Convert Named Character Vector to data.frame

Question

I have a named character vector returned from xmlAttrs like this:

testVect <- structure(c("11.2.0.3.0", "12.89", "12.71"), .Names = c("db_version", 
             "elapsed_time", "cpu_time"))

I would like to convert it to a data frame that looks like this:

testDF <- data.frame("db_version"="11.2.0.3.0","elapsed_time"=12.89,"cpu_time"=12.71)
head(testDF)
  db_version elapsed_time cpu_time
1 11.2.0.3.0        12.89    12.71

Solution 1

It's as simple as data.frame(as.list(testVect)). Or if you want sensible data types for your columns, data.frame(lapply(testVect, type.convert), stringsAsFactors=FALSE).

Solution 2

The answers from @MatthewPlourde and @JackRyan work, but if you have a long named vector it is annoying to have a data frame with one row and many columns. If you'd rather have a "key" column and a "value" column with many rows, any of the following should work:

data.frame(keyName=names(testVect), value=testVect, row.names=NULL)

##        keyName      value
## 1   db_version 11.2.0.3.0
## 2 elapsed_time      12.89
## 3     cpu_time      12.71


## Suggested by @JWilliman
tibble::enframe(testVect)

## # A tibble: 3 x 2
##   name         value
##   <chr>        <chr>
## 1 db_version   11.2.0.3.0
## 2 elapsed_time 12.89
## 3 cpu_time     12.71


## Suggested by @Joe
stack(testVect)
##       values          ind
## 1 11.2.0.3.0   db_version
## 2      12.89 elapsed_time
## 3      12.71     cpu_time

Solution 3

I'm going to take a stab at this:

test.vector <- as.data.frame(t(testVect))
class(test.vector)

Solution 4

I used to use the functions suggested in these answers (as.list, as_tibble, t, enframe, etc.) but have since found out that dplyr::bind_rows now works to do exactly what the original question asks with a single function call.

library(dplyr)
testVect <- structure(c("11.2.0.3.0", "12.89", "12.71"), .Names = c("db_version", "elapsed_time", "cpu_time"))
testVect %>% bind_rows
#> # A tibble: 1 x 3
#>   db_version elapsed_time cpu_time
#>   <chr>      <chr>        <chr>   
#> 1 11.2.0.3.0 12.89        12.71

^{Created on 2019-11-10 by the reprex package (v0.3.0)}

As shown in tidyverse - prefered way to turn a named vector into a data.frame/tibble

Solution 5

named vector %>% as_tibble(.,rownames="column name of row.names")

View more solutions

Comments

[JelenaČuklina]

sad that there's no one-liner

[CoderGuy123]

Oddly, the tibble analog of this does not work: data_frame(as.list(testVect)) return a 5 row data frame.

[JWilliman]

@Deleet tibble will work with as_tibble(as.list(testVect)) or as_data_frame(as.list(testVect)) (as_data_frame is an alias for as_tibble).

[JWilliman]

Can also use tibble::enframe(testVect).

[merv]

In line with comments by @Deleet and @JWillliman, data.table(as.list(...)) does not work, but instead as.data.table(as.list(...)) does.

[Joe]

stack(testVect) also does this but leaves the values as characters.

[Agile Bean]

@Jelena-bioinf as one-liner with dplyr syntax, you can use testVect %>% as.list %>% as.data.frame %>% tidyr::gather() This actually produces the 'key' and 'value' columns which @dnlbrky referred to.

[tjebo]

Or even shorter, simply data.frame(t(testVect))

[AMS]

@Matthew Plourde Whether stringsAsFactors True or False, it gives the same data type. How to not change the data type?

[stevec]

stack(), what an underrated function!

[harmonica141]

Please add some explanation around here so everybody can learn. As of now a one-liner is a bit shallow.

[Amer]

Thank you, that was sueful!

[Esmu Igors]

Note to novices like me: this creates a df with 2 columns, one named "name" (with names as data, not row.names) and one named "value". I understand that OP meant precisely this but I came here searching for another but similar problem and spent a long time trying to figure out why this method produced strange results...

[tjebo]

stack is amazing, but it fails when your vector is not a vector ... ;) stackoverflow.com/questions/72398317/…