R问题中的整数64到日期时间转换？答案

【问题标题】：integer64 to datatime conversion in R issue?R问题中的整数64到日期时间转换？
【发布时间】：2019-03-07 14:47:53
【问题描述】：

给定以下dataframe 的integer64 unix 纪元：

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()）但出现错误：

    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：不知道如何转换 'time_stamp' 到类“POSIXct”

请告知如何处理integer64 epoch times。

【问题讨论】：

似乎正确。你的问题是什么？您需要定义origin，以便as.POSIXct 可以将您的整数添加到该原点，从而获得日期
哦，等等。你得到不同的错误......我运行你的最后一段代码，我得到了：Error in mutate_impl(.data, dots) : Evaluation error: 'origin' must be supplied.
这些看起来像纳秒。这似乎给出了一个合理的日期，但我不知道它是否正确：as.POSIXct("1970-01-01 00:00:00", tz = "UTC") + data_df$time_stamp / 1e9。（除法过程中会损失精度。如果这很重要，您需要使用高精度数字而不是浮点数。）
所以每次我都需要弄清楚除以什么常数并将原点设置为 1970-01-01？ @Sotos
您不需要弄清楚。您应该从创建这些时间的系统的规范中知道。

标签： r datetime unix-timestamp lubridate

【解决方案1】：

请原谅直接的语言，但你的问题没有意义。获取数据集的第一个元素：0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396。这在您列出的任何数据类型中根本无法表示。包括integer64。句号。

现在，恰好我的nanotime 包以最佳可用分辨率执行此操作，该分辨率以 64 整数表示的纳秒。并且 64 位整数允许自纪元以来的纳秒增量，精度约为 19 位。不是您要求的 100 多位数字。没有（小内存）变量可以。

至于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 包的integer64 类型。以示例为基础：

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>

我将这种纳秒粒度的双重表示与 POSIXct 表示一起用于 R 使用，包括整天绘图。（请注意，有一个格式错误，它在 UTC 中显示 nanotime / integer64 列，但底层表示是正确的，因为 pt 转换为 POSIXct 显示。目前在我的时区下午 3 点之后.)

【讨论】：

当时间为纳米结构时，无法在动物园聚合。 dput（头（采样器，10））结构（c（“134775”，“142110”，“79900”，“73893”，“91030”，“113602”，“129546”，“134999”，“131465”，“ 136492"), index = new("nanotime", .S3Class = "integer64", structure(c(2.69467041777383e-201, 2.69467041777383e-201, 2.69467041783873e-201, 2.69467042886863e-201, 2.69467042955947e-201, 2.69467043387697e -201, 2.69467043615511e-201, 2.69467043615511e-201, 2.69467045797641e-201, 2.69467045797641e-201), class= "integer64"))), 1234562)
是的。 zoo 被证明仅适用于 Date 或 POSIXct 类型的有序观察。将data.table 与nanotime 一起使用。

【解决方案2】：

在这种情况下，错误是描述性的。 as.POSIXct 不处理 integer64。这里有一些简单的命令表明这一点

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')

如果您可以忍受较低的精度（在 2038 年之前有一段时间），您可以转换为整数。

【讨论】：