英文:
Create a dummy variable based on two variables x1 and x2 (dummy=x1 only if at least one adjacent x2=yes)
问题
以下是用于演示生成数据框的代码：
d <- data.frame(x1 = c(rep("no", 5), rep("yes", 4), rep("no", 2), rep("yes", 3), rep("no", 2), rep("yes", 3)),x2 = c(rep("no", 6), rep("yes", 1), rep("no", 9), rep("yes", 2), rep("no", 1)),dummy = c(rep(0, 5), rep(1, 4), rep(0, 5), rep(0, 2), rep(1, 3)))
我有两个变量x1和x2。我想要一个名为'dummy'的虚拟变量，基于x1和x2的指标。具体来说，虚拟变量应该在捕获所有x1=yes值的情况下等于1，条件是其相邻的x2至少有一个为yes。如果x1=yes但其相邻的x2=no，则虚拟变量应为0。
要创建一个虚拟变量，当x1和x2都等于'yes'时取值为1，可以使用以下代码：
d$dummy = ifelse(d$x1 == "yes" & d$x2 == "yes", 1, 0)
但这不能捕获我希望实现的x1=yes的整个集群。
我希望的输出如下图所示：
有没有办法实现这个目标？
英文:
Below is the code to generate the dataframe for demonstration:
d<-data.frame(x1=c(rep("no",5),rep("yes",4),rep("no",2),rep("yes",3),rep("no",2),rep("yes",3)),x2=c(rep("no",6),rep("yes",1),rep("no",9),rep("yes",2),rep("no",1)),dummy=c(rep(0,5),rep(1,4),rep(0,5),rep(0,2),rep(1,3)))
I have two variables x1 and x2. What I want is a dummy variable, named as 'dummy', based on both x1 and x2 indicators. Specifically, the dummy should equal 1 by capturing all x1=yes values conditional that at least one of its adjacent x2=yes. If x1=yes but its adjacent x2=no, then the dummy should be 0.
It is easy to create a dummy variable taking value of 1 when both x1 and x2 equal 'yes', using
d$dummy=ifelse(d$x1=="yes" & d$x2=="yes",1,0)
But it would not be able to capture the whole cluster of x1=yes which is what I wish to do.
The desired output I am looking for is like this:
Any idea how this could be done?
答案1
得分: 1
你可以对consecutive_ids进行分组，然后如果x1 == "yes"且相邻的x2 == "yes"，则得到1。
library(dplyr) #1.1.0+d %>%group_by(cons = consecutive_id(x1)) %>%mutate(dummy = +(x1 == "yes" & any(x2 == "yes"))) %>%ungroup()x1 x2 dummy cons<chr> <chr> <int> <int>1 no no 0 12 no no 0 13 no no 0 14 no no 0 15 no no 0 16 yes no 1 27 yes yes 1 28 yes no 1 29 yes no 1 210 no no 0 311 no no 0 312 yes no 0 413 yes no 0 414 yes no 0 415 no no 0 516 no no 0 517 yes yes 1 618 yes yes 1 619 yes no 1 6
英文:
You can group_by consecutive_ids, and then get 1 if x1 == "yes", and any adjacent x2 is "yes".
library(dplyr) #1.1.0+d %>%group_by(cons = consecutive_id(x1)) %>%mutate(dummy = +(x1 == "yes" & any(x2 == "yes"))) %>%ungroup()x1 x2 dummy cons<chr> <chr> <int> <int>1 no no 0 12 no no 0 13 no no 0 14 no no 0 15 no no 0 16 yes no 1 27 yes yes 1 28 yes no 1 29 yes no 1 210 no no 0 311 no no 0 312 yes no 0 413 yes no 0 414 yes no 0 415 no no 0 516 no no 0 517 yes yes 1 618 yes yes 1 619 yes no 1 6
答案2
得分: 1
你可以使用 data.table::rleid() 来创建以 x1 为块的分组。
library(tidyverse)library(data.table)d %>% mutate(gp = rleid(x1)) %>%group_by(gp) %>%mutate(dummy2 = ifelse(x1 == "yes" & any(x2 == "yes"), 1, 0))# 一个 tibble: 19 × 5# Groups: gp [6]x1 x2 dummy gp dummy2<chr> <chr> <dbl> <int> <dbl>1 no no 0 1 02 no no 0 1 03 no no 0 1 04 no no 0 1 05 no no 0 1 06 yes no 1 2 17 yes yes 1 2 18 yes no 1 2 19 yes no 1 2 110 no no 0 3 011 no no 0 3 012 yes no 0 4 013 yes no 0 4 014 yes no 0 4 015 no no 0 5 016 no no 0 5 017 yes yes 1 6 118 yes yes 1 6 119 yes no 1 6 1
英文:
You can use data.table::rleid() to create groups with blocks of x1.
library(tidyverse)library(data.table)d %>% mutate(gp = rleid(x1)) %>%group_by(gp) %>%mutate(dummy2 = ifelse(x1 == "yes" & any(x2 == "yes") , 1, 0))# A tibble: 19 × 5# Groups: gp [6]x1 x2 dummy gp dummy2<chr> <chr> <dbl> <int> <dbl>1 no no 0 1 02 no no 0 1 03 no no 0 1 04 no no 0 1 05 no no 0 1 06 yes no 1 2 17 yes yes 1 2 18 yes no 1 2 19 yes no 1 2 110 no no 0 3 011 no no 0 3 012 yes no 0 4 013 yes no 0 4 014 yes no 0 4 015 no no 0 5 016 no no 0 5 017 yes yes 1 6 118 yes yes 1 6 119 yes no 1 6 1
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