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,这些变量是观察得分的重新编码值。我有一组变量,它们是将观察到的分数转换为新变量的“关键”,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得分= 1,第二个值对应于观察得分= 2,依此类推。
我需要创建四个与score1,score2,score3和score 4相对应的新变量。 我可以考虑手动进行重新编码,如下所示,但它非常缓慢而且乏味:
people$value1[person_score == 1] <- 10
people$value1[person_score == 2] <- 14
...等等得分1
people$value2[person_score == 1] <- 6
people$value2[person_score == 2] <- 9
...等等得分2
people$value3[person_score == 1] <- 11
people$value3[person_score == 2] <- 13
......等等得分3
people$value4[person_score == 1] <- 43
people$value4[person_score == 2] <- 44
......依此类推得分4
答案 0 :(得分:1)
我只想使用match
从分数data.frame
中找到正确的行...
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
答案 1 :(得分:0)
您可以使用qdap package's lookup
功能,如下所示:
## 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
答案 2 :(得分: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
" )