Question

我正在使用一些时间数据，而且我在将时差转换为数年和数月时遇到了问题。

我的数据或多或少看起来像这样，

dfn <- data.frame(
Today  = Sys.time(),
DOB  = seq(as.POSIXct('2007-03-27 00:00:01'), len= 26, by="3 day"),
Patient  = factor(1:26, labels = LETTERS))

首先，我从今天的数据（DOB）中减去出生数据（Today）。

dfn$ageToday <-  dfn$Today - dfn$DOB

这给了我Time difference in days。

dfn$ageToday
 Time differences in days
  [1] 1875.866 1872.866 1869.866 1866.866 1863.866
  [6] 1860.866 1857.866 1854.866 1851.866 1848.866
 [11] 1845.866 1842.866 1839.866 1836.866 1833.866
 [16] 1830.866 1827.866 1824.866 1821.866 1818.866
 [21] 1815.866 1812.866 1809.866 1806.866 1803.866
 [26] 1800.866
 attr(,"tzone")
 [1] ""

这是我提问的第一部分; 如何将此差异转换为年和月（四舍五入）？（即4.7,4.11等）

我阅读了?difftime手册页和?format，但我没弄明白。

任何帮助都将不胜感激。

此外，我想融化我的最终对象，如果我尝试使用此命令在上面的数据框上使用融合，

require(plyr)
require(reshape)
mdfn <- melt(dfn, id=c('Patient'))

我得到了这个奇怪的警告，我以前没有看到

Error in as.POSIXct.default(value) : 
  do not know how to convert 'value' to class "POSIXct"

所以，我的第二个问题是; 如何创建时间差异我可以melt与POSIXct变量一起使用？如果我在没有dfn$ageToday的情况下融化，那么一切都像魅力一样。

谢谢，Eric

Answer 1

lubridate软件包可以处理日期和时间，包括查找时间差异，非常简单。

library("lubridate")
library("reshape2")

dfn <- data.frame(
    Today  = Sys.time(),
    DOB  = seq(as.POSIXct('2007-03-27 00:00:01'), len= 26, by="3 day"),
    Patient  = factor(1:26, labels = LETTERS))

dfn$diff <- new_interval(dfn$DOB, dfn$Today) / duration(num = 1, units = "years")

mdfn <- melt(dfn, id=c('Patient'))
class(mdfn$value) # all values are coerced into numeric

new_interval()函数计算两个日期之间的时差。请注意，有一个函数today()可以替代您使用Sys.time。最后请注意duration()函数创建一个标准的ehm持续时间，您可以使用该持续时间将间隔除以标准单位的长度，在本例中为一年的单位。

如果您想要保留Today和DOB的内容，那么您可能希望先将所有内容转换为character并稍后重新转换...

library("lubridate")
library("reshape2")

dfn <- data.frame(
  Today  = Sys.time(),
  DOB  = seq(as.POSIXct('2007-03-27 00:00:01'), len= 26, by="3 day"),
  Patient  = factor(1:26, labels = LETTERS))

# Create standard durations for a year and a month
one.year <- duration(num = 1, units = "years")
one.month <- duration(num = 1, units = "months")

# Calculate the difference in years as float and integer
dfn$diff.years <- new_interval(dfn$DOB, dfn$Today) / one.year
dfn$years <- floor( new_interval(dfn$DOB, dfn$Today) / one.year )

# Calculate the modulo for number of months
dfn$diff.months <- round( new_interval(dfn$DOB, dfn$Today) / one.month )
dfn$months <- dfn$diff.months %% 12

# Paste the years and months together
# I am not using the decimal point so as not to imply this is
# a numeric representation of the diference
dfn$y.m <- paste(dfn$years, dfn$months, sep = '|')

# convert Today and DOB to character so as to preserve them in melting
dfn$Today <- as.character(dfn$Today)
dfn$DOB <- as.character(dfn$DOB)

# melt using string representation of difference between the two dates
dfn2 <- dfn[,c("Today", "DOB", "Patient", "y.m")]
mdfn2 <- melt(dfn2, id=c('Patient'))

# alternative melt using numeric representation of difference in years
dfn3 <- dfn[,c("Today", "DOB", "Patient", "diff.years")]
mdfn3 <- melt(dfn3, id=c('Patient'))

计算年龄和月份的年龄并融化数据

1 个答案: