Question

我有一个像这样的数据集：

 df<-data.frame(x=c("A","A","A","A", "B","B","B","B","B", 
                   "C","C","C","C","C","D","D","D","D","D"),  
                y= as.factor(c(rep("Eoissp2",4),rep("Eoissp1",5),"Eoissp1","Eoisp4","Automerissp1","Automerissp2","Acharias",rep("Eoissp2",3),rep("Eoissp1",2))))

我想为x的每个子集确定y中包含表达式Eois的完全重复的相应级别。因此，A，B和D将在向量中返回，因为A，B和D的每个级别都包含表达式Eois，而等级C由各种独特的等级组成（例如Eois，Automeris和Acharias）。对于此示例，输出将为：

   output<- c("A", "B", "D")

Answer 1

使用新的df：

> df %>% filter(str_detect(y,"Eois")) %>% group_by(x) %>% distinct(y) %>% 
    count() %>% filter(n==1) %>% select(x)
# A tibble: 2 x 1
# Groups:   x [2]
  x    
  <fct>
1 A    
2 B

（以下答案使用问题作者发布的原始df。）

使用magrittr＆amp;中的管道功能来自dplyr的函数：

> df %>% group_by(x) %>% distinct(y)
# A tibble: 7 x 2
# Groups:   x [3]
  x     y      
  <fct> <fct>  
1 A     plant1a
2 B     plant1b
3 C     plant1a
4 C     plant2a
5 C     plant3a
6 C     plant4a 
7 C     plant5a

然后你可以像这样汇总结果：

> results <- df %>% group_by(x) %>% distinct(y) %>% 
    count() %>% filter(n==1) %>% select(x)
> results
# A tibble: 2 x 1
# Groups:   x [2]
  x    
  <fct>
1 A    
2 B

如果您知道原始数据框始终按顺序附带x，则可以删除group_by部分。

Answer 2

基于dplyr的解决方案可以是：

library(dplyr)
df %>% group_by(x) %>%
  filter(grepl("Eoiss", y)) %>%
  mutate(y = sub("\\d+", "", y)) %>%
  filter(n() >1 & length(unique(y)) == 1) %>%
  select(x) %>% unique(.)

# A tibble: 3 x 1
# Groups: x [3]
#  x     
#  <fctr>
#1 A     
#2 B     
#3 D

数据

df<-data.frame(x=c("A","A","A","A", "B","B","B","B","B", "C","C","C","C","C","D","D","D","D","D"), y= as.factor(c(rep("Eoissp2",4), rep("Eoissp1",5),"Eoissp1","Eoisp4","Automerissp1","Automerissp2", "Acharias",rep("Eoissp2",3),rep("Eoissp1",2))))

识别仅包含表达式重复的子集

2 个答案: