filtering data.frame character column in r

Question

I have a variable term which includes loan term and has only two types: "36 months" and "60 months". I want to filter my dataset df and create two different subsets for these categories.

I have tried using the commands subset, filter,which etc but it didn't work.

codes that I have tried df1 –> df[which(df$term == "36 months"),], df1 –> filter(df, term == "36 months") df1–>df[df$term %in% c("36 months"), ] and other ways but always choices zero rows.

head(loan_data$term)

[1] " 36 months" " 60 months" " 36 months" " 36 months" " 60 months" " 36 months"

Any suggestion?

Thanks

can you add a minimal reproducible example please: https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example — Bulat, Jun 07 '18 at 21:09
Are you really use right assignment (`->`) or is that a typo in your question? — joran, Jun 07 '18 at 21:09
@Bulat thanks to you I saw the way the data was constructed. It had space in "_36 months" before 36 — Erick, Jun 07 '18 at 21:35

score 1 · Answer 1 · answered Jun 07 '18 at 21:12

I cannot reproduce your problem:

df <- data.frame(
  id = c(1, 2, 3),
  val = c("60 months", "36 months", "60 months")
)

df[df$val == "60 months", ]
# df$val == "60 months"
# 1  1 60 months
# 3  3 60 months
df[which(df$val == "60 months"),]
# df$val == "60 months"
# 1  1 60 months
# 3  3 60 months

rdfleay · Answer 2 · 2021-02-11T15:30:27.733

0

as more robustly explained here, using...

my.data.frame %>% filter(.data[[myName]] == 1) ...

df60 <- df %>% filter(.df[[term]] == "60 months")
df30 <- df %>% filter(.df[[term]] == "30 months")

..or here

ssP <- mutate( nino, PHASE = ifelse(PHASE == "E", "N","L"))

which for your df ..

df60 <- mutate( df, term = ifelse(term == "60 months"))
df30 <- mutate( df, term = ifelse(term == "30 months"))

edited Feb 11 '21 at 15:30

answered Feb 11 '21 at 15:15

rdfleay

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filtering data.frame character column in r

2 Answers2