Get rank position for every row in a dataframe column from a list-CodePudding

From the following list of values.

list_for_rank <- c(3,8,11)

Get the rank position of every row of column "value" as new column

df <- data.frame(id = c("xx1", "xx2","xx3","xx4","xx5"), value = c(10,6,3,7,9))

Expected result should be something like this new column "rank_position"

  df <- data.frame(id = c("xx1", "xx2","xx3","xx4","xx5"),
 value = c(10,6,3,7,9),
 rank_position=c(3,2,1,3,3))

Explanation: Value = 10 ranks 3th in list "list_for_rank" whereas value=6 ranks 2 on the list called "list_for_rank " and so on.

CodePudding user response：

df <- data.frame(id = c("xx1", "xx2","xx3","xx4","xx5"), value = c(10,6,3,7,9))

list_for_rank <- c(3,8,11)

using dplyr

library(dplyr)

df %>%
  group_by(value) %>%
  mutate(list_for_rank = rank(c(value, list_for_rank), ties.method = "first")[1])

# # A tibble: 5 x 3
# # Groups:   value [5]
#   id    value list_for_rank
#   <fct> <dbl>         <int>
# 1 xx1      10             3
# 2 xx2       6             2
# 3 xx3       3             1
# 4 xx4       7             2
# 5 xx5       9             3

using data.table

library(data.table)
setDT(df)

df[, list_for_rank := rank(c(value, list_for_rank), ties.method = "first")[1], by = value]

df

#     id value list_for_rank
# 1: xx1    10             3
# 2: xx2     6             2
# 3: xx3     3             1
# 4: xx4     7             2
# 5: xx5     9             3