将错误结果的数字分箱

Question

我对数值向量进行 bin 后，输出出现问题。

我正在尝试对预先使用 difftime 函数计算的停留时间进行分类。提供整个代码没有意义，因为这只是背景。然而，当我装箱时，我没有得到正确的答案。这是用 los 分配的逗留时间。

dput(los)
c(61.0416666666667, 61.0416666666667, 61.0416666666667, 2, 2, 3, 3)

这是我的休息时间。我在里面使用 na.rm 尝试了几种方法。我通过了 na.rm 和 TRUE，FALSE 并把它从我的休息中拿走了。

breaks <- c(0, 0.8, 0.16, 
           1.0, 1.8, 1.16, 
           2.0, 2.8, 2.16, 
           3.0, 3.8, 3.16, 
           4.0, 4.8, 4.16, 
           5.0, 5.8, 5.16, 
           6.0, 6.8, 6.16, 
           7.0, 14.0, 21.0, 28.0, max(los)) #, , na.rm = FALSE

尽管如此，下一个代码尝试了

dt_los$losbinned <- cut(dt_los$LOS, 
                         breaks = breaks,
                         labels = c("0hrs", "8hrs", "16hrs", "1 d", 
                                 "1 d 8hrs", "1 d 16hrs", "2 d", 
                                 "2 d 8hrs", "2 d 16hrs", "3 d", 
                                 "3 d 8hrs", "3 d 16hrs", "4 d", 
                                 "4 d 8hrs", "4 d 16hrs", "5 d", 
                                 "5 d 8hrs", "5 d 16hrs", "6 d",  
                                 "6 d 8hrs","6 d 16hrs", "7 - 14 d", 
                                 "14 - 21 d", "21 - 28 d", "> 28 d"), 
                         right = FALSE)#

为 'right' 传递了不同的参数给我这个：

当 right = FALSE 时，我没有得到 61.04 分类“> 28 d”的 LOS。 B但是请为其他 2.00 和 3.00 获取正确的垃圾箱。

structure(list(IDcol = 101:107, Admissions = structure(c(1539160200, 
1539160200, 1539160200, 1539154800, 1539154800, 1539154800, 1539154800
), class = c("POSIXct", "POSIXt"), tzone = "Europe/London"), 
    Discharges = structure(c(1544434200, 1544434200, 1544434200, 
    1539327600, 1539327600, 1539414000, 1539414000), class = c("POSIXct", 
    "POSIXt"), tzone = "Europe/London"), Admission_type = c("Elective", 
    "Emergency", "Emergency", "Elective", "Emergency", "Elective", 
    "Emergency"), LOS = c(61.0416666666667, 61.0416666666667, 
    61.0416666666667, 2, 2, 3, 3), Ward_code = c("DSN", "DSN", 
    "DNA", "NAS", "BAS", "BAS", "BAS"), Same_day_discharge = c(FALSE, 
    FALSE, FALSE, FALSE, FALSE, FALSE, FALSE), Spell_type = c("Elective", 
    "Emergency", "Emergency", "Elective", "Emergency", "Elective", 
    "Emergency"), Adm_period = c(TRUE, TRUE, TRUE, TRUE, TRUE, 
    TRUE, TRUE), losbinned = structure(c(NA, NA, NA, 7L, 7L, 
    10L, 10L), .Label = c("0hrs", "8hrs", "16hrs", "1 d", "1 d 8hrs", 
    "1 d 16hrs", "2 d", "2 d 8hrs", "2 d 16hrs", "3 d", "3 d 8hrs", 
    "3 d 16hrs", "4 d", "4 d 8hrs", "4 d 16hrs", "5 d", "5 d 8hrs", 
    "5 d 16hrs", "6 d", "6 d 8hrs", "6 d 16hrs", "7 - 14 d", 
    "14 - 21 d", "21 - 28 d", "> 28 d"), class = "factor")), row.names = c(NA, 
-7L), class = c("tbl_df", "tbl", "data.frame"))

当我通过 right = TRUE 时，61.04 的输出被合并到“>28 d”，这是所需的答案，但是，我没有得到 2.0 和 3.0 的正确容器，它们在 1 d 中被合并2.0 为 16 小时，3 为 2 d 16 小时。同样，这些应分别分为 2 和 3。

 structure(list(IDcol = 101:107, Admissions = structure(c(1539160200, 
1539160200, 1539160200, 1539154800, 1539154800, 1539154800, 1539154800
), class = c("POSIXct", "POSIXt"), tzone = "Europe/London"), 
    Discharges = structure(c(1544434200, 1544434200, 1544434200, 
    1539327600, 1539327600, 1539414000, 1539414000), class = c("POSIXct", 
    "POSIXt"), tzone = "Europe/London"), Admission_type = c("Elective", 
    "Emergency", "Emergency", "Elective", "Emergency", "Elective", 
    "Emergency"), LOS = c(61.0416666666667, 61.0416666666667, 
    61.0416666666667, 2, 2, 3, 3), Ward_code = c("DSN", "DSN", 
    "DNA", "NAS", "BAS", "BAS", "BAS"), Same_day_discharge = c(FALSE, 
    FALSE, FALSE, FALSE, FALSE, FALSE, FALSE), Spell_type = c("Elective", 
    "Emergency", "Emergency", "Elective", "Emergency", "Elective", 
    "Emergency"), Adm_period = c(TRUE, TRUE, TRUE, TRUE, TRUE, 
    TRUE, TRUE), losbinned = structure(c(25L, 25L, 25L, 6L, 6L, 
    9L, 9L), .Label = c("0hrs", "8hrs", "16hrs", "1 d", "1 d 8hrs", 
    "1 d 16hrs", "2 d", "2 d 8hrs", "2 d 16hrs", "3 d", "3 d 8hrs", 
    "3 d 16hrs", "4 d", "4 d 8hrs", "4 d 16hrs", "5 d", "5 d 8hrs", 
    "5 d 16hrs", "6 d", "6 d 8hrs", "6 d 16hrs", "7 - 14 d", 
    "14 - 21 d", "21 - 28 d", "> 28 d"), class = "factor")), row.names = c(NA, 
-7L), class = c("tbl_df", "tbl", "data.frame"))

实际和预期结果应该是为我的逗留时间分配的正确垃圾箱。对于 61.04 -> ">28d"，对于 2 -> "2 d"，对于 3 -> "3 d"。

如果这可以用 tidyverse 完成，那就太棒了。但尊重我分配的垃圾箱。但是，我知道这还没有完成。因此，可以接受我提出的更正代码，但已更正。

Answer 1

cut 函数的 bin 是独占的。

来自 cut 函数的帮助：因子水平标签构造为“(b1, b2]”、“(b2, b3]”等，对于 right = TRUE 和“[b1, b2)"

为了包含最低值（或本例中的最高值），需要 include.lowest=TRUE 选项。这将使第一个 bin 独占独占，“[b1, b2]”。

尝试：

labels<-c("0hrs", "8hrs", "16hrs", "1 d", 
          "1 d 8hrs", "1 d 16hrs", "2 d", 
          "2 d 8hrs", "2 d 16hrs", "3 d", 
          "3 d 8hrs", "3 d 16hrs", "4 d", 
          "4 d 8hrs", "4 d 16hrs", "5 d", 
          "5 d 8hrs", "5 d 16hrs", "6 d",  
          "6 d 8hrs","6 d 16hrs", "7 - 14 d", 
          "14 - 21 d", "21 - 28 d", "> 28 d")

dt_los$losbinned <- cut(los, breaks=breaks, labels=labels, right=FALSE, include.lowest = TRUE)

将错误结果的数字分箱

binning the numbers with wrong outcome

r

cut

binning