计算R的持续时间 [英] Calculate duration in R

问题描述

我想知道是否有一种简单的方法来计算持续时间。我有一个数据集，其中一个名为 m 的参数在值 -1 和 1 。我要计算：

I wonder if there is an easy way to calculate the duration. I have a dataset where the a parameter, called m, varies between the values -1 and 1 during time. I want to calculate:

1. m = -1 的情况下的总持续时间（以小时为单位）和 m = 1
   



2. 每段病例多长时间 m = -1 和 m = 1 分别是

1. The total duration (time in hours) of cases where m=-1 and m=1 respectively

2. How long is each period of cases where m=-1 and m=1 respectively is

m< -c（1,1,1,1，-1，-1，-1，-1,1,1,1,1,1，-1，-1，-1，-1，-1，-1， 1,1,1,1,1,1,1）

m<-c(1,1,1,1,-1,-1,-1,-1,1,1,1,1,1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1)

时间<-seq.POSIXt（as.POSIXct（Sys.Date（）），as。 POSIXct（Sys.Date（）+ 1），= = 1小时）

Time <- seq.POSIXt(as.POSIXct(Sys.Date()), as.POSIXct(Sys.Date()+1), by = "1 hour")

推荐答案

我将数据包data.table用于 split-apply-combine，并使用 cumsum 和 diff 标识运行。 code>：

I'd use package data.table for "split-apply-combine" and identify the runs using cumsum and diff:

DF <- read.table(text="Time,    m
2015-01-01 00:00,    -1
2015-01-01 01:00,    -1
2015-01-01 02:00,    -1
2015-01-01 03:00,    1
2015-01-01 04:00,    1
2015-01-01 05:00,    1
2015-01-01 06:00,    1
2015-01-01 07:00,    1
2015-01-01 08:00,    -1
2015-01-01 09:00,    -1
2015-01-01 10:00,    -1
2015-01-01 11:00,    -1
2015-01-01 12:00,    1
2015-01-01 13:00,    1
2015-01-01 14:00,    1
2015-01-01 15:00,    -1", header = TRUE, sep =",")

library(data.table)
setDT(DF)
DF[, Time := as.POSIXct(Time, format = "%Y-%m-%d %H:%M", tz = "GMT")]
DF[, run := cumsum(c(1, diff(m) != 0))]

DF1 <- DF[, list(m = unique(m), 
                 duration = difftime(max(Time), min(Time), unit = "min")), 
          by = run]
#   run  m duration
#1:   1 -1 120 mins
#2:   2  1 240 mins
#3:   3 -1 180 mins
#4:   4  1 120 mins
#5:   5 -1   0 mins

DF1[, sum(duration), by = m]
#    m  V1
#1: -1 300
#2:  1 360

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