R中的每小时组数据帧

Question

我有一个数据帧，它在日期列中包含DateTime值，在三个列中包含每个日期时间的计数。

我想对数据进行分组小时与三列

聚合函数适用于单一的列数，但我想这样做对整个数据帧。有小费吗？

aggregate(DateFreq$ColA,by=list((substr(DateFreq$Date,1,13))),sum) 

Answer 1

您可以使用aggregate的formula。但是你应该正确创建一个hour变量。

dat$hour <- as.POSIXlt(dat$Date)$hour 
aggregate(.~hour,data=dat,sum) 

这里一个例子：

Lines <- "Date,c1,c2,c3 
06/25/2013 12:01,0,1,1 
06/25/2013 12:08,-1,1,1 
06/25/2013 12:48,0,1,1 
06/25/2013 12:58,0,1,1 
06/25/2013 13:01,0,1,1 
06/25/2013 13:08,0,1,1 
06/25/2013 13:48,0,1,1 
06/25/2013 13:58,0,1,1 
06/25/2013 14:01,0,1,1 
06/25/2013 14:08,0,1,1 
06/25/2013 14:48,0,1,1 
06/25/2013 14:58,0,1,1" 

library(zoo) ## better to read/manipulate time series 
z <- read.zoo(text = Lines, header = TRUE, sep = ",", 
       index=0:1,tz='', 
       format = "%m/%d/%Y %H:%M") 


dat <- data.frame(Date = index(z),coredata(z)) 
dat$hour <- as.POSIXlt(dat$Date)$hour 
aggregate(.~hour,data=dat,sum) 

hour  Date c1 c2 c3 
1 12 5488624500 -1 4 4 
2 13 5488638900 0 4 4 
3 14 5488653300 0 4 4 

Answer 2

您可以使用dplyr使用dplyr::group_by和dplyr::summarise做聚合：

library(lubridate) 
library(anytime) 
library(tidyverse) 

Lines <- "Date,c1,c2,c3 
06/25/2013 12:01,0,1,1 
06/25/2013 12:08,-1,1,1 
06/25/2013 12:48,0,1,1 
06/25/2013 12:58,0,1,1 
06/25/2013 13:01,0,1,1 
06/25/2013 13:08,0,1,1 
06/25/2013 13:48,0,1,1 
06/25/2013 13:58,0,1,1 
06/25/2013 14:01,0,1,1 
06/25/2013 14:08,0,1,1 
06/25/2013 14:48,0,1,1 
06/25/2013 14:58,0,1,1" 

setClass("myDate") 
setAs("character","myDate", function(from) anytime(from)) 
df <- read.csv(text = Lines, header=TRUE, colClasses = c("myDate", "numeric", "numeric", "numeric")) 

df %>% 
    group_by(Date=floor_date(Date, "1 hour")) %>% 
    summarize(c1=sum(c1), c2=sum(c2), c3=sum(c3)) 
# A tibble: 3 × 4 
       Date c1 c2 c3 
       <dttm> <dbl> <dbl> <dbl> 
1 2013-06-25 12:00:00 -1  4  4 
2 2013-06-25 13:00:00  0  4  4 
3 2013-06-25 14:00:00  0  4  4