Question

我有一个包含10列的数据集，其中有10列有兴趣创建新的指标功能。功能是＆＃34; pT＆＃34;，＆＃34; pN＆＃34;，＆amp; ＆＃34; M＆＃34;他们都采取不同的价值观。关闭这3个特征所采用的所有值，需要在新变量中捕获9个唯一组合。

   PATHOT PATHON PATHOM
1       pT2    pN1     M0
4       pT1    pN1     M0
13      pT3    pN1     M0
161     pT1   *pN2     M0
391     pT1    pN1    *M1
810   *pTIS    pN1     M0
948     pT3   *pN2     M0
1043    pT2    pN1    *M1
1067   *pT4    pN1     M0

例如，新变量将具有值＆＃34; 1＆＃34;当PATHOT = pT2时，PATHON = pN1＆amp; PATHOM = M0，依此类推至值9.我已完成任务但在花费了近20行代码涉及所有独特组合的矢量化操作之后。

diag3_bs$sfd[diag3_bs$pathot=="pT2" & diag3_bs$pathon=="pN1" & 
               diag3_bs$pathom=="M0"] <- 1
diag3_bs$sfd[diag3_bs$pathot=="pT1" & diag3_bs$pathon=="pN1" & 
               diag3_bs$pathom=="M0"] <- 2
diag3_bs$sfd[diag3_bs$pathot=="pT3" & diag3_bs$pathon=="pN1" & 
               diag3_bs$pathom=="M0"] <- 3... so on upto 9.

我想问一下是否有更好的更自动化的方法来获得相同的结果？

dput(data.frame)在下面给出

 structure(list(F_STATUS = structure(c(1L, 1L, 1L, 1L, 1L, 1L,  1L, 1L,
 1L, 1L), .Label = "Y", class = "factor"), EVENT_ID = structure(c(1L, 
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "BASELINE", class =
 "factor"), 
     PAG_NAME = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
     1L), .Label = "BR2", class = "factor"), PTSIZE = c(3, 4, 
     2.7, 2, 0.9, 3, 3, 0.9, 3, 4.5), PTSIZE_U = structure(c(1L, 
     1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "CM", class = "factor"), 
     PT_SYM = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
     1L), .Label = c("", "-", "<", ">"), class = "factor"), PATHOT = structure(c(4L, 
     4L, 4L, 3L, 3L, 4L, 4L, 3L, 4L, 4L), .Label = c("*pT4", "*pTIS", 
     "pT1", "pT2", "pT3"), class = "factor"), PATHON = structure(c(2L, 
     2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("*pN2", "pN1"
     ), class = "factor"), PATHOM = structure(c(2L, 2L, 2L, 2L, 
     2L, 2L, 2L, 2L, 2L, 2L), .Label = c("*M1", "M0"), class = "factor"), 
     RSUBJID = 901000:901009, RUSUBJID = structure(1:10, .Label = c(
     "000301-000-901-251", "000301-000-901-252", "000301-000-901-253", 
     "000301-000-901-254", "000301-000-901-255", "000301-000-901-256", 
     "000301-000-901-257", "000301-000-901-258", "000301-000-901-259", 
     "000301-000-901-260", "000301-000-901-261", "000301-000-901-262")
, class = "factor")), .Names = c("F_STATUS",  "EVENT_ID", "PAG_NAME", "PTSIZE", "PTSIZE_U", "PT_SYM", "PATHOT", 
 "PATHON", "PATHOM", "RSUBJID", "RUSUBJID"), row.names = c(NA,  10L),
 class = "data.frame")

感谢。

Answer 1

我尝试编辑数据，因此它没有输入错误。还创建了可能组合的表格版本：

stg_tbl <- structure(list(PATHOT = structure(c(4L, 3L, 5L, 3L, 3L, 2L, 5L, 
4L, 1L), .Label = c("*pT4", "*pTIS", "pT1", "pT2", "pT3"), class = "factor"), 
    PATHON = structure(c(2L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 2L), .Label = c("*pN2", 
    "pN1"), class = "factor"), PATHOM = structure(c(2L, 2L, 2L, 
    2L, 1L, 2L, 2L, 1L, 2L), .Label = c("*M1", "M0"), class = "factor")), .Names = c("PATHOT", 
"PATHON", "PATHOM"), class = "data.frame", row.names = c("1", 
"4", "13", "161", "391", "810", "948", "1043", "1067"))

制作类别的文本等效矢量：

stg_lbls <- with(stg_tbl, paste(PATHOT, PATHON, PATHOM, sep="_") )

然后，使用这些级别创建的因子的as.numeric值将是所需的结果：

dat$stg <- with(dat, factor( paste(PATHOT, PATHON, PATHOM, sep="_"), levels=stg_lbls))
as.numeric(dat$stg)
#[1] 1 1 1 2 2 1 1 2 1 1

您可以按照常规方式分配这些值：

dat$sfd <- as.numeric(dat$stg)

Answer 2

我制作了一些新数据，这对您的问题非常有用。

k<-expand.grid(data.frame(a=letters[1:3],b=letters[4:6],c=letters[7:9]))
library(dplyr)
k %>% mutate(groups=paste0(a,b,c))->k2
k2$groups<-as.numeric(factor(k2$groups))
k2

这很粗糙，你不会选择哪个组合获得哪个数字，所以之后需要进行一些挖掘，但这很快。

基于多列

2 个答案: