Question

我只是想知道是否有人可以告诉我如何使用r来进行以下计算？

我有一个包含3列的一年一小时数据集，“date”“time”和“values”

例如：

'01/01/2000'     '08:00'     '10'     
'01/01/2000'     '09:00'     '30'
'01/01/2000'     '10:00'     '43'
'01/01/2000'     '11:00'     '55'
'01/01/2000'     '12:00'     '59'
'01/01/2000'     '13:00'     '45'
'01/01/2000'     '14:00'     '10'
'01/01/2000'     '15:00'     '15'
'01/01/2000'     '16:00'     '43'
'01/01/2000'     '17:00'     '45'
'01/01/2000'     '18:00'     '60'
'01/01/2000'     '19:00'     '10'

我想创建一个data.frame，用于计算值大于＆gt;的剧集的长度。 40，如果可能的话，用开始日期和时间表示，例如从上表中第一次出现超时是在上午10:00持续4小时，第二次出现是在16:00持续3小时，所以我想知道是否可以创建如下数据框？

     'date'      'time'    'Duration'  
'01/01/2000'     '10:00'       '4'
'01/01/2000'     '16:00'       '3'

等年度数据集

Answer 1

这是另一个依赖于plyr的解决方案：它可以更容易地计算其他数量在每个40以上的值的法术上，例如，平均值或最大值。

# Sample data
k <- 3
d <- data.frame( 
  date = rep( seq.Date( Sys.Date(), length=k, by="day" ), each=24 ),
  time = sprintf( "%02d:00", rep( 0:23, k ) ),
  value = round(200*runif(24*k))
)
d$timestamp <- as.POSIXct( paste( d$date, d$time ) )
d <- d[ order( d$timestamp ), ]
# Extract the spells above 40
n <- nrow(d)
d$inside <- d$value > 40
d$start  <- ! c(FALSE, d$inside[-n]) & d$inside
d$end    <- d$inside & ! c(d$inside[-1], FALSE)  # Not used
d$group  <- cumsum(d$start)  # Number the spells
d <- d[ d$inside, ]
library(plyr)
ddply( d, "group", summarize,
  start  = min(timestamp),
  end    = max(timestamp),
  length = length(value),
  mean   = mean(value)
)

超过40的值的法术可以持续数天：这可能是也可能不是你想要的。

Answer 2

另一种选择：

dat <- structure(list(date = c("01/01/2000", "01/01/2000", "01/01/2000", 
  "01/01/2000", "01/01/2000", "01/01/2000", "01/01/2000", "01/01/2000", 
  "01/01/2000", "01/01/2000", "01/01/2000", "01/01/2000"), 
  time = c("08:00", "09:00", "10:00", "11:00", "12:00", "13:00", "14:00", 
  "15:00", "16:00", "17:00", "18:00", "19:00"), value = c("10", "30", "43", 
  "55", "59", "45", "10", "15", "43", "45", "60", "10")), 
  .Names = c("date", "time", "values"), row.names = c(NA, -12L), 
  class = "data.frame")

run <- rle(dat$value > 40)
dat$exceeds <- rep(run$values, run$lengths)
dat$duration <- rep(run$lengths, run$lengths)
starts <- dat[head(c(1, cumsum(run$length) + 1), length(run$length)),]
result <- subset(starts, duration > 1 & exceeds)

result[, c(1, 2, 5)]

        date  time duration
3 01/01/2000 10:00        4
9 01/01/2000 16:00        3

Answer 3

一些数据

txt <- "'01/01/2000'     '08:00'     '10'     
'01/01/2000'     '09:00'     '30'
'01/01/2000'     '10:00'     '43'
'01/01/2000'     '11:00'     '55'
'01/01/2000'     '12:00'     '59'
'01/01/2000'     '13:00'     '45'
'01/01/2000'     '14:00'     '10'
'01/01/2000'     '15:00'     '15'
'01/01/2000'     '16:00'     '43'
'01/01/2000'     '17:00'     '45'
'01/01/2000'     '18:00'     '60'
'01/01/2000'     '19:00'     '10'"

tc <- textConnection(txt)

data <- read.table(tc,header=FALSE,as.is=TRUE)

功能

fun <- function(data,cutoff=40){
 data_above <- 1L*(data$V3>cutoff)
 id_start <- which(diff(c(0L,data_above))==1)
 id_end <- which(diff(c(data_above,0L))== -1)
 res <- cbind(data[id_start,1:2],Duration=id_end-id_start+1)
 return(res)
}

结果

fun(data)

          V1    V2 Duration
3 01/01/2000 10:00        4
9 01/01/2000 16:00        3

使用R计算剧集/事件的长度

3 个答案: