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?
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 at 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? > >______________________________________________ >R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide >http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code.-- Sent from my phone. Please excuse my brevity.
Hello, Here are two ways. The first keeps word4 as factor, the second coerces to character first, like Jeff said. df %>% filter(word1 == 'for' & word2 == 'the' & word3 == 'first') %>% pull(word4) %>% droplevels() df %>% mutate_if(is.factor, as.character) %>% filter(word1 == 'for' & word2 == 'the' & word3 == 'first') %>% pull(word4) Hope this helps, Rui Barradas ?s 12:41 de 27/03/20, Jeff Newmiller escreveu:> 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 at 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 >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >