如何根据其他变量的条件生成二进制变量? [英] How to generate binary variable according to the conditions of other variables?

问题描述

再一次，我为提出此类问题而道歉，但是R世界是如此之大，以至于我有时迷失了方向，即使我已经阅读了一些与R相关的最好的书. 我有以下数据库

Once again, I apologize for asking this type of question, but the R world is so large that sometime I feel lost, even if I have read some of the best book related with R. I have the following DB

ID=rep((1:3),3)
x<-as.Date("2013-1-1")
y<-as.Date("2013-1-2")
z<-as.Date("2013-1-3")
DATE<-c(x,x,x,y,x,y,z,z,z)
TRAP<-c(1,1,1,3,2,3,2,1,3)
IN<-data.frame(ID,DATE,TRAP)

并且我想根据以下条件生成一个二进制变量(RESULT):如果DATE和TRAP对于不同的ID相同，则RESULT> y否则为RESULT> n，就像这样

and I would like to produce a binary variable (RESULT) according to the following conditions: if the DATE and the TRAP is the same for the different ID, then RESULT>y otherwise RESULT>n, like this

RESULT<-c("y","y","y","y","n","y","n","n","n")
OUT<-cbind(IN,RESULT)

我认为应该使用ifelse函数，但是我不知道如何为每个ID显式声明相等控制的条件... 一如既往，每个建议都将不胜感激！

I think that the ifelse function should be used, but I don't know how to explicit the condition of equality controlling for each ID... As always, every suggestion is greatly appreciated!

推荐答案

这是使用plyr做到这一点的一种方法:

Here is a way to do it with plyr :

R> ddply(IN, .(DATE,TRAP), transform, RESULT=ifelse(length(ID)>1,"y","n"))
  ID       DATE TRAP RESULT
1  1 2013-01-01    1      y
2  2 2013-01-01    1      y
3  3 2013-01-01    1      y
4  2 2013-01-01    2      n
5  1 2013-01-02    3      y
6  3 2013-01-02    3      y
7  2 2013-01-03    1      n
8  1 2013-01-03    2      n
9  3 2013-01-03    3      n

请注意，行已重新排序.

Note that the rows have been reordered.

使用data.table的另一种解决方案:

Another solution with data.table :

R> DT <- data.table(IN)
R> DT[,RESULT:=ifelse(.N>1,"y","n"), by=list(DATE,TRAP)]
R> DT
   ID       DATE TRAP RESULT
1:  1 2013-01-01    1      y
2:  2 2013-01-01    1      y
3:  3 2013-01-01    1      y
4:  1 2013-01-02    3      y
5:  2 2013-01-01    2      n
6:  3 2013-01-02    3      y
7:  1 2013-01-03    2      n
8:  2 2013-01-03    1      n
9:  3 2013-01-03    3      n

这里没有重新排序.

或使用基础ave:

IN <- within(IN, { RESULT <- ave(TRAP, list(DATE, TRAP), 
               FUN= function(x) ifelse(length(x) > 1, "y", "n"))})
#   ID       DATE TRAP RESULT
# 1  1 2013-01-01    1      y
# 2  2 2013-01-01    1      y
# 3  3 2013-01-01    1      y
# 4  1 2013-01-02    3      y
# 5  2 2013-01-01    2      n
# 6  3 2013-01-02    3      y
# 7  1 2013-01-03    2      n
# 8  2 2013-01-03    1      n
# 9  3 2013-01-03    3      n

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