CODEWITHSUNDEEP
rdplyr
enjaku
enjaku

Reputation: 19

as_tibble only returns a single variable

I'm very new to R and coding, and this is my first reach-out to SO

I have a data frame (brfss2013) with 491775 obs and 330 variables. I want to take three of those variables ($qlmentl2, $misdeprd & $misnowork) and create a new data frame called "level_unhappy" with just those three variables

I've tried this code in R Studio, and the result is a data frame with 491775 obs (correct) and 1 variable. 

```{r create_level_unhappy, results='hide'}
level_unhappy <-  as_tibble(brfss2013$qlmentl2, brfss2013$misdeprd, brfss2013$misnowork, validate = FALSE)

```

I've also tried this...

level_unhappy <- as.data.frame(brfss2013$qlmentl2, brfss2013$misdeprd, brfss2013$misnowork)

... and received...

Error in !optional : invalid argument type
In addition: Warning message:
In as.data.frame.integer(brfss2013$qlmentl2, brfss2013$misdeprd,  :
  'row.names' is not a character vector of length 491775 -- omitting it. Will be an error!

What am I missing?

Cheers, -eric

Upvotes: 1

Views: 398

Answers (2)

Dij
Dij

Reputation: 1378

The first arguments of as.data.frame must be a list of objects you wish to concatenate into a data frame, because the second argument is looking for the row.names. So to avoid confusing R into thinking that your second data frame are the row.names, and since you want to join vectors, put them all in a cbind()

level_unhappy <- as.data.frame(cbind(brfss2013$qlmentl2, brfss2013$misdeprd, brfss2013$misnowork))

You can also avoid this headache by using R's other base command data.frame instead of as.data.frame, which provides some circumstantial advantages depending on what you are trying to do. However, in this case your code would have worked as written with data.frame:

level_unhappy <- data.frame(brfss2013$qlmentl2, brfss2013$misdeprd, brfss2013$misnowork)

data.frame doesn't assume any argument is anything other than data to be combined unelss you explicate as much using an optional argument call like row.names = "r1" for example.

Upvotes: 2

Marius
Marius

Reputation: 60080

as_tibble will do different things depending on what you pass as the first argument. By doing as_tibble(brfss2013$qlmentl2, ...), you're passing a vector as a first argument. The as_tibble method for vectors is not set up to accept multiple vectors, and the other vectors you pass end up getting used as row names etc. Instead, I think you want:

as_tibble(bfrss2013[, c("qlmentl2","misdeprd","misnowork")])

That way, you're passing a data frame as the first argument. as_tibble will convert the dataframe to tibble.

This only really applies if you specifically want to convert your data to tibble though. Tibbles mostly work like dataframes, with a couple added features. If all you want to do is separate out those columns into a separate variable, you can do:

new_df <- bfrss2013[, c("qlmentl2","misdeprd","misnowork")]

Upvotes: 1

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