customer_id] # customer_id MaxAmount #1: 1 110 #2: 2 27 #3: 3 98，获取split，找到list获取'MaxAmount'

sum

或使用max

setDT(df1)[, .(MaxAmount = max(unlist(lapply(split(amount,
                      order_number), sum)))), customer_id]
#   customer_id MaxAmount
#1:           1       110
#2:           2        27
#3:           3        98

选择单个订单中花费的最大金额

3 个答案: