在R的四分之一中未购买后将0 [英] put a 0 after not buying in a quarter in r
问题描述
我有一个变量ID，月份（或期间）和该月的收入的基数。我需要的是，如果客户在接下来的3个月内购买，则输入1；否则，则输入0，然后针对所有ID进行输入。
例如，如果我在第1个月内购物，并且在接下来的3个月内有购买交易，则在该行中为该客户放置1。
在最近的3个月内，将出现NA。
I have a base with the variables ID, month (or period) and the incomes of that month. What I need is to put a 1 if the client buys in the next 3 months or a 0 if not, and do it for all ID. For example, if I am in month 1 and there's a purchase in the next 3 months, then put a 1 in that row for that client. In the last periods as there will not be 3 months, an NA appears.
df<-tibble::tribble(
~ID, ~Month, ~Incomes,
1L, 1L, 5000L,
1L, 2L, 0L,
1L, 3L, 0L,
1L, 4L, 0L,
1L, 5L, 0L,
1L, 6L, 0L,
1L, 7L, 400L,
1L, 8L, 300L,
1L, 9L, 0L,
1L, 10L, 0L,
1L, 11L, 0L,
1L, 12L, 0L,
1L, 13L, 400L,
2L, 1L, 0L,
2L, 2L, 100L,
2L, 3L, 0L,
2L, 4L, 0L,
2L, 5L, 0L,
2L, 6L, 0L,
2L, 7L, 0L,
2L, 8L, 1500L,
2L, 9L, 0L,
2L, 10L, 0L,
2L, 11L, 0L,
2L, 12L, 100L,
2L, 13L, 750L,
3L, 1L, 0L,
3L, 2L, 0L,
3L, 3L, 0L,
3L, 4L, 0L,
3L, 5L, 700L,
3L, 6L, 240L,
3L, 7L, 100L,
3L, 8L, 0L,
3L, 9L, 0L,
3L, 10L, 0L,
3L, 11L, 0L,
3L, 12L, 500L,
3L, 13L, 760L
)
df<-as.data.frame(df)
# ID Month Incomes
# 1 1 5000
# 1 2 0
# 1 3 0
# 1 4 0
# 1 5 0
# 1 6 0
# 1 7 400
# 1 8 300
# 1 9 0
# 1 10 0
# 1 11 0
# 1 12 0
# 1 13 400
# 2 1 0
# 2 2 100
# 2 3 0
# 2 4 0
# 2 5 0
# 2 6 0
# 2 7 0
# 2 8 1500
# 2 9 0
# 2 10 0
# 2 11 0
# 2 12 100
# 2 13 750
# 3 1 0
# 3 2 0
# 3 3 0
# 3 4 0
# 3 5 700
# 3 6 240
# 3 7 100
# 3 8 0
# 3 9 0
# 3 10 0
# 3 11 0
# 3 12 500
# 3 13 760
我希望看起来像这样：
dffinal<- tibble::tribble(
~ID_RUT, ~Month, ~Incomes, ~Quarter,
1L, 1L, 5000L, 0L,
1L, 2L, 0L, 0L,
1L, 3L, 0L, 0L,
1L, 4L, 0L, 1L,
1L, 5L, 0L, 1L,
1L, 6L, 0L, 1L,
1L, 7L, 400L, 1L,
1L, 8L, 300L, 0L,
1L, 9L, 0L, 0L,
1L, 10L, 0L, 0L,
1L, 11L, 0L, NA,
1L, 12L, 0L, NA,
1L, 13L, 400L, NA,
2L, 1L, 0L, 1L,
2L, 2L, 100L, 0L,
2L, 3L, 0L, 0L,
2L, 4L, 0L, 0L,
2L, 5L, 0L, 1L,
2L, 6L, 0L, 1L,
2L, 7L, 0L, 1L,
2L, 8L, 1500L, 0L,
2L, 9L, 0L, 1L,
2L, 10L, 0L, 1L,
2L, 11L, 0L, NA,
2L, 12L, 100L, NA,
2L, 13L, 750L, NA,
3L, 1L, 0L, 0L,
3L, 2L, 0L, 1L,
3L, 3L, 0L, 1L,
3L, 4L, 0L, 1L,
3L, 5L, 700L, 1L,
3L, 6L, 240L, 1L,
3L, 7L, 100L, 0L,
3L, 8L, 0L, 0L,
3L, 9L, 0L, 1L,
3L, 10L, 0L, 1L,
3L, 11L, 0L, NA,
3L, 12L, 500L, NA,
3L, 13L, 760L, NA
)
# ID Month Incomes Quarterly
# 1 1 5000 0
# 1 2 0 0
# 1 3 0 0
# 1 4 0 1
# 1 5 0 1
# 1 6 0 1
# 1 7 400 1
# 1 8 300 0
# 1 9 0 0
# 1 10 0 0
# 1 11 0 NA
# 1 12 0 NA
# 1 13 400 NA
# 2 1 0 1
# 2 2 100 0
# 2 3 0 0
# 2 4 0 0
# 2 5 0 1
# 2 6 0 1
# 2 7 0 1
# 2 8 1500 0
# 2 9 0 1
# 2 10 0 1
# 2 11 0 NA
# 2 12 100 NA
# 2 13 750 NA
# 3 1 0 0
# 3 2 0 1
# 3 3 0 1
# 3 4 0 1
# 3 5 700 1
# 3 6 240 1
# 3 7 100 0
# 3 8 0 0
# 3 9 0 1
# 3 10 0 1
# 3 11 0 NA
# 3 12 500 NA
# 3 13 760 NA
有人吗？感谢您的宝贵时间
Does anyone how to do it? Thanks for your time
推荐答案
1）前滚前滚收入> ; 0 如果为TRUE，则返回TRUE，否则为FALSE。使用 + 将其转换为数字。 1：3表示使用从当前点开始的偏移量1，2，3，即下三个收入。如果要考虑每个收入末尾的下一个和下两个收入，请将 partial = TRUE 参数添加到 rollapply
1) rollapply Roll forward along Incomes > 0 returning TRUE if any are TRUE and FALSE otherwise. Convert that to numeric using +. 1:3 means use offsets 1, 2, 3 from the current point, i.e. the next three incomes. Add the partial=TRUE argument to rollapply if you want to consider the next and next two incomes near the end of each group where there are not three left.
library(dplyr)
library(zoo)
df %>%
group_by(ID) %>%
mutate(Quarter = +rollapply(Incomes > 0, list(1:3), any, fill = NA)) %>%
ungroup
2）SQL 一种SQL解决方案会是：
2) SQL An SQL solution would be:
library(sqldf)
over <- "partition by ID rows between 1 following and 3 following"
fn$sqldf("select
*,
(max(Incomes > 0) over ($over)) +
(case when (count(*) over ($over)) = 3 then 0 else Null end) as Quarter
from df")
如果可以处理少于3行的元素，可以简化此过程。 over 来自上面：
This can be simplified if it is OK to process elements for which there are fewer than 3 rows following. over is from above:
fn$sqldf("select *, (max(Incomes > 0) over ($over)) as Quarter from df")
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