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Home/ Questions/Q 7599825
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T22:45:21+00:00 2026-05-30T22:45:21+00:00

I have this nested data frame test <- structure(list(id = c(13, 27), seq =

  • 0

I have this nested data frame

test <- structure(list(id = c(13, 27), seq = structure(list(
`1` = c("1997", "1997", "1997", "2007"),
`2` = c("2007", "2007", "2007", "2007", "2007", "2007", "2007")), 
.Names = c("1", "2"))), .Names = c("penr", 
"seq"), row.names = c("1", "2"), class = "data.frame")

I want a list of all values in the second column, namely

result <- c("1997", "1997", "1997", "2007", "2007", "2007", "2007", "2007", "2007", "2007", "2007")

Is there an easy way to achieve this?

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  1. Editorial Team
    Editorial Team
    2026-05-30T22:45:22+00:00Added an answer on May 30, 2026 at 10:45 pm

    This line does the trick:

    do.call("c", test[["seq"]])
    

    or equivalent:

    c(test[["seq"]], recursive = TRUE)
    

    or even:

    unlist(test[["seq"]])
    

    The output of these functions is:

        11     12     13     14     21     22     23     24     25     26     27 
    "1997" "1997" "1997" "2007" "2007" "2007" "2007" "2007" "2007" "2007" "2007" 
    

    To get rid of the names above the character vector, call as.character on the resulting object:

    > as.character((unlist(test[["seq"]])))
     [1] "1997" "1997" "1997" "2007" "2007" "2007" "2007" "2007" "2007" "2007"
    [11] "2007"
    
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