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我正在尝试从 SPSS 转换为 R；我想将我的清理语法翻译为学习 R 的好方法，但我在 SPSS 中的简单命令方面遇到了很多麻烦。

以下代码在 SPSS 中有效：

IF  (n_race = 1 & (n_ethnicity = 7 | n_ethnicity = 8 | n_ethnicity = 99)) race_eth=1.
EXECUTE.
IF  (n_ethnicity = 3 & (n_race = 1 | n_race = 99)) race_eth=2.
EXECUTE.
IF  (n_race = 2) race_eth=3.
EXECUTE.
IF  (n_race = 5 & n_ethnicity NE 9) race_eth=4.
EXECUTE.
IF  ((n_ethnicity = 2 | n_ethnicity = 4 | n_ethnicity = 5) & (n_race = 1 |   n_race = 99)) race_eth=4.
EXECUTE.
IF  (n_ethnicity = 9 & (n_race = 1 | n_race = 5 | n_race = 99)) race_eth=5.
EXECUTE.
IF  (n_race = 4 | (n_ethnicity = 6 & (n_race = 1 | n_race = 99))) race_eth=6.
EXECUTE.
IF  (n_ethnicity = 1 & (n_race = 1 | n_race = 99)) race_eth=6.
EXECUTE.
IF (n_race=99 AND (n_ethnicity=7 OR n_ethnicity=99 OR n_ethnicity=8))    race_eth=99.
EXECUTE.

我认为这段代码在 R 中会做同样的事情，但是单元频率不一样（注意我在 R 中将 n_ethnicity 重命名为 n_eth）。我将不胜感激任何帮助！谢谢！

OMSOct14_Mar16$race_eth[n_race == 1 & (n_eth == 7 | n_eth == 8 | n_eth == 99)] <- 1
OMSOct14_Mar16$race_eth[n_eth == 3 & (n_race == 1 | n_race == 99)] <- 2
OMSOct14_Mar16$race_eth[n_race == 2] <- 3
OMSOct14_Mar16$race_eth[n_race == 5 & n_eth != 9] <- 4
OMSOct14_Mar16$race_eth[(n_eth == 2 | n_eth==4 | n_eth==5) & (n_race == 1 | n_race == 99)] <- 4
OMSOct14_Mar16$race_eth[n_eth == 9 & (n_race == 1 | n_race == 5 | n_race==99)] <- 5
OMSOct14_Mar16$race_eth[n_race == 4 | (n_eth == 6 & (n_race==1 | n_race == 99))] <- 6
OMSOct14_Mar16$race_eth[n_eth == 1 & (n_race == 1 | n_race == 99)] <- 6
OMSOct14_Mar16$race_eth[n_race == 99 & (n_eth == 7 | n_eth == 8 | n_eth==99)] <- 99
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1 回答 1

1

with你可以使用and来做这样的事情ifelse：

set.seed(99)
n_race<-c(1,99,2,5,4)
n_eth<-c(7,8,99,3,9,2,4,5,6,1)
OMSOct14_Mar16<-data.frame(n_race = sample(n_race, 20, replace = T),
                           n_eth = sample(n_eth, 20, replace = T))

OMSOct14_Mar16$race_eth<-
  with(OMSOct14_Mar16,ifelse(n_eth == 1 & n_eth %in% c(7,8,99), 1,
                             ifelse(n_eth == 3 & n_race %in% c(1,99), 2,
                             ifelse(n_race == 2, 3,
                             ifelse(n_race==5 & n_eth !=9, 4,
                             ifelse(n_eth %in% c(2,4,5) & n_race %in% c(1,99), 4,
                             ifelse(n_eth == 9 & n_race %in% c(1,5,99), 5,
                             ifelse(n_race ==4 | n_eth == 6 & (n_race %in% c(1,99)), 6,
                             ifelse(n_eth == 1 & n_race %in% c(1,99), 6,
                             ifelse(n_race == 99 & (n_eth %in% c(7,8)), 99, NA))))))))))

head(OMSOct14_Mar16,5)
n_race n_eth race_eth
2    99        3
1     7       NA
5     6        4
4     2        6
2     5        3
于 2016-06-29T23:21:42.217 回答