转换列类/类型时维护date / POSIXct列

Question

我有一个400列的数据框，其中散布了多个日期列。 在下面的代表性示例中，我想实现以下目标：

1. 将因子转换为数字或字符或POSIXct
2. 在合理的地方将字符转换为数字/整数

3. 将任何包含日期的列都转到POSIXct，而不管它是因子，字符还是日期

    set.seed(123) df1 <- data.frame( A = as.numeric(1:10), B = sample(seq(as.POSIXct('2000/01/01'), as.POSIXct('2018/01/01'), by="day"), size=10), C = as.numeric(sample(20:90, size = 10)), D = sample(c("yes", "no"), size=10, replace = TRUE), E = as.factor(sample(1000:2000, size = 10)), F = as.character(c("test","test2","test3","test4","test5","test6","test7","test8","test9","test10")), G = as.factor(c("test","test2","test3","test4","test5","test6","test7","test8","test9","test10")), H = as.character(sample(seq(as.POSIXct('2000/01/01'), as.POSIXct('2018/01/01'), by="day"), size=10)),stringsAsFactors=FALSE ) df1 ABCDEFGH 1 1 2005-03-06 00:00:00 87 no 1963 test test 2002-07-27 23:00:00 2 2 2014-03-11 00:00:00 51 no 1902 test2 test2 2007-06-17 23:00:00 3 3 2007-05-11 23:00:00 66 no 1690 test3 test3 2007-06-11 23:00:00 4 4 2015-11-22 00:00:00 58 no 1793 test4 test4 2006-08-20 23:00:00 5 5 2016-12-02 00:00:00 26 no 1024 test5 test5 2002-09-27 23:00:00 6 6 2000-10-26 00:00:00 79 no 1475 test6 test6 2002-06-30 23:00:00 7 7 2009-06-30 23:00:00 35 no 1754 test7 test7 2004-03-11 00:00:00 8 8 2016-01-19 00:00:00 22 no 1215 test8 test8 2008-05-17 23:00:00 9 9 2009-11-30 00:00:00 40 yes 1315 test9 test9 2004-10-12 00:00:00 10 10 2008-03-17 00:00:00 85 yes 1229 test10 test10 2015-06-03 23:00:00 unlist(lapply(df1, class)) A B1 B2 CDEFGH "numeric" "POSIXct" "POSIXt" "numeric" "character" "factor" "character" "factor" "character" 

到目前为止，我已经尝试了以下操作（但它没有保留POSIXct列B）或将字符日期列（H列）转换为POSIXct：

df1_clean <- df1 %>% mutate_all(funs(type.convert(as.character(.), as.is = TRUE)))
unlist(lapply(df1_clean, class))
      A           B           C           D           E           F            G           H 
      "integer" "character"   "integer" "character"   "integer" "character" "character" "character" 

对于这个小的数据集，我可以仅调用列，并使用lubridate将B和H转换为POSIXct，但我希望它能在整个数据框中自动显示。

任何帮助将非常感激！ 谢谢萌

Answer 1

这可能不是最优雅的方法-但它似乎对我有用。

#install.packages("tidyverse")
#install.packages("dataCompareR")
library("tidyverse")
library("dataCompareR")



# create reproducible df
set.seed(123)
df1 <- data.frame(
  A = as.numeric(1:10),
  B = sample(seq(as.POSIXct('2000/01/01', tz = "UTC"), as.POSIXct('2018/01/01', tz = "UTC"), by="day"), size=10),
  C = as.numeric(sample(20:90, size = 10)),
  D = sample(c("yes", "no"), size=10, replace = TRUE),
  E = as.factor(sample(1000:2000, size = 10)),
  F = as.character(c("test","test2","test3","test4","test5","test6","test7","test8","test9","test10")),
  G = as.factor(c("test","test2","test3","test4","test5","test6","test7","test8","test9","test10")),
  H = as.character(sample(seq(as.POSIXct('2000/01/01', tz = "UTC"), as.POSIXct('2018/01/01', tz = "UTC"), by="day"), size=10)),stringsAsFactors=FALSE
)
df1 #look at df

unlist(lapply(df1, class)) #look at df classes


df1_clean <- df1 %>% mutate_all(funs(type.convert(as.character(.), as.is = TRUE))) #reassign classes by running type.convert (input are all variables from the df but as.character)
unlist(lapply(df1_clean, class)) #look at df classes now

#check if a column is a Date - https://stackoverflow.com/questions/18178451/is-there-a-way-to-check-if-a-column-is-a-date-in-r
tmp=sapply(df1_clean, function(x) !all(is.na(as.Date(as.character(x),format="%Y-%m-%d", tz = "UTC")))) 

# if tmp is True, change according column to as.POSIXct 
for (i in 1:ncol(df1_clean)){
  if (tmp[i] == T){
    df1_clean[,i]<- as.POSIXct(df1_clean[,i], tz = "UTC")
  }
}

df1_clean #look at df
unlist(lapply(df1_clean, class)) #look at df classes


comp <- rCompare(df1, df1_clean) #compare your dfs before and after using the dataCompareR package
summary(comp) # check summary