具有相同数量变量的数据帧列表并删除一个变量内的重复项并在数据帧的 rest 中执行相同操作

Question

I have the following list of data frames and each data frame has 3 variables (a, b and c)

my.list <- list(d1, d2, d3, d4)

Inside my data frame, I have duplicated strings in "a" and I want to delete the rows with duplicated values

The current code i am using:

my.listnew <- lapply(my.list, function(x) unique(x["a"]))

The problem i have with this code is that the other 2 columns "b" and "c" are gone and I want to keep them, while the duplicated rows are deleted

Answer 1

Use duplicated to remove the duplicated values in column a while keeping other columns.

my.listnew <- lapply(my.list, function(x) x[!duplicated(x$a), ])

Answer 2

Just for reference, tidyverse style of doing it-

set.seed(1)
my.list <- list(d1 = data.frame(a = sample(letters[1:3], 5, T),
                                b = rnorm(5),
                                c = rnorm(5)), 
                d2 = data.frame(a = sample(letters[1:3], 5, T),
                                b = rnorm(5),
                                c = rnorm(5)), 
                d3 = data.frame(a = sample(letters[1:3], 5, T),
                                b = rnorm(5),
                                c = rnorm(5)))
library(tidyverse)
map(my.list, ~ .x %>% filter(!duplicated(a)) )
#> $d1
#>   a         b          c
#> 1 a 1.5952808  0.5757814
#> 2 c 0.3295078 -0.3053884
#> 3 b 0.4874291  0.3898432
#> 
#> $d2
#>   a          b         c
#> 1 b  0.2522234 0.3773956
#> 2 a -0.8919211 0.1333364
#> 
#> $d3
#>   a          b          c
#> 1 a -0.2357066  1.1519118
#> 2 c -0.4333103 -0.4295131
#> 3 b -0.6494716  1.2383041

^{Created on 2021-05-13 by the reprex package (v2.0.0)}

If you also want to combine the dataframes in output you may use map_dfr instead of map in above

Answer 3

We can use subset without any anonymous function

out <- lapply(my.list, subset, subset = !duplicated(a))

---

Or using data.table with unique

library(data.table)
out <- lapply(my.list, function(dat) unique(as.data.table(dat), by = 'a'))