r 将二进制响应折叠为单一响应

Question

这是我的数据集

 State   Response
 AL      NO
 AR      NO
 AK      NO
 AZ      NO
 CA      YES
 CA      NO
 CO      YES
 CO      NO
 FL      NO
 GA      NO

有些州既是又是否，例如 CA、CO.。我如何折叠每个州的是、否值并将是、否更改为仅是。这是我的预期输出

 State   Response
 AL      NO
 AR      NO
 AK      NO
 AZ      NO
 CA      YES
 CO      YES
 FL      NO
 GA      NO

非常感谢任何建议。谢谢

Answer 1

我们可以 arrange 'State' 的行和 'Response' 的逻辑向量，然后 filter 唯一的行

library(dplyr)
df1 %>%
   arrange(State, Response != "YES") %>%
   filter(!duplicated(State))

-输出

   State Response
1    AK       NO
2    AL       NO
3    AR       NO
4    AZ       NO
5    CA      YES
6    CO      YES
7    FL       NO
8    GA       NO

数据

df1 <- structure(list(State = c("AL", "AR", "AK", "AZ", "CA", "CA", 
"CO", "CO", "FL", "GA"), Response = c("NO", "NO", "NO", "NO", 
"YES", "NO", "YES", "NO", "NO", "NO")), class = "data.frame", row.names = c(NA, 
-10L))

Answer 2

我建议 distinct()（.keep_all = TRUE 保留所有列，但删除重复的行）来自 dplyr。例如：

State <- c("AL", "AR", "AK", "AZ", "CA", "CA", "CO", "CO", "FL", "GA")
Response <- c("NO", "NO", "NO", "NO", "YES", "NO", "YES", "NO", "NO", "NO")

data <- bind_cols(
  State, Response
) %>% 
  rename(
    State = `...1`,
    Response = `...2`
  ) %>% 
  distinct(State, .keep_all= TRUE)

data

输出：

> data
# A tibble: 8 × 2
  State Response
  <chr> <chr>   
1 AL    NO      
2 AR    NO      
3 AK    NO      
4 AZ    NO      
5 CA    YES     
6 CO    YES     
7 FL    NO      
8 GA    NO

Answer 3

另一种解决方案：

library(dplyr)
df %>% 
  count(State) %>% 
  mutate(Response = ifelse(n>1,"YES","NO")) %>% 
  select(-n)

已编辑：

library(dplyr)
df %>% 
  count(State) %>% 
  mutate(Response = ifelse(n>1,"YES",NA)) %>%
  replace(is.na(.),"NO") %>%
  select(-n)

已编辑 2 - 决赛：

library(dplyr)
df %>% 
  add_count(State) %>% 
  mutate(Response = ifelse(n>1,"YES",Response)) %>% 
  distinct(select(.,-n))

Answer 4

如果 any 状态中的值为“YES”，您可以 return “YES”。

library(dplyr)

df %>%
  group_by(State) %>%
  summarise(Response = if(any(Response == 'YES')) 'YES' else 'NO')

# State Response
#  <chr> <chr>   
#1 AK    NO      
#2 AL    NO      
#3 AR    NO      
#4 AZ    NO      
#5 CA    YES     
#6 CO    YES     
#7 FL    NO      
#8 GA    NO

因为 "YES" > "NO" 在这种情况下您可以使用 max。

df %>% group_by(State) %>% summarise(Response = max(Response))

或以 R 为基数 -

aggregate(Response~State, df, max)

r 将二进制响应折叠为单一响应

r collapse binary response to single response

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collapse

数据