Convert List to Named List in R

Convert list to named list in R

So you want your list to have names:

NAME <- paste0("Degrees", 1:length(Degrees_df1))

Any of the following is OK:

names(Degrees_df1) <- NAME

attr(Degrees_df1, "names") <- NAME

Degrees_df1 <- "names<-"(Degrees_df1, NAME)

Degrees_df1 <- setNames(Degrees_df1, NAME)

Degrees_df1 <- structure(Degrees_df1, names = NAME)

But I think, the best thing is to give names when you create the list. For example, if you do:

list(1, 2, 3, 4)

the resulting list has no names. While if you do

list(a = 1, b= 2, c = 3, d = 4)

the resulting list has names.

If I am applying paste function over there I am getting error like this:
Error in assign(names(paste0("Degrees_df", i)), paste0("Degrees", 1:length(get(paste0("Degrees_df", : invalid first argument
Sorry, I wanted to modify your code inside for loop using paste function.

You possibly need this (not efficient):

df_i <- get(paste0("Degrees_df", i))   ## a local variable
names(df_i) <- paste0("Degrees", 1:length(df_i))    ## modify local variable
assign(paste0("Degrees_df", i), df_i)   ## write back and update

or (better):

df_i <- get(paste0("Degrees_df", i))   ## a local variable
assign(paste0("Degrees_df", i),
       setNames(df_i, paste0("Degrees", 1:length(df_i))))
# assign(paste0("Degrees_df", i),
#        "names<-"(df_i, paste0("Degrees", 1:length(df_i))))
# assign(paste0("Degrees_df", i),
#        structure(df_i, names = paste0("Degrees", 1:length(df_i))))

assign is used to assign value (or another variable) to a variable. It looks like your error code tries to assign names attributes. Note, the names of a list / data.frame is "attributes", not a variable, so you can not use assign to change them.

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"

How to turn a named vector into a named list, while grouping entries with the same name into one list element?

We can do this with splitting the vector by the names of the vector

tmp_n <- split(unname(tmp_v), names(tmp_v))
identical(tmp_n, tmp_l)
#[1] TRUE

Converting the names of a list elements as a variable in a data.frame

When I think of iterating over a vector I think of the map functions in purrr. In this case, map2() iterates over multiple arguments simultaneously. I grabbed the names of the list elements and stored them in a vector, n:

n <- names(List)

I then iterated over List and n at the same time, calling cbind() to add a variable (i.e., column) to each data frame:

map2(List, n, ~ cbind(.x, Name = .y))

Output:


$bar1
  study sd Name
1     A  1 bar1

$bar2
  study sd Name
1     B  2 bar2
2     C  3 bar2

$bar3
  study sd Name
1     Z  4 bar3

$bar4
  study sd Name
1     H  5 bar4

We can collapse List to a single data frame with a call to bind_rows():

result <- map2(List, n, ~ cbind(.x, Name = .y))
bind_rows(result)

Output:


  study sd Name
1     A  1 bar1
2     B  2 bar2
3     C  3 bar2
4     Z  4 bar3
5     H  5 bar4

Taken as one statement that would be:

map2(List, names(List), ~ cbind(.x, Name = .y)) %>% 
  bind_rows()

Convert named vector to list in R

Like I said in the comment, you can use split to create a list.

a.list <- split(a, names(a))
a.list <- lapply(a.list, unname)

A one-liner would be

a.list <- lapply(split(a, names(a)), unname)
#$I
#[1] 1 2 3 4
#
#$II
#[1] 5 6 7 8

EDIT.
Then, thelatemail posted a simplification of this in his comment. I've timed it using Devin King's way and it's not only simpler it's also 25% faster.

a.list <- split(unname(a),names(a))

r data.frame to a named list

One option is to deframe to create a named vector and then use as.list. to convert each element as a list element

library(tibble)
library(dplyr)
deframe(d) %>%
   as.list

Or in base R with split

with(d, split(value, key))

R Unlist named list into one string with preserving list names

You can use paste() to create the "=" separator and collapse with ";". names() allows you to access the names in the list.

UPDATED with @Dason's suggestion.

paste(names(nl),nl,sep="=",collapse=";" )

[1] "case1=master2;case2=5;case3=master;case4=345"

Convert named list of lists to dataframe using tidy approach

We may use bind_rows which have the .id that creates a new column from the names of the list

library(dplyr)
 bind_rows(df, .id = "fitname")
# A tibble: 4 × 6
  fitname   term        estimate std.error statistic p.value
  <chr>     <chr>          <dbl>     <dbl>     <dbl>   <dbl>
1 e1m1_fit  (Intercept)      2.7      0.03      88.0    0.01
2 e1m1_fit  log10(q)        -0.1      0.01     -15.6    0.01
3 e2m2a_fit (Intercept)      2.7      0.03      79.8    0.01
4 e2m2a_fit log10(q)        -0.1      0.01     -15.5    0.01

In addition, if the df list was created by looping with map, the _dfr can return a single tibble/data.frame with the .id specified as the name of the list

library(purrr)
map_dfr(yourlist, ~ yourfun(.x), .id = "fitname")

convert named list with mixed content to data frame

We can use stack from base R

stack(my_list)

According to ?stack

The stack function is used to transform data available as separate columns in a data frame or list into a single column that can be used in an analysis of variance model or other linear model. The unstack function reverses this operation.

Or with enframe

library(tidyverse)
enframe(my_list) %>% # creates the 'value' as a `list` column
   mutate(value = map(value, as.character)) %>% # change to single type
   unnest

Turn a dataframe into a named list of dataframes with names from a column in R

We can use:

library(tibble)
as.list(deframe(df))

Convert List to Named List in R