1

我有几个变量，

qual_cols <- c("ExterQual", "ExterCond", "BsmtQual", "BsmtCond", "HeatingQC", "KitchenQual", "FireplaceQu", "GarageQual", "GarageCond", "PoolQC")

每列由相同的五个 chr 值组成：

grades <- c( "Po", "Fa", "TA", "Gd", "Ex")

我想数字化

"Po" = 1; "Fa" = 2; "TA" = 4  "Gd" = 6 "Ex" = 11

为了乘以创建新变量，如下所示：

combi$GarageGrade <- combi$GarageQual * combi$GarageCond
combi$ExterGrade <- combi$ExterQual * combi$ExterCond
combi$KitchenScore <- combi$KitchenAbvGr * combi$KitchenQual

有哪些方法可以实现这一目标？

注意：我是编程的初学者，所以夸大其词是值得赞赏的。

例子：

head(combi$ExterQual) # (Gd, TA, Gd, TA, Gd, Ta)

价值等价物将是 (6, 4, 6, 4, 6, 4)

head(combi$ExterCond) # (TA, TA, TA, TA, TA, TA)

价值等价物将是 (4, 4, 4, 4, 4, 4)

combi$ExterGrade <- combi$ExterQual * combi$ExterCond
head(combi$ExterGrade) # expected output: (24, 16, 24, 16, 24, 16)
4

2 回答 2

0

我们可以使用命名向量将值更改为数字

newdata <- combi[qual_cols]
newdata[] <- lapply(combi[qual_cols], function(x) 
         setNames(c(1, 2, 4, 6, 11), grades)[x])
nm1 <- grep("(Cond|Qual)$", names(newdata), value = TRUE)
nm2 <- sub("[A-Z][a-z]+$", "", nm1)
nm3 <- paste0(unique(nm2), 'Grade')
newdata[nm3] <- lapply(split.default(newdata[nm1], nm2), function(x) Reduce(`*`, x))

数据

set.seed(24)
combi <- as.data.frame(matrix(sample(grades, 10 * 5, replace = TRUE), 
   ncol = 10, dimnames = list(NULL, qual_cols)), stringsAsFactors = FALSE)
于 2018-12-01T18:10:30.980 回答
0

如果变量名不一致（即不仅仅是 Qual 和 Cond），这是一种更灵活的方法：

用 Po、Fa、TA、Gd、Ex 值识别变量

qual_cols <- c("ExterQual", "ExterCond", "BsmtQual", "BsmtCond", "HeatingQC", "KitchenQual", "FireplaceQu", "GarageQual", "GarageCond", "PoolQC")

用数值替换等级类别。

numeric_quals = sapply(combi[qual_cols], function(x) ifelse(x == 'Po', 1, 
                                        ifelse(x =='Fa', 2,
                                        ifelse(x =='TA', 4,
                                        ifelse(x == 'Gd', 6,
                                        ifelse(x == 'Ex', 11, 0))))) )

替换（更改）数据框中的变量。

combi = combi %>% select(-qual_cols) %>% cbind(numeric_quals)

执行乘法

房子的整体质量

combi$OverallGrade <- combi$OverallQual * combi$OverallCond
head(combi$OverallGrade)

泳池总分

combi$PoolScore <- combi$PoolArea * combi$PoolQC

车库总尺寸

combi$AllGarage <- combi$GarageCars * combi$GarageArea

浴室总数

combi$TotalBath <- combi$BsmtFullBath + (0.5 * combi$BsmtHalfBath) + 
  combi$FullBath + (0.5 * combi$HalfBath)

房屋竣工总面积（包括地下室）

combi$AllSF <- combi$GrLivArea + combi$TotalBsmtSF

门廊总 SF

combi$AllPorchSF <- combi$OpenPorchSF + combi$EnclosedPorch + 
  combi$X3SsnPorch + combi$ScreenPorch

组合添加

combi$Additions <- combi$YearRemodAdd + as.numeric(combi$GarageYrBlt)
于 2018-12-05T20:24:20.157 回答