Question

我有这个数据集：

Pars <-structure(list(Pars.PeopleSoftID = c(11, 18, 18, 22, 39, 181), 
    Pars..Central.1. = c(0, 1, 1, 1, 1, 1), Pars..Central.2. = c(0, 
    1, 1, 1, 1, 1), Pars.External = c(3, 2, 2, 1, 2, 5), Pars.Minimum = c(15, 
    2, 1.71, 1, 1, 4.39), Pars.Maximum = c(15, 2.76, 2, 1, 1, 
    5.48)), .Names = c("Pars.PeopleSoftID", "Pars..Central.1.", 
"Pars..Central.2.", "Pars.External", "Pars.Minimum", "Pars.Maximum"
), row.names = c(NA, 6L), class = "data.frame")

我尝试添加一个列，根据Central 1，Central 2和External列中的内容计算一个数字，并将该数字乘以Pars $ Minimum中的值。

逻辑是：

如果Central 1 = 0，Central 2 = 0，External = 0 THEN＆lt; - 0 * Pars $ Minimum

如果Central 1 = 1，Central 2 = 0，External = 0 THEN＆lt; - 1 * Pars $ Minimum

如果Central 1 = 1，Central 2 = 1，External = 0 THEN＆lt; - 0.7 * Pars $ Minimum

如果Central 1 = 1，Central 2 = 1，External＆gt; 0 THEN＆lt; - 0.7 * Pars $ Minimum

如果Central 1 = 1，Central 2 = 0，External＆gt; 0 THEN＆lt; - 0.8 * Pars $ Minimum

我刚试过这个，但似乎只是为了一切返回NA的值：

 Pars$Cen1 <- with(Pars, if(`Central 1` == 0 , 0, ifelse(`Central 1` == 1 & `Central 2` == 0 & External == 0, 1, ifelse(`Central 1` == 1 & `Central 2` == 1 & External == 0, .7, ifelse(`Central 1` == 1 & `Central 2` == 1 & External > 0, .7, ifelse(`Central 1` == 1 & `Central 2` == 0 & External > 0, .8)

Answer 1

我也重命名了一些列，不知道为什么需要调用每个列名称＆＃34; Pars。＆＃34;如果该表已被称为Pars，但这可以在这里做你想要的，应该给你一个格式使用。

Pars <-structure(list(Pars.PeopleSoftID = c(11, 18, 18, 22, 39, 181), 
    Pars..Central.1. = c(0, 1, 1, 1, 1, 1), Pars..Central.2. = c(0, 
    1, 1, 1, 1, 1), Pars.External = c(3, 2, 2, 1, 2, 5), Pars.Minimum = c(15, 
    2, 1.71, 1, 1, 4.39), Pars.Maximum = c(15, 2.76, 2, 1, 1, 
    5.48)), .Names = c("PeopleSoftID", "Central.1", 
"Central.2", "External", "Minimum", "Maximum"
), row.names = c(NA, 6L), class = "data.frame")

Pars$Cen1 <- with(Pars,ifelse(Central.1 == 0 & Central.2 == 0 & External == 0,0*Minimum,
                       ifelse(Central.1 == 1 & Central.2 == 0 & External == 0,1*Minimum, 
                       ifelse(Central.1 == 1 & Central.2 == 1 & External >= 0,0.7*Minimum,
                       ifelse(Central.1 == 1 & Central.2 == 0 & External > 0,.8*Minimum,0)))))

Answer 2

另一种可能性是制作一个值映射并查找粘贴在那里的条件：

  valuemap = c("000" = 0, "001"= 0, "100"= 1, "110"= 0.7, "111" = 0.7, "101" = 0.8)

  Pars$Cen1 <- Pars$Pars.Minimum * 
    valuemap[paste0(Pars$Pars..Central.1.,Pars$Pars..Central.2., 
                    1 * (Pars$Pars.External > 0))]

Answer 3

library(dplyr)
mutate(Pars, Cen1 = case_when(
    `Central 1` == 0 & `Central 2` == 0 & External == 0 ~ 0 * Minimum,
    `Central 1` == 1 & `Central 2` == 0 & External == 0 ~ 1 * Minimum,
    `Central 1` == 1 & `Central 2` == 1 & External == 0 ~ 0.7 * Minimum,
    `Central 1` == 1 & `Central 2` == 1 & External > 0 ~ 0.7 * Minimum,
    `Central 1` == 1 & `Central 2` == 0 & External > 0 ~ 0.8 * Minimum))

Answer 4

你的最终ifelse陈述只有两个参数，所以如果你试图分配那个＆＃39; no＆＃39;声明，您将收到错误。

这似乎对我有用：

Pars$Cen1 <- ifelse(Pars$Pars..Central.1. == 0, 0, 
                ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 0 & Pars$Pars.External == 0, 1, 
                       ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 1 & Pars$Pars.External == 0, .7, 
                              ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 1 & Pars$Pars.External > 0, .7, 
                                     ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 0 & Pars$Pars.External > 0, .8, "Missing")))))

如果具有四个标准的Else声明返回NA

4 个答案: