CODEWITHSUNDEEP
rlapplysapply
hbabbar
hbabbar

Reputation: 967

How to apply a function to dataframe using list

I have a df

id a1 a2
1  x1 y1
2  x2 y2

and another dataframe df2

id name1 name2
1  a1    b1
1  a2    b2
2  a3    b3
3  a4    b4
3  a5    b5

df2 could contain multiple records of unique id's from df1.

I need to join the dataframes in such a way that for each row of df1, i should have one column from first record of df2 and if it exists, second column from second record.

To explain, the output should be like :

id a1 a2 n1 n2
1  x1 y1 a1  a2
2  x2 y2 a3 NA

For doing this I have split df2 on id using split

s <- split(df2, df2$id)

but i'm unsure how to use sapply over that. Any pointers for this

Upvotes: 0

Views: 42

Answers (1)

akrun
akrun

Reputation: 887118

If we are not taking the 'name2' column

library(dplyr)
df2 %>% 
   filter(id %in% df$id) %>% 
   select(-name2) %>% 
   group_by(id) %>% 
   mutate(rn = paste0("n", row_number())) %>%
   spread(rn, name1) %>%
   left_join(df, .)
#  id a1 a2 n1   n2
#1  1 x1 y1 a1   a2
#2  2 x2 y2 a3 <NA>

data

df <- structure(list(id = 1:2, a1 = c("x1", "x2"), a2 = c("y1", "y2"
)), .Names = c("id", "a1", "a2"), class = "data.frame", row.names = c(NA, 
 -2L))


df2 <- structure(list(id = c(1L, 1L, 2L, 3L, 3L), name1 = c("a1", "a2", 
 "a3", "a4", "a5"), name2 = c("b1", "b2", "b3", "b4", "b5")), .Names = c("id", 
 "name1", "name2"), class = "data.frame", row.names = c(NA, -5L))

Upvotes: 1

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