Question

我有一个带有日期列的dataframe，该列名为dateDecision。它们的格式为1970/01/01。我正在尝试将日期分为year，month和date列。我使用了以下代码

df %>% mutate(year = lubridate::year(dateDecision), 
                month = lubridate::month(dateDecision), 
                day = lubridate::day(dateDecision))

我收到一条错误消息，指出as.POSIXlt.character(x, tz = tz(x)) : character string is not in a standard unambiguous format中的错误

当我使用dput（head（df，10））时，这是输出

structure(list(term = c("1791", "1791", "1791", "1791", "1791", 
"1792", "1792", "1792", "1792", "1792"), dateDecision = c("8/3/1791", 
"8/3/1791", "8/3/1791", "8/3/1791", "8/3/1791", "8/11/1792", 
"8/11/1792", "8/11/1792", "8/11/1792", "8/11/1792"), decisionType = c("6", 
"6", "6", "6", "6", "8", "8", "8", "8", "8"), dateArgument = c("8/2/1791", 
"8/2/1791", "8/2/1791", "8/2/1791", "8/2/1791", "8/9/1792", "8/9/1792", 
"8/9/1792", "8/9/1792", "8/9/1792")), row.names = c(NA, -10L), class = c("tbl_df", 
"tbl", "data.frame"))

我该如何解决？

Answer 1

您首先需要使用lubridate将dateDecision列转换为日期

df %>% mutate(dateDecision=lubridate::mdy(dateDecision),
              year = lubridate::year(dateDecision), 
              month = lubridate::month(dateDecision), 
              day = lubridate::day(dateDecision))

Answer 2

首先使用anytime::anydate()库将您的dateDecision列转换为日期：

library(anytime)
    date <- "1970/01/01"
    anytime::anydate(date)
    [1] "1970-01-01"

然后运行您的变异。您也可以使用tk_augment_timeseries_signature()中的timetk，并且仅选择year，month和date列。

library(timetk)
library(tidyverse)

# make some reproducible data
dates <-
    seq(as.Date('2019-01-01'), as.Date('2019-12-31'), by = 'days') 

dates <- as_tibble(dates)

#add the time signature features including year, month, day, day of week, day of month, #day of year, hour, minute, second to the input data

dates %>% 
tk_augment_timeseries_signature()

将数据转换为R中的lubridate的日期，月，年

2 个答案: