从单独的代码表重新编码答案

【问题标题】：Recoding from separate code table从单独的代码表重新编码
【发布时间】：2016-11-02 17:26:28
【问题描述】：

我有一个如下数据集：

dat1 <- read.table(header=TRUE, text="
                   ID  Age  Align  Weat
                   8645    15-24  A  1
                   6228    15-24  B  1
                   5830    15-24  A  3
                   1844    25-34  B  1
                   4461    35-44  B  2
                   2119    35-44  C  2
                   2115    45-54  A  1
                   ")
dat1
    ID   Age Align Weat
1 8645 15-24     A    1
2 6228 15-24     B    1
3 5830 15-24     A    3
4 1844 25-34     B    1
5 4461 35-44     B    2
6 2119 35-44     C    2
7 2115 45-54     A    1

Age、Align 和 Weat 列的属性在代码数据框中描述：

dat2 <- read.table(header=TRUE, text="
                   Code  Desc  Column
                   15-24    Young  Age
                   25-34    Young  Age
                   35-44    Middle  Age
                   45-54    Middle  Age
                   A    Straight  Align
                   B    Curve  Align
                   C    Hill  Align
                   1    Clear  Weat
                   2    Cloudy  Weat
                   3    Rain  Weat
                   ")
dat2
    Code     Desc Column
1  15-24    Young    Age
2  25-34    Young    Age
3  35-44   Middle    Age
4  45-54   Middle    Age
5      A Straight  Align
6      B    Curve  Align
7      C     Hill  Align
8      1    Clear   Weat
9      2   Cloudy   Weat
10     3     Rain   Weat

我想匹配代码数据框以获取我的数据集，如下所示：

    ID    Age    Align   Weat
1 8645  Young Straight  Clear
2 6228  Young    Curve  Clear
3 5830  Young Straight   Rain
4 1844  Young    Curve  Clear
5 4461 Middle    Curve Cloudy
6 2119 Middle     Hill Cloudy
7 2115 Middle Straight  Clear

我目前正在使用以下代码来完成我的任务，这对于具有 500 列和这些列的代码表的大型数据集来说效率不高。

age <- subset(dat2, Column=="Age")
age
   Code   Desc Column
1 15-24  Young    Age
2 25-34  Young    Age
3 35-44 Middle    Age
4 45-54 Middle    Age

align <- subset(dat2, Column=="Align")
align
  Code     Desc Column
5    A Straight  Align
6    B    Curve  Align
7    C     Hill  Align

weat <- subset(dat2, Column=="Weat")
weat
   Code   Desc Column
8     1  Clear   Weat
9     2 Cloudy   Weat
10    3   Rain   Weat

dat1$Age <- age$Desc[match(dat1$Age, age$Code)]
dat1$Align <- align$Desc[match(dat1$Align, align$Code)]
dat1$Weat <- weat$Desc[match(dat1$Weat, weat$Code)]

dat1
    ID    Age    Align   Weat
1 8645  Young Straight  Clear
2 6228  Young    Curve  Clear
3 5830  Young Straight   Rain
4 1844  Young    Curve  Clear
5 4461 Middle    Curve Cloudy
6 2119 Middle     Hill Cloudy
7 2115 Middle Straight  Clear

【问题讨论】：

?plyr::melt 会指向你想去的地方
等待.. 可能是 reshape2 而不是 plyr

标签： r match recode

【解决方案1】：

您可以在dat1 中的变量上使用for 循环：

# 'intersect' is needed to recode only those columns which have description
for (each_column in intersect(colnames(dat1), dat2$Column)){
    curr_dict = dat2$Column %in% each_column
    code = dat2$Code[curr_dict]
    descr = dat2$Desc[curr_dict]
    dat1[[each_column]] = descr[match(dat1[[each_column]], code)]
}

【讨论】：

【解决方案2】：

尝试一个简单的 for 循环：

varnames <- unique(dat2$Column)
dat3 <- dat1
for (i in varnames)
{   startvars <- names(dat3)[!names(dat3) %in% i]
    dat3 <- merge(dat3, subset(dat2, Column==i),
                        by.x=i, by.y="Code")[,c(startvars, "Desc")]
    colnames(dat3)[names(dat3) %in% "Desc"] <- i 
}

结果：

    ID    Age    Align   Weat
1 8645  Young Straight  Clear
2 2115 Middle Straight  Clear
3 6228  Young    Curve  Clear
4 1844  Young    Curve  Clear
5 4461 Middle    Curve Cloudy
6 2119 Middle     Hill Cloudy
7 5830  Young Straight   Rain

这显然不是超高效的，带有一些 dcast 的 data.table 解决方案可能是合适的，但我将把它留给其他人思考。

PS：必须通过将stringsAsFactors= F, colClasses= rep("character",4)) 添加到read.table 来稍微重新格式化第一个数据集

【讨论】：