找出一个组中的所有列在R中是否都是一致的

Question

我有一个数据框架，df

df <- structure(list(ID = structure(c("ID961", "ID961", 
"ID961", "ID961", "ID726", 
"ID726", "ID726", "ID864", 
"ID864", "ID864"), label = "ID"), 
    TYPE = structure(c("blind", "blind", "blind", "blind", 
    "blind", "blind", "blind", "blind", "blind", "notblind"
    ), label = "blind or not"), AGE = structure(c(50, 
    50, 50, 50, 67, 67, 67, 35, 35, 35), label = "Age"), AGEU = structure(c("YEARS", 
    "YEARS", "YEARS", "YEARS", "YEARS", "YEARS", "YEARS", "YEARS", 
    "YEARS", "YEARS"), label = "Age Units"), AVISIT = structure(c("26", 
    "46", "46", "36", "66", 
    "64", "67", "37", "37", 
    "67"), label = "treat")), row.names = c(NA, 
-10L), class = c("tbl_df", "tbl", "data.frame"), label = "df")

> df
# A tibble: 10 x 5
   ID    TYPE    AGE AGEU  AVISIT
   <chr> <chr> <dbl> <chr> <chr> 
 1 ID961 blind    50 YEARS 26    
 2 ID961 blind    50 YEARS 46    
 3 ID961 blind    50 YEARS 46    
 4 ID961 blind    50 YEARS 36    
 5 ID726 blind    67 YEARS 66    
 6 ID726 blind    67 YEARS 64    
 7 ID726 blind    67 YEARS 67    
 8 ID864 blind    35 YEARS 37    
 9 ID864 blind    35 YEARS 37    
10 ID864 notblind 35 YEARS 67 

对于每个ID，我想知道匹配该ID的每个列条目是否是统一的。

例如：

ID961 - TYPE是一样的，AGE是一样的，AGEU是一样的，但是AVISIT是不一样的。

ID726 - TYPE是一样的，AGE是一样的，AGEU是一样的，但是AVISIT是不一样的。

ID864 - AGE是一样的，AGEU是一样的，但是AVISIT是不一样的，TYPE是不一样的。

因此，我希望返回的是AGE和AGEU在ID中都是统一的，例如：

uniform
[1] "AGE"  "AGEU"

我不知道该怎么做-我知道我可以用

match <- df %>% group_by(ID)

但那就不知道怎么继续下去了。

Answer 1

也许你能做到：

library(dplyr)

df %>%
  group_by(ID) %>%
  summarise(across(everything(), ~ n_distinct(.x) == 1), .groups = "drop")

# Or deprecated way
df %>%
  group_by(ID) %>%
  summarise_all(~n_distinct(.x) == 1)

# A tibble: 3 x 5
  ID    TYPE  AGE   AGEU  AVISIT
  <chr> <lgl> <lgl> <lgl> <lgl> 
1 ID726 TRUE  TRUE  TRUE  FALSE 
2 ID864 FALSE TRUE  TRUE  FALSE 
3 ID961 TRUE  TRUE  TRUE  FALSE 

如果你想要列名，你可以：

df %>%
  group_by(ID) %>%
  summarise(across(everything(), ~ n_distinct(.x) == 1), .groups = "drop") %>%
  rowwise() %>%
  transmute(ID, uniform = toString(names(.)[which(c_across(cols = -ID)) + 1]))

# Or ...
df %>%
  group_by(ID) %>%
  summarise_all(~n_distinct(.x) == 1)  %>%
  transmute(ID, 
            uniform = pmap_chr(.[-1], ~ toString(names(df)[c(FALSE, ...)])))  

# A tibble: 3 x 2
# Rowwise: 
  ID    uniform        
  <chr> <chr>          
1 ID726 TYPE, AGE, AGEU
2 ID864 AGE, AGEU      
3 ID961 TYPE, AGE, AGEU