计算年龄和月份的年龄并融化数据

Eri*_*ail 4 time r

我正在处理一些时间数据,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))
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首先,我从DOB今天的数据(Today)中减去birth()形式的数据.

dfn$ageToday <-  dfn$Today - dfn$DOB
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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] ""
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这是我提问的第一部分; 如何将此差异转换为年和月(四舍五入到几个月)?(即4.7,4.11等)

我阅读了?difftime手册页?format,但是我没弄明白.

任何帮助,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'))
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我收到了以前没见过的这个奇怪的警告

Error in as.POSIXct.default(value) : 
  do not know how to convert 'value' to class "POSIXct"
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所以,我的第二个问题是; 如何创建时间差异我可以meltPOSIXct变量一起使用?如果我融化,dfn$ageToday一切都像魅力一样.

谢谢,埃里克

gau*_*den 5

lubridate软件包使得处理日期和时间,包括找到时间差异,非常容易.

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

如果你想保留的内容TodayDOB,那么你可能需要的一切转化为character第一,后来再改...

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'))
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