Tidyverse - Prefered Way to Turn a Named Vector into a Data.Frame/Tibble

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

This is now directly supported using bind_rows (introduced in dplyr 0.7.0):

library(tidyverse)) 
vec <- c("a" = 1, "b" = 2)

bind_rows(vec)
#> # A tibble: 1 x 2
#>       a     b
#>   <dbl> <dbl>
#> 1     1     2

This quote from https://cran.r-project.org/web/packages/dplyr/news.html explains the change:

bind_rows() and bind_cols() now accept vectors. They are treated as rows by the former and columns by the latter. Rows require inner names like c(col1 = 1, col2 = 2), while columns require outer names: col1 = c(1, 2). Lists are still treated as data frames but can be spliced explicitly with !!!, e.g. bind_rows(!!! x) (#1676).

With this change, it means that the following line in the use case example:

txt %>% map(read_xml) %>% map(xml_attrs) %>% map_df(~t(.) %>% as_tibble)

can be rewritten as

txt %>% map(read_xml) %>% map(xml_attrs) %>% map_df(bind_rows)

which is also equivalent to

txt %>% map(read_xml) %>% map(xml_attrs) %>% { bind_rows(!!! .) }

The equivalence of the different approaches is demonstrated in the following example:

library(tidyverse)
library(rvest)

txt <- c('<node a="1" b="2"></node>',
         '<node a="1" c="3"></node>')

temp <- txt %>% map(read_xml) %>% map(xml_attrs)

# x, y, and z are identical
x <- temp %>% map_df(~t(.) %>% as_tibble)
y <- temp %>% map_df(bind_rows)
z <- bind_rows(!!! temp)

identical(x, y)
#> [1] TRUE
identical(y, z)
#> [1] TRUE

z
#> # A tibble: 2 x 3
#>       a     b     c
#>   <chr> <chr> <chr>
#> 1     1     2  <NA>
#> 2     1  <NA>     3

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

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).

Converting grouped tibble to named list

We can use split

with(my_data, split(list_values,
     factor(list_names, levels = unique(list_names))))
$Ford
[1] "Ranger"   "F150"     "Explorer"

$Chevy
[1] "Equinox"

$Dodge
[1] "Caravan" "Ram"   

---

Or with unstack

unstack(my_data, list_values ~ list_names)
$Chevy
[1] "Equinox"

$Dodge
[1] "Caravan" "Ram"    

$Ford
[1] "Ranger"   "F150"     "Explorer"

Create empty tibble/data frame with column names coming from a vector

You can create a named vector, vec, where the first argument sets the type of column you want. The rep("", 3) line says I want three character columns. Then the second argument is the vector of column names.

Use dplyr::bind_rows to convert this into tibble with one row. Then [0, ] selects zero rows, leaving it empty.

With this method, you can control the data type for each column easily.

library(dplyr)

vec <- setNames(rep("", 3), letters[1:3])
bind_rows(vec)[0, ]

# A tibble: 0 x 3
# ... with 3 variables: a <chr>, b <chr>, c <chr>

You can also use as_tibble if you transpose the named vector. I guess I use bind_rows because I usually have dplyr loaded but not tibble.

library(tibble)

vec <- setNames(rep("", 3), letters[1:3])
as_tibble(t(vec))[0, ]

# A tibble: 0 x 3
# ... with 3 variables: a <chr>, b <chr>, c <chr>

If you know all of the columns are of a single type (e.g., character), you can do something like this.

vec <- letters[1:3]
df <- bind_rows(setNames(rep("", length(vec)), vec))[0, ]

How to create a one-row data frame from a vector in R?

We could use unnest_wider after returning the output in a list in summarise

library(dplyr)
library(tidyr)
mtcars %>%
    group_by(cyl) %>%
    summarise(out = list(boxplot.stats(wt)$stats)) %>% 
    unnest_wider(out) %>% 
    rename_at(-1, ~ str_replace(., '\\.+', 'x'))
# A tibble: 3 x 6
#    cyl    x1    x2    x3    x4    x5
#   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1     4  1.51  1.88  2.2   2.62  3.19
#2     6  2.62  2.82  3.22  3.44  3.46
#3     8  3.17  3.52  3.76  4.07  4.07

---

Or if we want to use the OP's method, then set the names for that vector and use as_tibble_row

library(purrr)
library(stringr)
mtcars %>% 
  group_by(cyl) %>% 
  group_map(~ tibble(cyl = first(.x$cyl), 
          setNames(boxplot.stats(.$wt)$stats, str_c('x', 1:5)) %>% 
             as_tibble_row) , .keep = TRUE) %>%
  bind_rows
# A tibble: 3 x 6
#    cyl    x1    x2    x3    x4    x5
#  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1     4  1.51  1.88  2.2   2.62  3.19
#2     6  2.62  2.82  3.22  3.44  3.46
#3     8  3.17  3.52  3.76  4.07  4.07

---

As the output of group_map is always a list, it may be better to use group_modify to return a tbl thus avoiding the last map_dfr/bind_rows

mtcars %>% 
  group_by(cyl) %>% 
  group_modify(~ setNames(boxplot.stats(.$wt)$stats, str_c('x', 1:5)) %>%
         as_tibble_row , .keep = TRUE) %>% 
  ungroup
# A tibble: 3 x 6
#    cyl    x1    x2    x3    x4    x5
#  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1     4  1.51  1.88  2.2   2.62  3.19
#2     6  2.62  2.82  3.22  3.44  3.46
#3     8  3.17  3.52  3.76  4.07  4.07

What is the tidyverse way to apply a function designed to take data.frames as input across a grouped tibble in R?

We can use the OP's function in group_modify

library(dplyr)
MyDF %>% 
   group_by(Fruit) %>% 
   group_modify(~ .x %>% 
       summarise(MyVal = myFun(.x))) %>%
   ungroup

-output

# A tibble: 2 × 2
  Fruit  MyVal
  <chr>  <int>
1 Apple  42925
2 Mango 295425

---

Or in group_map where the .y is the grouping column

MyDF %>% 
   group_by(Fruit) %>%
   group_map(~ bind_cols(.y, MyVal = myFun(.))) %>%
   bind_rows
# A tibble: 2 × 2
  Fruit  MyVal
  <chr>  <int>
1 Apple  42925
2 Mango 295425

Does the pipe opperator turn a data frame into a tibble? If so, how can I prevent it?

I think you just need to convert the object to data.frame before you can use renames_to_column:

gss_cat %>%
  group_by(race, partyid) %>%
  summarise(Freq = n()) %>%
  pivot_wider(names_from = race, values_from = Freq) %>%
  drop_na() %>%
  column_to_rownames("partyid") %>%
  chisq.test(.) %>%
  `[[`("residuals") %>% 
  as.data.frame() %>% 
  rownames_to_column("partyid")
#              partyid     Other       Black      White
# 1          No answer  2.923648   2.8649510  -2.262282
# 2        Other party -2.311635  -4.6581430   2.834128
# 3  Strong republican -8.950509 -15.3086303   9.782022
# 4 Not str republican -7.246500 -16.8275514   9.856563
# 5       Ind,near rep -3.546661 -10.4554526   5.793770
# 6        Independent 12.196368  -4.4484394  -2.272619
# 7       Ind,near dem  3.783087  -0.6287861  -1.033036
# 8   Not str democrat  5.478600   8.9946078  -5.823392
# 9    Strong democrat -5.339841  32.7172447 -12.447571

Convert any number of vectors into a dataframe whilst preserving data types and using vector names as column names in R

This works fine with data.frame. You just need to add the argument, stringsAsFactors=FALSE.

df = data.frame(var_a, var_b, var_c, stringsAsFactors = FALSE)
sapply(df, class)
      var_a       var_b       var_c 
"character"   "numeric"    "factor" 

Replacement of column values based on a named vector

You could use col :

df$col1 <- vec[as.character(df$col)]

Or in mutate :

library(dplyr)
df %>% mutate(col1 = vec[as.character(col)])
#   col col1 
#   <int> <chr>
# 1     1 a    
# 2     1 a    
# 3     1 a    
# 4     1 a    
# 5     2 b    
# 6     2 b    
# 7     3 c    
# 8     3 c    
# 9     3 c    
#10     3 c    
#11     3 c    

---

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