示例函数在 ifelse 中重复相同的值

Question

我有以下数据框：

structure(list(Store = c("vpm", "vpm", 
"vpm"), Date = structure(c(18042, 18042, 18042), class = "Date"), 
    UniqueImageId = c("vp3_523", "vp3_668", "vp3_523"), EntryTime = structure(c(1558835514, 
    1558834942, 1558835523), class = c("POSIXct", "POSIXt")), 
    ExitTime = structure(c(1558838793, 1558838793, 1558839824
    ), class = c("POSIXct", "POSIXt")), Duration = c(3279, 3851, 
    4301), Age = c(35L, 35L, 35L), EntryPoint = c("Entry2Side", 
    "Entry2Side", "Entry2Side"), ExitPoint = c("Exit2Side", "Exit2Side", 
    "Exit2Side"), AgeNew = c("15_20", "25_32", "15_20"), GenderNew = c("Female", 
    "Male", "Female")), row.names = 4:6, class = c("data.table", 
"data.frame"))

我正在尝试为列 AgeNew 填充一个随机数，并且我正在使用带有 ifelse 条件的 sample 函数。

我尝试了以下方法

d$AgeNew <- ifelse(d$AgeNew == "0_2",   sample(0:2,  1,replace = TRUE), 
            ifelse(d$AgeNew == "15_20", sample(15:20,1,replace = TRUE), 
            ifelse(d$AgeNew == "25_32", sample(25:36,1,replace = TRUE), 
            ifelse(d$AgeNew == "38_43", sample(36:43,1,replace = TRUE), 
            ifelse(d$AgeNew == "4_6",   sample(4:6,  1,replace = TRUE), 
            ifelse(d$AgeNew == "48_53", sample(48:53,1,replace = TRUE), 
            ifelse(d$AgeNew == "60_Inf",sample(60:65,1,replace = TRUE), 
                                        sample(8:13, 1,replace = TRUE))))))))

但是我得到了重复的相同值。例如，对于 0_2 年龄组，我只填充了 2 个。我尝试使用 set.seed

set.seed(123)

然后运行 ifelse 仍然重复相同的值。

Answer 1

一个更简单的选择是将 _ 替换为 :，然后 evaluate 和 sample 该范围内的元素

library(data.table)
d[, AgeNew := sapply(sub("_", ":", sub('Inf', '65', AgeNew)),
           function(x) sample(eval(parse(text = x)), 1))]
d[is.na(AgeNew), AgeNew := sample(8:13, 1)]
d
#  Store       Date UniqueImageId           EntryTime            ExitTime Duration Age EntryPoint ExitPoint AgeNew GenderNew
#1:   vpm 2019-05-26       vp3_523 2019-05-25 21:51:54 2019-05-25 22:46:33     3279  35 Entry2Side Exit2Side     15    Female
#2:   vpm 2019-05-26       vp3_668 2019-05-25 21:42:22 2019-05-25 22:46:33     3851  35 Entry2Side Exit2Side     30      Male
#3:   vpm 2019-05-26       vp3_523 2019-05-25 21:52:03 2019-05-25 23:03:44     4301  35 Entry2Side Exit2Side     17    Female

或 tidyverse

的另一个选项

library(tidyverse)
d %>% 
   mutate(AgeNew = str_replace(AgeNew, "Inf", "65")) %>%
   separate(AgeNew, into = c('start', 'end'), convert = TRUE) %>% 
   mutate(AgNew = map2_int(start, end, ~ sample(.x:.y, 1)))

或者另一种选择是按 _ 拆分，然后采样

d[, AgeNew := unlist(lapply(strsplit(sub('Inf', '65', AgeNew),  "_"), function(x)
            sample(as.numeric(x[1]):as.numeric(x[2]), 1)))]

请注意，我们不需要任何嵌套 ifelse 来更改此处。没有任何 ifelse

更容易做到这一点

注意 2：OP 以 data.table 为例，这里我们展示了 data.table 方法

注意3：使用嵌套ifelse是非常低效的

注意 4：基于 strsplit 的方法首次发布于此

关于为什么ifelse的工作方式不同，在?ifelse

的文档中已经提到了

If yes or no are too short, their elements are recycled. yes will be evaluated if and only if any element of test is true, and analogously for no.

Answer 2

这已经在某处讨论过（目前找不到来源）。它之所以这样，是因为 ifelse 在一个条件下只运行一次，因此该值被回收。考虑这个例子，

x <- c(1, 2, 1, 2, 1, 2)

ifelse(x == 1, sample(1:10, 1), sample(20:30, 1))
#[1]  1 26  1 26  1 26
ifelse(x == 1, sample(1:10, 1), sample(20:30, 1))
#[1] 10 28 10 28 10 28
ifelse(x == 1, sample(1:10, 1), sample(20:30, 1))
#[1]  9 24  9 24  9 24

正如我们所见，它给出了相同的数字，该数字在两种情况下都被回收。为避免这种情况，我们需要将 size of sample 指定为 ifelse

中 test 条件的长度

ifelse(x == 1, sample(1:10, length(x)), sample(20:30, length(x)))
#[1]  7 23  1 26 10 24
ifelse(x == 1, sample(1:10, length(x)), sample(20:30, length(x)))
#[1]  3 23  5 26  6 22 
ifelse(x == 1, sample(1:10, length(x)), sample(20:30, length(x)))
#[1]  2 30  9 27  1 29

Answer 3

您将需要处理 Inf。根据您的示例，我假设您要在出现 Inf 时添加 +5。所以基于这个假设，我们可以做，

sapply(strsplit(d$AgeNew, '_'), function(i){
                  sample(i[1]:replace(i[2], i[2] == 'Inf', as.numeric(i[1]) + 5), 1)
                  })

#[1] 60 32 19

注意： 我将 AgeNew 的第一个条目更改为 60_Inf 以测试

示例函数在 ifelse 中重复相同的值

Sample function repeats same value within ifelse

if-statement

r

sample