Maximum slope for a given interval each day

问题

I have a set of time series data with ground surface temperatures measured every 10 minutes over multiple days (actually 2 years of data) from three different locations. What I am interested in calculating is the maximum slope (rate of temperature increase) for any 60 minute interval for each day for each site.

So essentially I would like to work through each day, 10 minutes at a time, with a 60 minute window and calculate the slope for each window, and then determine the maximum slope and when during the day it occurred. I would then like to apply this function to every day in the data set. The date/time is in the following format (%m/%d/%y %H:%M).

I am imagining something using ddply and the zoo package and function rollapply, to do something like this pseudocode

ddply(data, .(location, day), function(d) max(rollapply(slope(d$temp~d$time, data=d)))

Where "time" is the time within each day (every 10 min) and "day" is simply the date so the function can be applied across all dates. Obviously, "slope" is not an R function and would have to be written to calculate actual slopes.

Does anyone have more experience with zoo and rollapply or can think of another way to solve this problem?

I've included some sample data here from a single location (so the location column has been removed) https://gist.github.com/natemiller/42eaf45747f31a6ccf9a

Thanks for any assistance, Nate

EDIT: I have since used a combination of geektrader's Joshua Ulrich's answers from below, and used basic algebra to convert the values back to units of ºC per hour

    CperH<-dat$Temp-(dat$Temp/(1+dat$ROC))

Works well.

回答1:

You can use xts timeseries package which is very good for timeseries analysis. Combined with TTR package, you can get what you want quite easily.

require(xts)
require(TTR)
dat <- read.csv("https://gist.github.com/natemiller/42eaf45747f31a6ccf9a/raw/916443cfb353d82e8af6cdebdd80b2e956317b24/sampleTempData.csv")

dat.xts <- .xts(x = dat$Temp, index = as.POSIXct(strptime(dat$Date, format = "%m/%d/%y %H:%M")))
names(dat.xts) <- "Temp"
head(dat.xts)
##                     Temp
## 2011-04-11 03:48:00  9.5
## 2011-04-11 03:58:00  9.5
## 2011-04-11 04:08:00  9.5
## 2011-04-11 04:18:00  9.5
## 2011-04-11 04:28:00  9.5
## 2011-04-11 04:38:00  9.5


dat.xts$ROC <- ROC(dat.xts, n = 6)
head(dat.xts, 10)
##                     Temp ROC
## 2011-04-11 03:48:00  9.5  NA
## 2011-04-11 03:58:00  9.5  NA
## 2011-04-11 04:08:00  9.5  NA
## 2011-04-11 04:18:00  9.5  NA
## 2011-04-11 04:28:00  9.5  NA
## 2011-04-11 04:38:00  9.5  NA
## 2011-04-11 04:48:00  9.5   0
## 2011-04-11 04:58:00  9.5   0
## 2011-04-11 05:08:00  9.5   0
## 2011-04-11 05:18:00  9.5   0

dat.xts[which.max(dat.xts$ROC), ]
##                     Temp       ROC
## 2011-04-12 09:48:00 14.5 0.5340825


# If you want to do analysis on per day basis.
dat.xts <- .xts(x = dat$Temp, index = as.POSIXct(strptime(dat$Date, format = "%m/%d/%y %H:%M")))
names(dat.xts) <- "Temp"
head(dat.xts)
##                     Temp
## 2011-04-11 03:48:00  9.5
## 2011-04-11 03:58:00  9.5
## 2011-04-11 04:08:00  9.5
## 2011-04-11 04:18:00  9.5
## 2011-04-11 04:28:00  9.5
## 2011-04-11 04:38:00  9.5


ll <- split.xts(dat.xts, f = "days")


ll <- lapply(ll, FUN = function(x) {
    x$ROC <- ROC(x, 6)
    return(x)
})

max.ll <- lapply(ll, function(x) x[which.max(x$ROC), ])

max.ll
## [[1]]
##                     Temp       ROC
## 2011-04-11 13:38:00 20.5 0.4946962
## 
## [[2]]
##                     Temp       ROC
## 2011-04-12 09:48:00 14.5 0.5340825
## 
## [[3]]
##                     Temp       ROC
## 2011-04-13 10:18:00 15.5 0.4382549
## 
## [[4]]
##                     Temp       ROC
## 2011-04-14 10:38:00 14.5 0.3715636

来源：https://stackoverflow.com/questions/15552200/maximum-slope-for-a-given-interval-each-day

标签

plyr

apply

zoo