从R中的字符日期中删除分和秒

我有这个时间戳向量：

c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04", 
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36", 
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39", 
"01/09/2019 10:52:20")

我想去掉字符向量中的分和秒，这样我就有了

01/09/2019 9

和

01/09/2019 10

最有效的方法是什么？

这里有一个

datevec <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04", 
      "01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36", 
      "01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39", 
      "01/09/2019 10:52:20")

format(as.POSIXct(datevec, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H")

# Result
 [1] "01/09/2019 09" "01/09/2019 09" "01/09/2019 09" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
 [7] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"

---

datevec您想要的输出类是什么？这个怎么样：

v <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04", 
  "01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36", 
  "01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39", 
  "01/09/2019 10:52:20")


strptime(v, "%m/%d/%Y %H")

v另一个：

dates <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04", 
                  "01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36", 
                  "01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39", 
                  "01/09/2019 10:52:20")
unlist(lapply(dates,function(x) strsplit(x,":")[[1]][1]))

这看起来不错

unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)]

（在来自的帮助下制作）

替代方案可能是

sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1)

使用一些基准测试和Ronak最近的评论，fixed=TRUE使方法更快，我们看到方法four（上述方法）是最快的

mystring <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04", 
              "01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36", 
              "01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39", 
              "01/09/2019 10:52:20")

microbenchmark(one = sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1),
           two = unlist(lapply(mystring,function(x) strsplit(x,":", fixed=TRUE)[[1]][1])),
           three = strptime(mystring, "%m/%d/%Y %H"),
           four = unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)],
           five = format(as.POSIXct(mystring, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H"), 
           six = gsub("(.*?):.*", "\\1", mystring),
           seven = str_extract(mystring, ".+(?=:.+:)"),
           times = 100000)



    Unit: microseconds
  expr     min      lq      mean  median       uq        max neval
   one  42.792  49.471  85.63742  52.572  57.1310  669280.96 1e+05
   two  64.637  70.618 114.16364  73.252  77.6840  582466.94 1e+05
 three 129.456 134.771 156.82308 136.188 139.2030  339715.94 1e+05
  four  12.860  15.641  22.75699  17.254  18.5440  305703.52 1e+05
  five 482.888 505.647 633.15388 512.880 552.1155  551274.28 1e+05
   six  37.889  43.121  52.79030  45.567  49.1880   32954.59 1e+05
 seven  53.432  59.051  88.05015  62.326  69.9320 1180361.17 1e+05

mystring这里是另一个使用
gsub的

通过
（）
和
\\1
捕获模式要引用捕获的组，需要
？
使正则表达式变懒，因为有多个
：

gsub("(.*?):.*", "\\1", dates)

您还可以使用
stru extract
from
stringr
：

date_strings <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04", 
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36", 
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39", 
"01/09/2019 10:52:20")

str_extract(date_strings, ".+(?=:.+:)")

 [1] "01/09/2019 9"  "01/09/2019 9"  "01/09/2019 9"  "01/09/2019 10"
 [5] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
 [9] "01/09/2019 10" "01/09/2019 10"

date\u字符串是的。我认为
fixed=TRUE
使它更快。非常有趣，实际上4，fixed=TRUE是最快的，改变了基准来显示这一点。有趣的公认答案，你对效率的定义是什么？我对
tidyverse
软件包有很大的偏见：）啊，我明白了，我不怪你：）下次你可能应该把它放在问题里。。。
date_strings <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04", 
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36", 
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39", 
"01/09/2019 10:52:20")

str_extract(date_strings, ".+(?=:.+:)")

 [1] "01/09/2019 9"  "01/09/2019 9"  "01/09/2019 9"  "01/09/2019 10"
 [5] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
 [9] "01/09/2019 10" "01/09/2019 10"