Convertendo o carimbo de data / hora no formato R correto usando chron () - r

Desde já, obrigado!

Aqui está uma amostra (n = 10) das datas em que estoutrabalhando com. Estes são de um excel .csv. Eu preciso convertê-los em um formato R reconhecível, a fim de realizar aritmética básica (dias entre etc). Esta é a saída de dput ().

help.data <- structure(c(55L, 49L, 50L, 52L, 51L, 114L, 53L, 59L, 63L, 48L
), .Label = c("1/15/2012 8:00", "1/16/2012 6:00", "1/18/2012 10:00",
"1/5/2012 4:00", "1/5/2012 8:00", "1/7/2012 2:00", "1/8/2012 22:01",
"1/9/2012 16:01", "10/1/2011 4:00", "10/12/2011 8:00", "10/18/2011 0:00",
"10/19/2011 6:01", "10/2/2011 6:00", "10/21/2011 4:00", "10/21/2011 6:01",
"10/23/2011 6:01", "10/25/2011 14:00", "10/29/2011 4:01", "10/3/2011 4:00",
"10/30/2011 6:01", "10/31/2011 6:00", "10/5/2011 2:00", "10/8/2011 2:00",
"11/1/2011 0:00", "11/16/2011 10:00", "11/19/2011 10:00", "11/19/2011 6:00",
"11/25/2011 10:01", "11/27/2011 10:00", "11/9/2011 14:01", "12/1/2011 2:01",
"12/1/2011 8:00", "12/10/2011 4:03", "12/13/2011 2:01", "12/15/2011 4:00",
"12/16/2011 4:00", "12/18/2011 14:01", "12/22/2011 6:00", "12/25/2011 0:01",
"12/25/2011 6:00", "12/27/2011 22:00", "12/5/2011 8:01", "12/9/2011 8:01",
"2/3/2012 2:00", "2/3/2012 8:01", "2/5/2012 8:00", "2/9/2012 6:01",
"3/11/2011 8:00", "3/13/2011 8:00", "3/14/2011 0:02", "3/14/2011 10:00",
"3/14/2011 8:01", "3/15/2011 20:02", "3/17/2011 8:00", "3/18/2011 4:01",
"3/19/2011 10:01", "3/19/2011 12:00", "3/21/2011 4:00", "3/24/2011 10:00",
"3/26/2011 4:01", "3/27/2011 6:00", "3/29/2011 12:00", "3/4/2011 6:00",
"4/1/2011 0:00", "4/12/2011 8:00", "4/14/2011 10:00", "4/17/2011 10:00",
"4/21/2011 18:01", "4/29/2011 6:00", "4/3/2011 8:00", "4/4/2011 6:02",
"4/6/2011 8:00", "5/10/2011 12:00", "5/13/2011 6:00", "5/16/2011 8:00",
"5/19/2011 14:00", "5/2/2011 14:00", "5/20/2011 6:00", "5/23/2011 4:00",
"5/24/2011 16:02", "5/24/2011 4:00", "5/26/2011 4:01", "5/27/2011 4:01",
"5/28/2011 14:00", "5/31/2011 16:01", "5/31/2011 4:01", "5/9/2011 4:01",
"5/9/2011 8:00", "6/10/2011 12:00", "6/12/2011 12:00", "6/13/2011 6:02",
"6/15/2011 16:01", "6/15/2011 4:02", "6/17/2011 2:01", "6/19/2011 6:00",
"6/21/2011 8:01", "6/23/2011 20:01", "6/23/2011 22:01", "6/23/2011 4:02",
"6/26/2011 10:01", "6/26/2011 4:01", "6/27/2011 2:01", "6/28/2011 2:02",
"6/29/2011 10:00", "6/29/2011 6:01", "6/3/2011 10:01", "6/4/2011 6:00",
"6/7/2011 20:02", "6/7/2011 22:01", "7/1/2011 4:01", "7/10/2011 20:01",
"7/10/2011 6:01", "7/12/2011 6:01", "7/13/2011 8:01", "7/16/2011 4:00",
"7/16/2011 8:00", "7/17/2011 8:01", "7/18/2011 14:00", "7/2/2011 10:01",
"7/2/2011 2:01", "7/21/2011 4:01", "7/23/2011 6:01", "7/24/2011 16:02",
"7/27/2011 4:00", "7/28/2011 4:01", "7/31/2011 8:00", "7/4/2011 4:01",
"7/7/2011 0:00", "7/8/2011 4:00", "7/9/2011 12:01", "8/10/2011 6:01",
"8/11/2011 2:00", "8/12/2011 2:02", "8/15/2011 6:01", "8/18/2011 4:00",
"8/2/2011 8:00", "8/20/2011 4:00", "8/20/2011 8:00", "8/24/2011 4:01",
"8/25/2011 18:01", "8/29/2011 6:01", "8/3/2011 6:02", "8/30/2011 4:01",
"8/4/2011 4:00", "8/5/2011 6:00", "8/6/2011 4:00", "8/8/2011 6:01",
"9/1/2011 4:01", "9/10/2011 2:00", "9/14/2011 8:02", "9/19/2011 6:02",
"9/2/2011 10:01", "9/20/2011 4:01", "9/20/2011 8:00", "9/22/2011 10:01",
"9/22/2011 16:00", "9/27/2011 4:00", "9/28/2011 2:00", "9/29/2011 4:00",
"9/3/2011 10:00", "9/3/2011 8:01", "9/30/2011 6:00", "9/4/2011 10:01",
"9/7/2011 8:01", "9/9/2011 8:00"), class = "factor")
help.data

Executar o código acima deve resultar em… meus dados de amostra.

 [1] 3/18/2011 4:01  3/13/2011 8:00  3/14/2011 0:02  3/14/2011 8:01
[5] 3/14/2011 10:00 7/13/2011 8:01  3/15/2011 20:02 3/24/2011 10:00
[9] 3/4/2011 6:00   3/11/2011 8:00

Seguindo o exemplo no arquivo de ajuda chron (), executei o seguinte código

help.data <- as.character(help.data)
help = t(as.data.frame(strsplit(help.data," ")))
row.names(help) = NULL

Isso tudo parece funcionar bem, mas quando eu corro oúltimo bit de código (abaixo) eu continuamente recebo erros retornados em relação ao formato. Como meus formatos são diferentes dos exemplos, mudei o código, mas ainda retornei erros.

library(chron)
help.done = chron(dates=help[,1],times=help[,2],
format=c("m/d/y" , "h:m"))

O formato acima ("m / d / y", "h: m") é como eu quero que os dados sejam exibidos. Qualquer ajuda é muito apreciada!!

Respostas:

Ou você poderia simplesmente usar as.POSIXct(help.data, format="%m/%d/%Y %H:%M", tz="GMT").

Ou:

library(lubridate)
mdy_hm(as.character(help.data))

Os tempos devem estar no formato h: m: s. Isso vai funcionar:

chron(dates=dates(help[,1]),times=times(paste0(help[,2], ":00")),
format=c("m/d/Y" , "h:m:s"))