Question

我有一个数据，有一个名为Date的列，我将数据输入到R. 这是我的数据：

unique(data$Date)
 [1] ""           "2016/12/20" "2016/12/27" "2017/1/7"   "2017/1/27"  "2017/2/1"   "2017/2/2"   "2017/2/5"   "2017/2/6"   "2017/2/7"  
[11] "2017/2/8"   "2017/2/10"  "2017/2/11"  "2017/2/13"  "2017/2/14"  "2017/2/15"  "2017/2/17"  "2017/2/16"  "2017/2/24"  "2017/2/19" 
[21] "2017/2/21"  "2017/2/20"  "2017/2/26"  "2017/2/22"  "2017/3/2"   "2017/2/25"  "2017/2/28"  "2017/3/1"   "2017/3/4"   "2017/3/5"  
[31] "2017/3/6"   "2017/3/10"  "2017/3/8"   "2017/3/9"   "2017/3/11"  "2017/3/12"  "2017/3/13"  "2017/3/15"  "2017/3/29"  "2017/5/13" 
[41] "2015/10/5"  "2016/2/22"  "2015/3/6"   "2015/3/7"   "2015/10/15" "2015/3/9"   "2016/1/30"  "2015/10/29" "2015/10/24" "2015/10/17"
[51] "2016/1/8"   "2015/9/24"  "2016/2/15"  "2015/12/8"  "2015/12/10" "2016/2/6"   "2015/11/29" "2016/1/23"  "2015/10/11" "2016/2/16" 
[61] "2015/9/28"  "2016/1/29"  "2015/11/27" "2015/10/12" "2015/11/1"  "2015/11/16" "2015/10/10" "2015/11/30" "2016/1/2"   "2016/1/21" 
[71] "2016/4/22"  "2015/10/21" "2015/11/12" "2015/12/28" "2015/12/30" "2015/11/6"  "2015/10/8"  "2015/12/6"  "2016/1/24"  "2016/1/17" 
[81] "2016/2/26"  "2016/3/6"   "2016/2/17"  "2016/1/11"  "2015/12/3"  "2016/2/11"  "2015/11/22" "2015/10/2"  "2015/10/3"  "2015/11/4" 
[91] "2016/2/10"  "2015/12/9"  "2015/10/9"  "2015/12/1"  "2016/2/25"  "2016/1/19"  "2016/1/18"  "2015/12/13" "2016/2/14"  "2016/3/10" 

class(data$Date)
[1] "character"

我使用character将此date更改为as.Date()格式：

data$Date <- as.Date(data$Date)
Error in charToDate(x) : 
character string is not in a standard unambiguous format

我不知道怎么弄明白。我认为数据中的问题是""。我还有另一个名为Date2的列，但此列中不包含"" 有什么建议吗？

另外，如果我想同时使用两列进行as.Date并定义指定的格式，如`as.Date（x，“％Y /％m /％d”），我该怎么办？？

data[,c("Date", "Date2") := lapply(.SD, as.Date), .SDcols = c("Date", "Date2")]

Answer 1

您可以使用ymd()包中的lubridate函数转换日期，空字符串将转换为NA。例如，

> library(lubridate)
> (newdates <- ymd(dates))
   [1] NA           "2016-12-20" "2016-12-27" "2017-01-07" "2017-01-27" "2017-02-01" "2017-02-02" "2017-02-05"
   [9] "2017-02-06" "2017-02-07" "2017-02-08" "2017-02-10" "2017-02-11" "2017-02-13" "2017-02-14" "2017-02-15"
   [17] "2017-02-17" "2017-02-16" "2017-02-24" "2017-02-19" "2017-02-21" "2017-02-20" "2017-02-26" "2017-02-22"
   [25] "2017-03-02" "2017-02-25" "2017-02-28" "2017-03-01" "2017-03-04" "2017-03-05" "2017-03-06" "2017-03-10"
   [33] "2017-03-08" "2017-03-09" "2017-03-11" "2017-03-12" "2017-03-13" "2017-03-15" "2017-03-29" "2017-05-13"
   [41] "2015-10-05" "2016-02-22" "2015-03-06" "2015-03-07" "2015-10-15" "2015-03-09" "2016-01-30" "2015-10-29"
   [49] "2015-10-24" "2015-10-17" "2016-01-08" "2015-09-24" "2016-02-15" "2015-12-08" "2015-12-10" "2016-02-06"
   [57] "2015-11-29" "2016-01-23" "2015-10-11" "2016-02-16" "2015-09-28" "2016-01-29" "2015-11-27" "2015-10-12"
   [65] "2015-11-01" "2015-11-16" "2015-10-10" "2015-11-30" "2016-01-02" "2016-01-21" "2016-04-22" "2015-10-21"
   [73] "2015-11-12" "2015-12-28" "2015-12-30" "2015-11-06" "2015-10-08" "2015-12-06" "2016-01-24" "2016-01-17"
   [81] "2016-02-26" "2016-03-06" "2016-02-17" "2016-01-11" "2015-12-03" "2016-02-11" "2015-11-22" "2015-10-02"
   [89] "2015-10-03" "2015-11-04" "2016-02-10" "2015-12-09" "2015-10-09" "2015-12-01" "2016-02-25" "2016-01-19"
   [97] "2016-01-18" "2015-12-13" "2016-02-14" "2016-03-10"
> is.Date(newdates)
 [1] TRUE

Answer 2

正如评论中所确定的，答案是您需要指定要转换的日期的格式。在您的情况下，"%Y/%m/%d"

data$Date <- as.Date(data$Date, "%Y/%m/%d")

说明

您需要执行此操作的原因是因为您向量中的第一个条目是""，并且您尚未指定格式。

as.Date函数应用于字符时，首先检查format参数是否缺失。如果是，它会尝试根据向量的第一个元素猜测格式。

通过执行

测试"%Y-%m-%d"和"%Y/%m/%d"格式

xx <- ""
strptime(xx, "%Y-%m-%d")
NA
strptime(xx, "%Y/%m/%d")
NA

更具体地说，它使用以下测试（其中xx是向量的第一个元素）

if(is.na(xx) || 
    !is.na(strptime(xx, f <- "%Y-%m-%d", tz = "GMT")) || 
    !is.na(strptime(xx, f <- "%Y/%m/%d", tz = "GMT"))){
  print("success!")     ## I added this print statement for illustration purposes
}else{
  stop("character string is not in a standard unambiguous format")
}

正如您所看到的，xx在所有FALSE条件下评估为if，因此该函数必须输入stop方法。

要演示，请参阅这些陈述的结果

as.Date(c("2015/10/5", ""))  
# [1] "2015-10-05" NA
## SUCCESS, because it can 'guess' the first entry's format

as.Date(c("", "2015/10/5"))  
## ERROR: can't 'guess' the first entry's format

as.Date(c("2015/10/5", ""), format = "%Y/%m/%d") 
# [1] "2015-10-05" NA
## SUCCESS, because you've specified the format

as.Date(c("2015-10/5", "")) 
## ERROR: you haven't specified the format, 
## AND it's not one of the 'guessed' options ("%Y-%m-%d", "%Y/%m/%d")

字符格式无法更改为日期格式

2 个答案:

说明