Finding ALL duplicate rows, including "elements with smaller subscripts"

duplicated has a fromLast argument. The "Example" section of ?duplicated shows you how to use it. Just call duplicated twice, once with fromLast=FALSE and once with fromLast=TRUE and take the rows where either are TRUE.

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Some late Edit: You didn't provide a reproducible example, so here's an illustration kindly contributed by @jbaums

vec <- c("a", "b", "c","c","c") 
vec[duplicated(vec) | duplicated(vec, fromLast=TRUE)]
## [1] "c" "c" "c"

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Edit: And an example for the case of a data frame:

df <- data.frame(rbind(c("a","a"),c("b","b"),c("c","c"),c("c","c")))
df[duplicated(df) | duplicated(df, fromLast=TRUE), ]
##   X1 X2
## 3  c  c
## 4  c  c

Duplicated rows in a dataframe could be obtained with dplyr by doing

library(tidyverse)
df = bind_rows(iris, head(iris, 20)) # build some test data
df %>% group_by_all() %>% filter(n()>1) %>% ungroup()

To exclude certain columns group_by_at(vars(-var1, -var2)) could be used instead to group the data.

If the row indices and not just the data is actually needed, you could add them first as in:

df %>% add_rownames %>% group_by_at(vars(-rowname)) %>% filter(n()>1) %>% pull(rowname)

You need to assemble the set of duplicated values, apply unique, and then test with %in%. As always, a sample problem will make this process come alive.

> vec <- c("a", "b", "c","c","c")
> vec[ duplicated(vec)]
[1] "c" "c"
> unique(vec[ duplicated(vec)])
[1] "c"
>  vec %in% unique(vec[ duplicated(vec)]) 
[1] FALSE FALSE  TRUE  TRUE  TRUE

I've had the same question, and if I'm not mistaken, this is also an answer.

vec[col %in% vec[duplicated(vec$col),]$col]

Dunno which one is faster, though, the dataset I'm currently using isn't big enough to make tests which produce significant time gaps.