data.frame using raw data instead of filtered data

Asked Aug 20 '15 at 14:12

Active Aug 20 '15 at 14:46

Viewed 28 times

I have got that issue I can't my head around:

library(dplyr)

d<-data.frame(c1=c("a","a","b","b"),c2=c(1,1,3,4))

dFiltered <-  d %>%
  filter(c1 == "a") %>%
  select(c1)

> dFiltered
  c1
1  a
2  a

so then I want to create a data.frame to get the frequency of c1 based on the filtered data

> data.frame(table(dFiltered))
  dFiltered Freq
1         a    2
2         b    0

I don't understand why the value b appears as well even though it shows 0, I don't expect to find it here because if I run the same but selecting a value from c2

> dFiltered <-  d %>%
   filter(c2 == "1") %>%
   select(c2)
> data.frame(table(dFiltered))

then I get what I expect

  dFiltered Freq
1         1    2

I am sure there is a rational explanation to this

edited Aug 20 '15 at 14:46

asked Aug 20 '15 at 14:12

Guillaume Lombard

2

Because its a factor. See `?droplevels` Or just do `d %>% filter(c1 == "a") %>% count(c1)`. Btw, `dplyr` isn't base R. – David Arenburg Aug 20 '15 at 14:18
and also of [this](http://stackoverflow.com/questions/26826865/how-to-drop-unused-levels-after-filtering-by-factor) – jeremycg Aug 20 '15 at 14:19
Thanks Jeremy, why does it matter that dplyr is not a base R? – Guillaume Lombard Aug 20 '15 at 14:35
It's just when you creating a reproducible example you should add `library(dplyr)` as it isn't being exported with base R – David Arenburg Aug 20 '15 at 14:45
I will make sure I do. Thanks guys – Guillaume Lombard Aug 20 '15 at 14:46

data.frame using raw data instead of filtered data

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