1

我有一个数据集，其中包含一组人的观察分数，如下所示：

person_id <- c(1:50)
person_score <- rep(1:10,5)
people <- data.frame(person_id, person_score)

我需要创建一组新变量，这些变量是观察分数的重新编码值。我有一组变量，它们是将观察到的分数转换为新变量的“键”，如下所示：

observed <- c(1,2,3,4,5,6,7,8,9,10)
score1 <- c(10,14,17,18,20,21,22,26,28,31)
score2 <- c(6,9,11,14,17,18,20,24,25,26)
score3 <- c(11,13,15,17,19,21,23,25,27,29)
score4 <- c(43,44,45,46,47,48,49,50,51,52)
scores <- data.frame(observed,score1,score2, score3, score4)

...其中第一个值对应于观察到的分数 = 1，第二个值对应于观察到的分数 = 2，依此类推。

我需要创建四个新变量，分别对应 score1、score2、score3 和 score 4。我可以考虑手动重新编码，如下所示，但它非常缓慢且繁琐：

people$value1[person_score == 1] <- 10
people$value1[person_score == 2] <- 14

...等等 score1 

people$value2[person_score == 1] <- 6
people$value2[person_score == 2] <- 9

...等等 score2

people$value3[person_score == 1] <- 11
people$value3[person_score == 2] <- 13

...等等 score3

people$value4[person_score == 1] <- 43
people$value4[person_score == 2] <- 44

...等等 score4

4

3 回答 3

1

我只是用来match从分数中找到正确的行data.frame......

idx <- match( people$person_score , scores$observed )

people_new <- cbind( people , scores[ idx , -1 ] )

head(people_new)
#  person_id person_score score1 score2 score3 score4
#1         1            1     10      6     11     43
#2         2            2     14      9     13     44
#3         3            3     17     11     15     45
#4         4            4     18     14     17     46
#5         5            5     20     17     19     47
#6         6            6     21     18     21     48
于 2013-07-15T20:29:16.090 回答
0

您可以按如下方式使用qdap 包的功能： lookup

## person_id <- c(1:50)
## person_score <- rep(1:10,5)
## people <- data.frame(person_id, person_score)
## 
## observed <- c(1,2,3,4,5,6,7,8,9,10)
## score1 <- c(10,14,17,18,20,21,22,26,28,31)
## score2 <- c(6,9,11,14,17,18,20,24,25,26)
## score3 <- c(11,13,15,17,19,21,23,25,27,29)
## score4 <- c(43,44,45,46,47,48,49,50,51,52)
## scores <- data.frame(observed,score1,score2, score3, score4)

library(qdap)
people[, 3:6] <- lapply(scores[, -1], function(x) lookup(people$person_score, scores[, 1], x))

people
##    person_id person_score score1 score2 score3 score4
## 1          1            1     10      6     11     43
## 2          2            2     14      9     13     44
## 3          3            3     17     11     15     45
## 4          4            4     18     14     17     46
## 5          5            5     20     17     19     47
## 6          6            6     21     18     21     48
## 7          7            7     22     20     23     49
.
.
.
## 50        50           10     31     26     29     52
于 2013-07-15T20:26:35.790 回答
0

它只是两个 data.frames 的连接：您可以使用merge

merge( people, scores, by.x = "person_score", by.y = "observed", all.x = TRUE )

或sqldf。

library(sqldf)
sqldf( "
  SELECT    *
  FROM      people
  LEFT JOIN scores
  ON        people.person_score = scores.observed
" )
于 2013-07-15T20:39:55.630 回答