Convert Named Character Vector to Data.Frame

Convert Named Character Vector to data.frame

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).

How do I change a named vector to a data frame retaining the names?

As said by @Axeman in the comments, you have a named vector. The easiest way to transform that into a dataframe is by using the stack-function from base R.

Some example data:

output <- setNames(c(92.08,90.68,54.09,92.87,97.40), LETTERS[1:5])

which looks like:

> output
    A     B     C     D     E 
92.08 90.68 54.09 92.87 97.40

Transform into a dataframe with:

df <- stack(output)

which gives:

> df
  values ind
1  92.08   A
2  90.68   B
3  54.09   C
4  92.87   D
5  97.40   E

To get the columnnames and column order as specified in the question:

df <- setNames(stack(output)[2:1], c('Category','Value'))

which gives:

  Category Value
1        A 92.08
2        B 90.68
3        C 54.09
4        D 92.87
5        E 97.40

converting a vector into a dataframe columnwise

You can transpose the vector and convert it into dataframe/tibble.

t(x) %>% as_tibble()
t(x) %>% data.frame()

#  estimate ci.low ci.up
#1    0.595   0.11 2.004

create a data frame from named vector in R

We can use

vc <- c(1,2,5,6,7,9,12,15,18,19,27)
quartiles <- quantile(vc, probs = seq(0,1,0.01))
df <- data.frame(names(quartiles), quartiles, row.names = NULL)
colnames(df) <- c("quartile", "pvalue")
df$quartile <- as.numeric(gsub("%","", df$quartile))
> df
    quartile pvalue
1          0    1.0
2          1    1.1
3          2    1.2
4          3    1.3
5          4    1.4
6          5    1.5

Convert named vector to data frame using attribute values

You can use indexing by row/column value to do this efficiently:

row.nums <- as.numeric(sapply(strsplit(names(a), "\\."), "[", 1))
col.nums <- as.numeric(sapply(strsplit(names(a), "\\."), "[", 2))
mat <- matrix(NA, max(row.nums), max(col.nums))
mat[cbind(row.nums, col.nums)] <- a
mat
#      [,1]  [,2] 
# [1,] "foo" "bar"
# [2,] "dog" "cat"

Convert named vector to df with bind_rows

We can also use stack to convert named vector to dataframe and then use rename_all

stack(x)[2:1] %>% dplyr::rename_all(~c("Feature","Gain"))

#                Feature        Gain
#1              spend_7d 0.704766041
#2        d7_utility_sum 0.168417670
#3  recent_utility_ratio 0.101030461
#4                   IOS 0.009606189
#5  is_publisher_organic 0.008247149
#6 is_publisher_facebook 0.007932490

data

x <- c(spend_7d = 0.704766041126329, d7_utility_sum = 0.168417670114147, 
recent_utility_ratio = 0.101030461314979, IOS = 0.0096061893294683, 
is_publisher_organic = 0.008247148545793, is_publisher_facebook = 0.0079324895692841)

Convert character vector to data.frame after str_extract()

Maybe you can use gsub to remove double quotes from the string and "\n".

objeto <- gsub('"|\n|,', '', objeto)

which returns :

objeto
#[1] "CONTRATAÇÃO DE EMPRESA PARA A PRESTAÇÃO DE SERVIÇOS ESPECIALIZADOS DE AFERIÇÃO DO DESEMPENHO E DA QUALIDADE DA CONCESSIONÁRIA MINAS ARENA GESTÃO DE INSTALAÇÕES ESPORTIVAS S"

This can also be written as :

objeto <- gsub('["\n,]', '', objeto)

Or using str_remove_all from stringr library

stringr::str_remove_all(objeto, '[\n",]')

which is a shortcut for str_replace_all

stringr::str_replace_all(objeto, '[\n",]', '')

Rename a named vector when name has a match in dataframe

Simple

names(list1)=df$new_name[match(names(list1),df$ID)]

         3         49        163        223        360        429 
0.17492311 0.11435087 0.07615158 0.10201893 0.14140281 0.13680918

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