integer64到R问题中的数据转换？

Question

给定以下dataframe of integer64 unix时代：

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))

我想将它转换为日期时间(as.POSIXct或anytime())，但是我得到了一个错误：

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

给予：

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:不知道如何将“time_stamp”转换为“POSIXct”类

请提供如何处理integer64的建议。

Answer 1

恕我直言，但你的问题毫无意义。接受数据集的第一个元素：0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396。这在您列出的任何数据类型中都是无法表示的。包括integer64。句号。

现在，我的nanotime包以最好的可用分辨率(以64整数表示的纳秒)实现了这一功能。64位整数允许自那个时代以来的纳秒增量，精度约为19位。不是你要求的100+数字。没有(小内存)变量可以。

至于nanotime，example()展示了一些用途，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支持这里使用的bit64包的integer64类型。在这个例子的基础上：

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> 

我使用这种纳秒粒度的双重表示和R的POSIXct表示，包括一整天的绘图。(请注意，有一个格式错误，它显示了UTC中的nanotime / integer64列，但是底层表示形式是正确的，就像pt到POSIXct的转换显示的那样。目前我的时区在下午3点刚过。)

Answer 2

在这种情况下，错误是描述性的。as.POSIXct不处理integer64。下面是一些简单的命令，它们表示如下

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')

如果您可以接受较低的精度(在2038年之前有一段时间)，则可以转换为整数。