R 如何将字符时间数据放入因子中

dataframe（Mydata）包含值

Delivery.Time   Customer.Name              Product.Name
11:22:00        Grace Wangen               Insalatone pollo
19:46:47        ystein Skagestad           CLASSIC CHEESEBURGER
08:54:20        Hans Hallseth              Latte
15:25:21        Thomas Stlen               SUSHI MORIAWASE     
14:45:00        Ingvild Berg               SUSHI CHIISAI
12:45:33        Christopher Frenning       Funghi

Mydata["Delivery.Labels"] <- NA   ## Added new columns

deliBreaks <- c("19:00:00","05:00:00","11:00:00","12:00:00","15:00:00")


Mydata$Delivery.Labels <-c("Dinner", "Breakfast", "Brunch", "Lunch", "Snackstime")[ findInterval(chron(times=Mydata$Delivery.Time), chron(times=deliBreaks))]

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下面是一个简单的例子。 首先将数据转换为时间数据。 然后使用cut（）函数将数据划分为时间块

time<- strptime(c("11:22:00", "19:46:47", "08:54:20", "15:25:21", "14:45:00", "12:45:33"),  format= "%T") 


Delivery.Labels <-    cut(time, breaks = strptime(c("00:00:01", "11:00:00","12:00:00","15:00:00", "19:00:00", "23:59:59"),  "%T"), 
        labels = c("Breakfast", "Brunch", "Lunch", "Snackstime", "Dinner"), 
       right = TRUE,
        include.lowest=TRUE
        )

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time你能写出问题并用一些句子解释你想要什么吗？错误消息（'vec'必须按非递减顺序排序）@elin确保你输入的是
deliBreaks
作为时间而不是字符串。已更改但没有正确的结果@艾琳，请帮帮我。非常感谢你的帮助。
time<- strptime(c("11:22:00", "19:46:47", "08:54:20", "15:25:21", "14:45:00", "12:45:33"),  format= "%T") 


Delivery.Labels <-    cut(time, breaks = strptime(c("00:00:01", "11:00:00","12:00:00","15:00:00", "19:00:00", "23:59:59"),  "%T"), 
        labels = c("Breakfast", "Brunch", "Lunch", "Snackstime", "Dinner"), 
       right = TRUE,
        include.lowest=TRUE
        )