R - 创建的函数不能正确输出

Question

我试图找到两个日期之间的差异，然后将该值分组到因子水平中。我以前用其他数值而不是日期做过这个，无法弄清楚我做错了什么。我在创建函数时没有遇到任何错误，但尝试了两种不同的方法来应用它。

我最初是按天计算的，因为稍后我需要一个天的值。以周为单位分组，是为了以后可视化创建一些层次。

#created Lead_time column to calculate how far in advance appointment was booked
#formated in days
df7$Lead_Time <- difftime(df7$Appointment_Date_Time, df7$appt_create_date, units = "days")
#to negate when negatives are created due to the appointment being created after the start time
df7$Lead_Time <- as.integer(df7$Lead_Time)

#group Lead_Time by weeks
group_Lead_Time <- function(Lead_Time){
  if (Lead_Time <= 28){
    return('0-4 Weeks')
  }else if(Lead_Time > 29 & Lead_Time <= 56){
    return('5-8 Weeks')
  }else if (Lead_Time > 57 & Lead_Time <= 84){
    return('8-12 Weeks')
  }else if (Lead_Time > 85 & Lead_Time <= 112){
    return('12-16 Weeks')
  }else if (Lead_Time > 113 & Lead_Time <=140){
    return('16-20 Weeks')  
  }else if (Lead_Time > 141 & Lead_Time <=168){
    return('20-24 Weeks')  
  }else if (Lead_Time > 168){
    return('24+ Weeks')
  }
}
df7$Lead_Time_Grouped <- as.factor(group_Lead_Time(df7$Lead_Time))
df7$Lead_Time_Grouped <- sapply(df7$Lead_Time,group_Lead_Time)

如果有人有更好的方法来处理负值，我也愿意接受。这些是我收到的错误消息：

> df7$Lead_Time_Grouped <- as.factor(group_Lead_Time(df7$Lead_Time))
Warning messages:
1: In if (Lead_Time <= 28) { :
  the condition has length > 1 and only the first element will be used
2: In if (Lead_Time > 29 & Lead_Time <= 56) { :
  the condition has length > 1 and only the first element will be used
3: In if (Lead_Time > 57 & Lead_Time <= 84) { :
  the condition has length > 1 and only the first element will be used
4: In if (Lead_Time > 85 & Lead_Time <= 112) { :
  the condition has length > 1 and only the first element will be used
> df7$Lead_Time_Grouped <- sapply(df7$Lead_Time,group_Lead_Time)
Error in if (Lead_Time <= 28) { : missing value where TRUE/FALSE needed

UPDATE/EDIT：感谢您为我指明了 ifelse 的方向。能够使用以下代码解决我的挑战。

#group Lead_Time by weeks
group_Lead_Time <- function(appt_lead_time){
  ifelse (appt_lead_time <= 28,'0-4 Weeks',
          ifelse (appt_lead_time > 29 & appt_lead_time <= 56, '5-8 Weeks',
                  ifelse (appt_lead_time > 57 & appt_lead_time <= 84, '8-12 Weeks',
                          ifelse (appt_lead_time > 85 & appt_lead_time <= 112, '12-16 Weeks',
                                  ifelse (appt_lead_time > 113 & appt_lead_time <=140, '16-20 Weeks',
                                      ifelse (appt_lead_time > 141 & appt_lead_time <=168, '20-24 Weeks',
                                              '24+ Weeks'))))))
                                                    
  }

df7$appt_lead_time_weeks <- group_Lead_Time(df7$appt_lead_time)

Answer 1

这是一个小玩具示例，可能也会帮助您使用 cut()。这假设您已经将提前期计算为一个整数。

# simulate some lead times between -10 and 200 days
set.seed(1234)
lead_times <- as.integer(round(runif(100, -10, 200), 0))

# convert all negatives to 0 (i.e., no lead time)
# you don't actually have to do this, but thought I would show it
lead_times_adj <- pmax(lead_times, 0)

# convert to weeks
lead_times_weeks <- lead_times_adj / 7

# use cut to break the data into factors by number of weeks
group_lead_time <- cut(lead_times_weeks, breaks = c(-Inf, 4, 8, 12, 16, 20, 24, Inf),
                       labels = c("0-4 Weeks", "4-8 Weeks", "8-12 Weeks",
                                  "12-16 Weeks", "16-20 Weeks", "20-24 Weeks",
                                  "24+ Weeks"))

# create a data frame to show an example result
data.frame(Lead_Time = lead_times,
           Lead_Time_Grouped = group_lead_time)

所以你会得到这样的东西...

    Lead_Time Lead_Time_Grouped
1          14         0-4 Weeks
2         121       16-20 Weeks
3         118       16-20 Weeks
4         121       16-20 Weeks
5         171         24+ Weeks
6         124       16-20 Weeks
7          -8         0-4 Weeks
8          39         4-8 Weeks
9         130       16-20 Weeks
...

Answer 2

在评论的帮助下，我得出了以下解决方案：

#group Lead_Time by weeks
group_Lead_Time <- function(appt_lead_time){
  ifelse (appt_lead_time <= 28,'0-4 Weeks',
  ifelse (appt_lead_time > 29 & appt_lead_time <= 56, '5-8 Weeks',
  ifelse (appt_lead_time > 57 & appt_lead_time <= 84, '8-12 Weeks',
  ifelse (appt_lead_time > 85 & appt_lead_time <= 112, '12-16 Weeks',
  ifelse (appt_lead_time > 113 & appt_lead_time <=140, '16-20 Weeks',
  ifelse (appt_lead_time > 141 & appt_lead_time <=168, '20-24 Weeks',
                                              '24+ Weeks'))))))
  }

df7$appt_lead_time_weeks <- group_Lead_Time(df7$appt_lead_time)

R - 创建的函数不能正确输出

R - created function does not output correctly

r

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if-statement

datediff