Question

我对将数据分组到特定类别有疑问。

通常情况下，如果我有一个因子变量，我会执行类似下面的操作来将数据转储/重新编码为首选模式：

educ = NA
educ[educ2 %in% levels(educ2)[c(5,8)]] <- "HS or Some College"
educ[educ2 %in% levels(educ2)[2:3]] <- "College Degree"
educ[educ2 %in% levels(educ2)[c(4,6)]] <- "Advanced Degree" 
educ[educ2 %in% levels(educ2)[c(1,7,9)]] <- NA
educ = factor(educ)

然而，我正在努力重新组合一个因子变量TIME，它有10,000 +级别。数据结构如下：

> levels(wj$time)
    [1] "0:00:05"  "0:00:07"  "0:00:08"  "0:00:10"  "0:00:13"  "0:00:15"  "0:00:18"  "0:00:23"  "0:00:31"  "0:00:34"  "0:00:36" 
   [12] "0:00:39"  "0:00:41"  "0:00:47"  "0:00:48"  "0:00:54"  "0:00:55"  "0:00:56"  "0:00:59"  "0:01:01"  "0:01:02"  "0:01:03" 
   [23] "0:01:13"  "0:01:17"  "0:01:31"  "0:01:33"  "0:01:41"  "0:01:44"  "0:01:48"  "0:01:50"  "0:01:52"  "0:01:53"  "0:01:55" 
   [34] "0:02:08"  "0:02:12"  "0:02:13"  "0:02:21"  "0:02:26"  "0:02:27"  "0:02:30"  "0:02:32"  "0:02:33"  "0:02:36"  "0:02:37" 
   [45] "0:02:38"  "0:02:43"  "0:02:45"  "0:02:53"  "0:02:56"  "0:03:07"  "0:03:15"  "0:03:19"  "0:03:21"  "0:03:22"  "0:03:24" 
   [56] "0:03:30"  "0:03:36"  "0:03:39"  "0:03:41"  "0:03:49"  "0:03:56"  "0:03:59"  "0:04:02"  "0:04:04"  "0:04:07"  "0:04:10" 
   [67] "0:04:11"  "0:04:12"  "0:04:14"  "0:04:16"  "0:04:17"  "0:04:19"  "0:04:22"  "0:04:27"  "0:04:28"  "0:04:30"  "0:04:37" 
   [78] "0:04:39"  "0:04:41"  "0:04:49"  "0:04:51"  "0:04:52"  "0:04:53"  "0:04:54"  "0:05:05"  "0:05:06"  "0:05:20"  "0:05:22"

当有很多因子水平时，我只是不确定如何快速将数据存入特定括号。我想将它们分为0:12:00 to 0:05:00和0:05:01 to 0:10:00等等。有这么多因素水平，我只是在如何确定何时开始和结束分组方面有点迷失。有人可以提供任何帮助吗？拥有10,000多个桶，这成为我传统做事的一个问题。

谢谢！

Answer 1

您可以将时间戳拆分为其组件：这些桶很容易计算。

# Sample data
n <- 10
d <- data.frame(
  time = paste( 
    sample(0:23, n, replace=TRUE), 
    sample(0:59, n, replace=TRUE), 
    sample(0:59, n, replace=TRUE), 
    sep=":" 
  ),
  value = rnorm(n)
)

# Split the "time" column into its components
d$time <- as.character( d$time )
times <- strsplit( d$time, ":" )
times <- lapply( times, as.numeric )
times <- do.call(rbind, times)
colnames(times) <- c("hour", "minute", "second")
d <- cbind(times, d)

# Build the buckets
d$bucket <- paste(
  sprintf( "%02d:%02d:00", d$hour, floor( d$minute / 5 ) * 5 ),
  sprintf( "%02d:%02d:59", d$hour, floor( d$minute / 5 ) * 5 + 4 ),
  sep=" to "
)

Answer 2

您遇到的问题是您有一个有效的连续变量，您以特定字符格式表示该变量存储为因子。这里的因素并不合适，因为这些级别只表示数据中出现的值，而不是预定义的一组可能值。它是字符向量的事实是因为它表示格式化数据类型的特定约定，即时间。我猜想它是几小时：分钟：秒，但考虑到你的例子，它可能是几天（？）：小时：分钟。如果是小时：分钟：秒，那么最好将这些时间表示为times包中的chron对象。如果这样做，那么问题就变成了如何将连续变量分类为离散组。这是通过cut函数完成的。

Answer 3

结合@Brian Diggs＆amp;的答案/代码@Vincent Zoonekynd，我会推荐一些功能：

?strptime
?POSIXlt
?cut.POSIXt


#create factorized time vector within data frame
n <- 10
d <- data.frame(
  time =  as.factor(paste( 
    sample(0:23, n, replace=TRUE), 
    sample(0:59, n, replace=TRUE), 
    sample(0:59, n, replace=TRUE), 
    sep=":" 
  )),
  value = rnorm(n)
)

#convert to time format, then apply cuts per hour
(d$time<- cut.POSIXt(strptime(d$time, format="%H:%M:%S"), breaks="hour"))

如果你不想每小时休息，你可以使用“day”或其他东西。您也可以查看我们的this链接，查看您的问题的答案，我通过查找“将字符串转换为时间”找到了该答案。

HTH。

将值分组到括号中

3 个答案: