Don't use factors in the first place? Use character data.
On March 27, 2020 4:17:57 AM PDT, Elahe chalabi via R-help
<r-help@r-project.org> wrote:
>Hello everyone,
>
>I have the following dataframe
>
>
> library(dplyr)
> dput(df)
> structure(list(Freq = c(19L, 19L, 18L, 15L, 14L, 13L, 13L, 12L,
> 11L, 11L, 11L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 9L), word1 =
>structure(c(3L,
> 11L, 5L, 6L, 11L, 3L, 7L, 10L, 8L, 11L, 13L, 1L, 1L, 4L, 5L,
> 9L, 12L, 14L, 2L), .Label = c("a", "art", "at", "by", "for",
> "for ", "i", "is", "on", "said", "the", "there", "to", "when"
> ), class = "factor"), word2 = structure(c(13L, 12L, 13L, 13L,
> 6L, 13L, 5L, 8L, 11L, 4L, 2L, 3L, 10L, 13L, 13L, 13L, 1L, 9L,
> 7L), .Label = c("are", "be", "bit", "bottom", "dont", "end",
> "hotel", "in", "it", "little", "one", "rest", "the"), class =
>"factor"),
> word3 = structure(c(5L, 9L, 6L, 7L, 9L, 11L, 13L, 1L, 9L,
> 9L, 2L, 9L, 3L, 5L, 6L, 10L, 12L, 4L, 8L), .Label = c("a",
> "able", "bit", "comes", "end", "first", "first ", "florence",
> "of", "other", "same", "so", "want"), class = "factor"),
> word4 = structure(c(5L, 9L, 10L, 8L, 9L, 10L, 11L, 7L, 9L,
> 9L, 11L, 1L, 5L, 5L, 6L, 2L, 4L, 11L, 3L), .Label = c("a",
> "hand", "italy", "many", "of", "place", "statement", "states",
> "the", "time", "to"), class = "factor")), class = "data.frame",
>row.names = c(NA,
> -19L))
>
>
>Is there a way to modify the following command so that the filter
>function doesn't return all the Levels?
>
>
>
> filter(df,(word1 == 'for' & word2 == 'the' & word3 == 'first')) $
>word4
> [1] time place
> Levels: a hand italy many of place statement states the time to
>
>
>
>Thanks a lot!
>Elahe
>
>______________________________________________
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--
Sent from my phone. Please excuse my brevity.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
