R 随时间超前和滞后（移位）

Question

我尝试在数据框的列上使用滞后，但是当涉及到时间时它就不起作用了。我试过 shift、lag 和 tlag。

示例：

y = strptime(sprintf("%s:%s:%s", 4, 20, 10), "%H:%M:%S")
yy = strptime(sprintf("%s:%s:%s", 10, 20, 10), "%H:%M:%S")
lag(c(y,yy))

Error in format.POSIXlt(x, usetz = usetz) : invalid component [[10]] in "POSIXlt" should be 'zone'

tlag(c(y,yy))

Error in n_distinct_multi(list(...), na.rm) : argument "time" is missing, with no default

shift(c(y,yy))
[[1]]
[1] NA 10

[[2]]
[1] NA 20

[[3]]
[1] NA  4

[[4]]
[1] NA  4

[[5]]
[1] NA  6

[[6]]
[1]  NA 117

[[7]]
[1] NA  2

[[8]]
[1]  NA 184

[[9]]
[1] NA  1

[[10]]
[1] NA    "BST"

[[11]]
[1]   NA 3600

我不想要任何时差，我只想要我的数据框中上一行的值，我认为这是滞后造成的："Lead and lag are useful for comparing values offset by a constant (e.g. the previous or next value)"。时间甚至不重要，它应该只从以前的位置选择任何 numeric/character/time 。我该如何解决这个问题，或者是否有一个不同的功能可以完成我想要的功能 - 我不想涉及任何循环，因为速度很重要并且数据帧很大。

来自我的数据框的示例：

structure(list(sec = c(52, 53, 54, 55, 56, 57, 58, 59, 0, 1), 
    min = c(50L, 50L, 50L, 50L, 50L, 50L, 50L, 50L, 51L, 51L), 
    hour = c(11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L
    ), mday = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), mon = c(6L, 
    6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), year = c(117L, 117L, 
    117L, 117L, 117L, 117L, 117L, 117L, 117L, 117L), wday = c(2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), yday = c(184L, 184L, 
    184L, 184L, 184L, 184L, 184L, 184L, 184L, 184L), isdst = c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), zone = c("BST", "BST", 
    "BST", "BST", "BST", "BST", "BST", "BST", "BST", "BST"), 
    gmtoff = c(NA_integer_, NA_integer_, NA_integer_, NA_integer_, 
    NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, 
    NA_integer_)), .Names = c("sec", "min", "hour", "mday", "mon", 
"year", "wday", "yday", "isdst", "zone", "gmtoff"), class = c("POSIXlt", 
"POSIXt"))

Answer 1

对于像下面这样的data.frame

  index                time
1     1 2017-07-04 04:20:10
2     2 2017-07-04 10:20:10

你可以使用dplyr

dplyr::lag(df$time, 1)
[1] NA                         "2017-07-04 04:20:10 CEST"

dplyr::lead(df$time, 1)
[1] "2017-07-04 10:20:10 CEST" NA

要将 lead/lag 列添加到 data.frame，您可以使用

dplyr::mutate(df, lead_1 = dplyr::lead(time, 1), lag_1 = dplyr::lag(time, 1))
  index                time              lead_1               lag_1
1     1 2017-07-04 04:20:10 2017-07-04 10:20:10                <NA>
2     2 2017-07-04 10:20:10                <NA> 2017-07-04 04:20:10

R 随时间超前和滞后（移位）

R lead and lag (shift) with times

time

r

shift

lag

lead