R lag/lead irregular time series data

女生的网名这么多〃 提交于 2019-12-19 04:47:11

问题


I have irregular time series data frame with time (seconds) and value columns. I want to add another column, value_2 where values are lead by delay seconds. So value_2 at time t equals to value at time t + delay or right after that.

ts=data.frame(
  time=c(1,2,3,5,8,10,11,15,20,23),
  value=c(1,2,3,4,5,6,7,8,9,10)
)

ts_with_delayed_value <- add_delayed_value(ts, "value", 2, "time")

> ts_with_delayed_value
   time value value_2
1     1     1       3
2     2     2       4
3     3     3       4
4     5     4       5
5     8     5       6
6    10     6       8
7    11     7       8
8    15     8       9
9    20     9      10
10   23    10      10

I have my own version of this function add_delayed_value, here it is:

add_delayed_value <- function(data, colname, delay, colname_time) {
  colname_delayed <- paste(colname, sprintf("%d", delay), sep="_")
  data[colname_delayed] <- NaN

  for (i in 1:nrow(data)) {
    time_delayed <- data[i, colname_time] + delay
    value_delayed <- data[data[colname_time] >= time_delayed, colname][1]
    if (is.na(value_delayed)) {
      value_delayed <- data[i, colname]
    }
    data[i, colname_delayed] <- value_delayed
  }

  return(data)
}

Is there a way to vectorize this routine to avoid the slow loop?

I'm quite new to R, so this code probably has lots of issues. What can be improved about it?


回答1:


You could try:

library(dplyr)
library(zoo)
na.locf(ts$value[sapply(ts$time, function(x) min(which(ts$time - x >=2 )))])
[1]  3  4  4  5  6  8  8  9 10 10



回答2:


This should work for your data. If you want to make a general function, you'll have to play around with lazyeval, which honestly might not be worth it.

library(dplyr)
library(zoo)

carry_back = . %>% na.locf(na.rm = TRUE, fromLast = FALSE)


data_frame(time = 
             with(ts, 
                  seq(first(time), 
                      last(time) ) ) ) %>%
  left_join(ts) %>%
  transmute(value_2 = carry_back(value),
            time = time - delay) %>%
  right_join(ts) %>%
  mutate(value_2 = 
           value_2 %>%
           is.na %>%
           ifelse(last(value), value_2) )



回答3:


What you want is not clear, give a pseudo code or a formula. It looks like this is what you want... From what I understand from you the last value should be NA

library(data.table)
setDT(ts,key='time')
ts_delayed = ts[,.(time_delayed=time+2)]
setkey(ts_delayed,time_delayed)
ts[ts_delayed,roll=-Inf]


来源:https://stackoverflow.com/questions/36753810/r-lag-lead-irregular-time-series-data

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