Question

给出以下dataframe Unix纪元中的以下integer64：

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()

礼物：

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”分类为“ POSIXct”

请建议如何处理integer64个时代。

Answer 1

请原谅直接的语言，但您的问题毫无道理。取数据集中的第一个元素：0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000282505613660396。在您列出的 any 数据类型中，这根本无法表示。包括integer64。句号。

现在，碰巧我的nanotime软件包以最佳可用分辨率完成了，该分辨率是用64整数表示的纳秒。自纪元以来，64位整数允许以纳秒为单位递增，精度约为19位。不是您要求的100多个数字。没有（小内存）变量。

对于nanotime，example()显示了一些用法，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integer64包的bit64类型的软件包。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>

我将这种纳秒级粒度的双重表示与POSIXct表示一起用于R使用，包括整日绘制。（请注意，有一个格式错误会在UTC中显示nanotime / integer64列，但基本表示是正确的，如pt转换为POSIXct所示。它目前是我所在时区的下午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。这是一些简单的命令，可以这样表示

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年之前花了一些时间），则可以转换为整数。

R中将integer64转换为datatime的问题？

2 个答案: