如何检查一个文档中的所有记录是否都包含在另一个文档中

Question

我有df1和df2，我想检查一下df1和df2是否都包含在df3$lbtest中。检查并标记包含的和缺失的是一个好方法吗?

df1<- tructure(list(CHMTEST = c("Sodium", "Potassium", "Chloride", 
"Bicarbonate (CO2)", "Calcium", "Glucose")), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame"))
df2<-structure(list(COATEST = c("PT", "aPTT")), row.names = c(NA, 
-2L), class = c("tbl_df", "tbl", "data.frame"))
df3<- structure(list(source = c("LBCH", "LBCH", "LBCH", "LBCH", "LBCO", 
"LBHE", "LBUA"), lbcat = c("Chemistry", "Chemistry", "Chemistry", 
"Chemistry", "Coagulation", "Hematology", "Urinalysis"), lbtest = c("Bicarbonate (CO2)", 
"Glucose", "Potassium", "Sodium", "PT", "WBC", "Glucose")), row.names = c(NA, 
-7L), class = c("tbl_df", "tbl", "data.frame"))

Answer 1

将从'df1'， 'df2'数据集中提取的列连接到一个向量中，使用%in%检查元素是否存在于'df3$lbtest'列中，并使用all包装

all(c(df1$CHMTEST, df2$COATEST) %in% df3$lbtest)
[1] FALSE

返回FALSE，因为某些元素仍然不存在，即

c(df1$CHMTEST, df2$COATEST) %in% df3$lbtest
[1]  TRUE  TRUE FALSE  TRUE FALSE  TRUE  TRUE FALSE

---

如果要返回存在而不存在的元素，则创建单个向量，生成逻辑索引并将其用于

子集

v1 <- c(df1$CHMTEST, df2$COATEST)
i1 <-  v1 %in% df3$lbtest
v1[i1] # present
[1] "Sodium"            "Potassium"         "Bicarbonate (CO2)" "Glucose"           "PT"   
v1[!i1] # not present
[1] "Chloride" "Calcium"  "aPTT" 

---

如果我们想从'df3'检查这一点，使用subset或dplyr::filter

library(dplyr)
df3 %>% 
filter(lbtest %in% c(df1$CHMTEST, df2$COATEST))
# A tibble: 6 x 3
source lbcat       lbtest           
<chr>  <chr>       <chr>            
1 LBCH   Chemistry   Bicarbonate (CO2)
2 LBCH   Chemistry   Glucose          
3 LBCH   Chemistry   Potassium        
4 LBCH   Chemistry   Sodium           
5 LBCO   Coagulation PT               
6 LBUA   Urinalysis  Glucose          
df3 %>% 
filter(!lbtest %in% c(df1$CHMTEST, df2$COATEST))
# A tibble: 1 x 3
source lbcat      lbtest
<chr>  <chr>      <chr> 
1 LBHE   Hematology WBC   

---

或者另一种选择是使列名相同，将'df1'， 'df2'绑定到单个数据集，并使用inner_join/anti_join

library(purrr)
> list(df1, df2) %>%
map_dfr(~ .x %>%
rename_with(~ 'lbtest', 1)) %>% 
inner_join(df3, .)
Joining, by = "lbtest"
# A tibble: 6 x 3
source lbcat       lbtest           
<chr>  <chr>       <chr>            
1 LBCH   Chemistry   Bicarbonate (CO2)
2 LBCH   Chemistry   Glucose          
3 LBCH   Chemistry   Potassium        
4 LBCH   Chemistry   Sodium           
5 LBCO   Coagulation PT               
6 LBUA   Urinalysis  Glucose                
> list(df1, df2) %>% 
map_dfr(~ .x %>% 
rename_with(~ 'lbtest', 1)) %>%
anti_join(df3, .)
Joining, by = "lbtest"
# A tibble: 1 x 3
source lbcat      lbtest
<chr>  <chr>      <chr> 
1 LBHE   Hematology WBC