R中相邻时间戳之间的时间差
问题描述:
我正在寻找一种有效的方法来计算给定日期的时间差。我的数据如下所示:R中相邻时间戳之间的时间差
Time_formatted Date Hours_diff
1/20/2016 19:19 1/20/2016 0:46
1/20/2016 18:33 1/20/2016 2:43
1/20/2016 15:50 1/20/2016 1:28
1/20/2016 14:22 1/20/2016 1:50
1/20/2016 12:32 1/20/2016 4:52
1/20/2016 7:40 1/20/2016 0
1/19/2016 23:23 1/19/2016
1/19/2016 23:06 1/19/2016
1/19/2016 22:37 1/19/2016
1/19/2016 21:56 1/19/2016
1/19/2016 21:05 1/19/2016
1/19/2016 17:53 1/19/2016
1/19/2016 17:39 1/19/2016
1/19/2016 17:01 1/19/2016
1/19/2016 15:31 1/19/2016
我首先想到的是使用ave
功能R:
ave(data$Time_formatted, data$Date, FUN=difftime)
,但一个明显的错误,会出现使用此功能Error in as.POSIXct(time2) : argument "time2" is missing, with no default
。由于数据非常大,循环效率低下。 任何想法来解决这类问题?
答
首先,我准备数据。我要确保日期和时间由列Time_formatted
存储在正确的格式和排序的数据帧:
# convert times to POSIXct, dates to Date
data$Time_formatted <- as.POSIXct(data$Time_formatted, format = "%m/%d/%Y %H:%M")
data$Date <- as.Date(data$Date, format = "%m/%d/%Y")
# sort
data <- data[order(data$Time_formatted), ]
然后我结合使用tapply()
与diff()
来计算以分钟为单位的差异。请注意,我以占首次每一天,其中的时间差不确定增加一个额外的零:
my_diff <- function(x, ...) {
c(0, diff(x, ...))
}
diffs <- unlist(tapply(data$Time_formatted, data$Date, my_diff))
的最后一步是从分钟的时间差转换为%H:%M
如下(见this answer的使用formatC()
):
mins2hm <- function(min) {
h <- min %/% 60
m <- min %% 60
hm <- paste(h, formatC(m, width = 2, flag = 0), sep = ":")
}
data$diffs <- mins2hm(diffs)
data
## Time_formatted Date diffs
## 15 2016-01-19 15:31:00 2016-01-19 0:00
## 14 2016-01-19 17:01:00 2016-01-19 1:30
## 13 2016-01-19 17:39:00 2016-01-19 0:38
## 12 2016-01-19 17:53:00 2016-01-19 0:14
## 11 2016-01-19 21:05:00 2016-01-19 3:12
## 10 2016-01-19 21:56:00 2016-01-19 0:51
## 9 2016-01-19 22:37:00 2016-01-19 0:41
## 8 2016-01-19 23:06:00 2016-01-19 0:29
## 7 2016-01-19 23:23:00 2016-01-19 0:17
## 6 2016-01-20 07:40:00 2016-01-20 0:00
## 5 2016-01-20 12:32:00 2016-01-20 4:52
## 4 2016-01-20 14:22:00 2016-01-20 1:50
## 3 2016-01-20 15:50:00 2016-01-20 1:28
## 2 2016-01-20 18:33:00 2016-01-20 2:43
## 1 2016-01-20 19:19:00 2016-01-20 0:46
+0
谢谢,我喜欢在这个例子中集成'tapply'! –
答
步骤和输出如下:
> df<-read.csv("data.txt",header=T,stringsAsFactors=F)
> df$Time_formatted<-strptime(a$Time_formatted,"%m/%d/%Y %H:%M")
> df$Date <-strptime(a$Date,"%m/%d/%Y")
> df<-df[order(df$Time_formatted,decreasing=T),] #Make sure it is ordered
> df
Time_formatted Date
1 2016-01-20 19:19:00 2016-01-20
2 2016-01-20 18:33:00 2016-01-20
3 2016-01-20 15:50:00 2016-01-20
4 2016-01-20 14:22:00 2016-01-20
5 2016-01-20 12:32:00 2016-01-20
6 2016-01-20 07:40:00 2016-01-20
7 2016-01-19 23:23:00 2016-01-19
8 2016-01-19 23:06:00 2016-01-19
9 2016-01-19 22:37:00 2016-01-19
10 2016-01-19 21:56:00 2016-01-19
11 2016-01-19 21:05:00 2016-01-19
12 2016-01-19 17:53:00 2016-01-19
13 2016-01-19 17:39:00 2016-01-19
14 2016-01-19 17:01:00 2016-01-19
15 2016-01-19 15:31:00 2016-01-19
> df$Hours_diff<-c(-diff(df$Time_formatted),0) # calculate time difference
> df[which(diff(df$Date)!=0),"Hours_diff"]<-0 # set the last timepoint in day to 0
> df$Hours_diff<-ifelse(df$Hours_diff>0,paste(floor(df$Hours_diff/60),df$Hours_diff%%60,sep=":"),0)
> df
Time_formatted Date Hours_diff
1 2016-01-20 19:19:00 2016-01-20 0:46
2 2016-01-20 18:33:00 2016-01-20 2:43
3 2016-01-20 15:50:00 2016-01-20 1:28
4 2016-01-20 14:22:00 2016-01-20 1:50
5 2016-01-20 12:32:00 2016-01-20 4:52
6 2016-01-20 07:40:00 2016-01-20 0
7 2016-01-19 23:23:00 2016-01-19 0:17
8 2016-01-19 23:06:00 2016-01-19 0:29
9 2016-01-19 22:37:00 2016-01-19 0:41
10 2016-01-19 21:56:00 2016-01-19 0:51
11 2016-01-19 21:05:00 2016-01-19 3:12
12 2016-01-19 17:53:00 2016-01-19 0:14
13 2016-01-19 17:39:00 2016-01-19 0:38
14 2016-01-19 17:01:00 2016-01-19 1:30
15 2016-01-19 15:31:00 2016-01-19 0
我不知道,我知道你想要什么。你想为每个日期在相邻时间戳之间的所有时间差异?或者每天只有一个号码? – Stibu
嗨,你在第一个假设中是正确的,我编辑过这个帖子来显示一个硬编码的例子。最大的诀窍是要处理一天中没有时间差异的第一时间...... –