根据R中第二个数据帧中的日期范围将数据帧中的数据分组

时间:2018-07-03 14:27:35

标签: r

我有以下两个数据帧:

Date <- seq(as.Date("2013/1/1"), by = "day", length.out = 46)
x <-data.frame(Date)
x$discharge <- c("1000","1100","1200","1300","1400","1200","1300","1300","1200","1100","1200","1200","1100","1400","1200","1100","1400","1000","1100","1200","1300","1400","1200","1300","1300","1200","1100","1200","1200","1100","1400","1200","1100","1400","1000","1100","1200","1300","1400","1200","1300","1300","1200","1100","1200","1200")
x$discharge <- as.numeric(x$discharge)

Date_from <- c("2013-01-01","2013-01-15","2013-01-21","2013-02-10")
Date_to <- c("2013-01-07","2013-01-20","2013-01-25","2013-02-15")
y <- data.frame(Date_from,Date_to)
y$concentration <- c("1.5","2.5","1.5","3.5")
y$Date_from <- as.Date(y$Date_from)
y$Date_to <- as.Date(y$Date_to)
y$concentration <- as.numeric(y$concentration)

我正在尝试根据日期范围xy,根据数据帧Date_from中每一行的数据,从数据帧Date_to中的日排放中计算出平均排放量在数据帧y中。请注意,2013年1月1日至2013年1月14日之间和2013年1月26日至2013年2月9日之间,数据帧y中的度量存在差距。该差距是由于在此期间未进行任何测量而造成的。由于我使用以下代码来计算y中每个日期范围的平均排放量,因此这一差距让我头疼。

rng <- cut(x$Date, breaks=c(y$Date_from, max(y$Date_to), 
                    include.lowest=T))
range<-cbind(x,rng)
discharge<-aggregate(cbind(mean=x$discharge)~rng, FUN=mean)

但是,如果您在数据框range中检查范围,则将2013-01-01至2013-01-07的范围扩展到2013-01-14,但我只需要将其扩展到2013-01 -07,然后稍作休息,直到下一个范围从2013-01-15开始。

2 个答案:

答案 0 :(得分:4)

您可以尝试使用tidyverse

library(tidyverse)
y %>% 
  split(seq_along(1:nrow(.))) %>% 
  map(~filter(x, between(Date, .$Date_from, .$Date_to)) %>% 
        summarise(Mean=mean(discharge))) %>% 
  bind_rows() %>% 
  bind_cols(y,.)
   Date_from    Date_to concentration     Mean
1 2013-01-01 2013-01-07           1.5 1214.286
2 2013-01-15 2013-01-20           2.5 1166.667
3 2013-01-21 2013-01-25           1.5 1300.000
4 2013-02-10 2013-02-15           3.5 1216.667

仅使用此代码,您可以看到值和组。

y %>% 
  split(seq_along(1:nrow(.))) %>% 
  map(~filter(x, between(Date, .$Date_from, .$Date_to))) 

答案 1 :(得分:2)

这是一个base的答案:

helper <- merge(x, y)
helper <- helper[helper$Date >= helper$Date_from & helper$Date <= helper$Date_to, ]
aggregate(helper$discharge,
          list(Date_from = helper$Date_from,
               Date_to = helper$Date_to),
          FUN = 'mean')

   Date_from    Date_to        x
1 2013-01-01 2013-01-07 1214.286
2 2013-01-15 2013-01-20 1166.667
3 2013-01-21 2013-01-25 1300.000
4 2013-02-10 2013-02-15 1216.667