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

Matthew Plourde · Accepted Answer · 2013-05-29T14:19:24.493

100

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).

edited May 29 '13 at 14:19

answered May 29 '13 at 14:09

Matthew Plourde

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Oddly, the **tibble** analog of this does not work: `data_frame(as.list(testVect))` return a 5 row data frame. – CoderGuy123 Mar 19 '18 at 21:34
6

@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 Apr 02 '18 at 22:11
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In line with comments by @Deleet and @JWillliman, `data.table(as.list(...))` does not work, but instead `as.data.table(as.list(...))` does. – merv Jul 03 '18 at 20:57
@Matthew Plourde Whether stringsAsFactors True or False, it gives the same data type. How to not change the data type? – AMS Jul 19 '20 at 19:10

dnlbrky · Answer 2 · 2018-11-13T14:21:11.693

76

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

edited Nov 13 '18 at 14:21

answered Dec 20 '14 at 22:56

dnlbrky

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sad that there's no one-liner – JelenaČuklina Jan 31 '18 at 16:29
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Can also use `tibble::enframe(testVect)`. – JWilliman Apr 02 '18 at 22:18
4

`stack(testVect)` also does this but leaves the values as characters. – Joe Nov 10 '18 at 19:15
@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. – Agile Bean Nov 28 '18 at 03:46
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`stack()`, what an underrated function! – stevec Jul 25 '20 at 15:41
Thank you, that was sueful! – Amer Dec 02 '20 at 22:25
stack is amazing, but it fails when your vector is not a vector ... ;) https://stackoverflow.com/questions/72398317/stack-is-failing-on-date-vectors/72398318#72398318 – tjebo May 26 '22 at 21:43

score 22 · Answer 3 · answered May 29 '13 at 14:10

22

I'm going to take a stab at this:

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

answered May 29 '13 at 14:10

Jack Ryan

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Or even shorter, simply `data.frame(t(testVect))` – tjebo Mar 04 '19 at 18:06

Arthur Yip · Answer 4 · 2020-07-28T02:59:50.270

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

score 4 · Answer 5 · answered Oct 31 '20 at 08:43

4

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

answered Oct 31 '20 at 08:43

BatmanFan

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Please add some explanation around here so everybody can learn. As of now a one-liner is a bit shallow. – harmonica141 Oct 31 '20 at 18:22

score 2 · Answer 6 · answered Feb 21 '21 at 23:20

2

Here's an example using tibble:

named_vector_df = tibble(name = names(named_vector), value = named_vector)

answered Feb 21 '21 at 23:20

Oliver Oliver

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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... – Esmu Igors Oct 13 '21 at 21:17

Convert Named Character Vector to data.frame

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