A和B组之间共有10个项目,每个项目的开始和结束日期都不同。对于给定时段内的每一天,都需要计算outputX和outputY的总和。我设法对所有项目都这样做,但是如何按组划分结果呢?
我已经使用lapply()和purrr:map()进行了几次尝试,也查看了过滤器和拆分,但无济于事。下面是一个不区分组的示例。
library(tidyverse)
library(lubridate)
df <- data.frame(
project = 1:10,
group = c("A","B"),
outputX = rnorm(2),
outputY = rnorm(5),
start_date = sample(seq(as.Date('2018-01-3'), as.Date('2018-1-13'), by="day"), 10),
end_date = sample(seq(as.Date('2018-01-13'), as.Date('2018-01-31'), by="day"), 10))
df$interval <- interval(df$start_date, df$end_date)
period <- data.frame(date = seq(as.Date("2018-01-08"), as.Date("2018-01-17"), by = 1))
df_sum <- do.call(rbind, lapply(period$date, function(x){
index <- x %within% df$interval;
list("X" = sum(df$outputX[index]),
"Y" = sum(df$outputY[index]))}))
outcome <- cbind(period, df_sum) %>% gather("id", "value", 2:3)
outcome
最终它应该是40x4的表格。一些建议非常感谢!
答案 0 :(得分:0)
如果我对您的理解正确,则需要使用内部联接。因此可以建议我们使用sqldf
。参见https://stackoverflow.com/a/11895368/9300556
利用您的数据,我们可以做到这样。无需计算df$interval
,但我们需要在ID
上加上period
,否则sqldf
无效。
df <- data.frame(
project = 1:10,
group = c("A","B"),
outputX = rnorm(2),
outputY = rnorm(5),
start = sample(seq(as.Date('2018-01-3'), as.Date('2018-1-13'), by="day"), 10),
end = sample(seq(as.Date('2018-01-13'), as.Date('2018-01-31'), by="day"), 10))
# df$interval <- interval(df$start_date, df$end_date)
period <- data.frame(date = seq(as.Date("2018-01-08"), as.Date("2018-01-17"), by = 1)) %>%
mutate(id = 1:nrow(.))
然后我们可以使用sqldf
sqldf::sqldf("select * from period inner join df
on (period.date > df.start and period.date <= df.end) ") %>%
as_tibble() %>%
group_by(date, group) %>%
summarise(X = sum(outputX),
Y = sum(outputY)) %>%
gather(id, value, -group, -date)
# A tibble: 40 x 4
# Groups: date [10]
date group id value
<date> <fct> <chr> <dbl>
1 2018-01-08 A X 3.04
2 2018-01-08 B X 2.34
3 2018-01-09 A X 3.04
4 2018-01-09 B X 3.51
5 2018-01-10 A X 3.04
6 2018-01-10 B X 4.68
7 2018-01-11 A X 4.05
8 2018-01-11 B X 4.68
9 2018-01-12 A X 4.05
10 2018-01-12 B X 5.84
# ... with 30 more rows