Question

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的参数在时间值-1和1之间变化。我想算一下：

m=-1和m=1个案的总持续时间（以小时为单位）分别
m=-1和m=1分别为
的每个案例有多长时间
M＆LT; -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＆lt; - seq.POSIXt（as.POSIXct（Sys.Date（）），as.POSIXct（Sys.Date（）+ 1），by =“1小时”）

Answer 1

我使用package data.table for＆＃34; split-apply-combine＆＃34;并使用cumsum和diff确定运行：

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

用R计算持续时间

1 个答案: