将命名字符向量转换为 data.frame

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最佳答案

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).

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I'm going to take a stab at this:

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

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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

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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

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

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Here's an example using tibble:

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