Integer64到r问题中的Datatime转换

Question

给定以下integer64 unix epoch的dataframe：

data_df <- structure(list(time_stamp = structure(c(0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396
), class = "integer64")), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -10L))

我想将其转换为日期时间（as.POSIXct或anytime()），但我会有一个错误：

    data_df %>%
  dplyr::select(time_stamp) %>% 
  head(10) %>%
  dplyr::mutate(dt = anytime(time_stamp)) %>% dput()

给出：

structure(list(time_stamp = structure(c(0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396
    ), class = "integer64"), dt = structure(c(0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396, 
    0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396
    ), class = c("POSIXct", "POSIXt"), tzone = "Etc/UTC")), class = c("tbl_df", 
    "tbl", "data.frame"), row.names = c(NA, -10L))
data_df %>%
  dplyr::select(time_stamp) %>% 
  head(10) %>%
  dplyr::mutate(dt = as.POSIXct(time_stamp))

as.posixct.default（time_stamp）中的错误：不知道如何转换 'time_stamp'to类" posixct"

请建议如何处理integer64 Epoch Times。

Answer 1

赦免直接语言，但是您的问题没有意义。获取数据集的第一个元素：0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396。在您列出的数据类型的任何中，这根本不可用。包括integer64。完整停止。

现在，碰巧我的nanotime软件包以最佳可用分辨率执行此，该分辨率是64个智能者中代表的纳秒。自时期以来，自时代以来，有64位整数允许纳秒增量，以大约19位数字精度。不是您要求的100多位数字。没有（小内存）变量可以。

对于nanotime，example()显示了一些用途，包括解析：

R> library(nanotime)
R> example(nanotime)
nanotmR> x <- nanotime("1970-01-01T00:00:00.000000001+00:00")
nanotmR> print(x)
[1] "1970-01-01T00:00:00.000000001+00:00"
nanotmR> x <- x + 1
nanotmR> print(x)
[1] "1970-01-01T00:00:00.000000002+00:00"
nanotmR> format(x)
[1] "1970-01-01T00:00:00.000000002+00:00"
nanotmR> x <- x + 10
nanotmR> print(x)
[1] "1970-01-01T00:00:00.000000012+00:00"
nanotmR> format(x)
[1] "1970-01-01T00:00:00.000000012+00:00"
nanotmR> format(nanotime(Sys.time()) + 1:3)  # three elements each 1 ns apart
[1] "2019-03-10T20:06:53.534292001+00:00" "2019-03-10T20:06:53.534292002+00:00" 
[3] "2019-03-10T20:06:53.534292003+00:00"
R> 

最重要的是，data.table支持此处使用的bit64软件包的integer64类型。在示例上构建：

R> library(data.table)
data.table 1.12.0  Latest news: r-datatable.com
R> dt <- data.table(ns = nanotime(Sys.time()) + 1:3)
R> dt[]
                                    ns
1: 2019-03-10T20:08:48.165136001+00:00
2: 2019-03-10T20:08:48.165136002+00:00
3: 2019-03-10T20:08:48.165136003+00:00
R> dt[, pt := as.POSIXct(ns)]
R> dt[]
                                    ns                         pt
1: 2019-03-10T20:08:48.165136001+00:00 2019-03-10 15:08:48.165136
2: 2019-03-10T20:08:48.165136002+00:00 2019-03-10 15:08:48.165136
3: 2019-03-10T20:08:48.165136003+00:00 2019-03-10 15:08:48.165136
R> 

我使用纳秒粒度的双重表示，并使用posixct表示r使用，包括整天绘制。（请注意，有一个格式的不事件，它显示了UTC中的nanotime/integer64列，但基础表示形式是合理而正确的，因为pt转换为POSIXct显示。目前在我的时区下午3点之后。）

Answer 2

在这种情况下，错误是描述性的。as.POSIXct不处理Integer64。以下是一些简单的命令，例如这样

library(bit64)
i <- 1
i64 <- as.integer64(i)
as.POSIXct(i, tz='UTC', origin='1970-01-01')
## You will get an error here
as.POSIXct(i64, tz='UTC', origin='1970-01-01')

如果您可以使用较低的精度（在2038年之前有一些时间），则可以转换为整数。