Converting a List of Data Frames into Individual Data Frames in R

Combine a list of data frames into one data frame by row

Use bind_rows() from the dplyr package:

bind_rows(list_of_dataframes, .id = "column_label")

Split a list into separate data frame in R

We can use imap to get the names and then use set_names

library(purrr)
library(dplyr)
library(stringr)
imap(list_a, ~ set_names(tibble(.x), .y)) %>%
        set_names(str_c("DF", 1:3)) %>% 
        list2env(.GlobalEnv)

DF1
# A tibble: 1 x 1
#  Banana
#   <dbl>
#1    8.7
DF2
# A tibble: 1 x 1
#  Strawberry
#       <dbl>
#1        2.3
DF3
# A tibble: 1 x 1
#  Apple
#  <dbl>
#1   3.5

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If we need separate columns

library(tibble)
enframe(list_a) %>% 
     unnest(c(value)) %>% 
     group_split(rn = row_number(), keep = FALSE) %>%
     set_names(str_c("DF", 1:3)) %>% 
     list2env(.GlobalEnv)
DF1
# A tibble: 1 x 2
#  name   value
#  <chr>  <dbl>
#1 Banana   8.7
DF2
# A tibble: 1 x 2
#  name       value
#  <chr>      <dbl>
#1 Strawberry   2.3
DF3
# A tibble: 1 x 2
#  name  value
#  <chr> <dbl>
#1 Apple   3.5

Combining a list of data frames into a new data frame in R

Note that in your list of dataframes (df_list) all the columns have different names (Area1, Area2, Area3) whereas in your output dataframe they all have been combined into one single column. So for that you need to change the different column names to the same one and bind the dataframes together.

library(dplyr)
library(purrr)

result <- map_df(df_list, ~.x %>% 
                  rename_with(~"Area", contains('Area')), .id = 'FileName')
result

#  FileName Area
#1 a1_areaX  100
#2 a2_areaX  200
#3 a3_areaX  300

converting list of data frames into single data frame

I am sure there is a better solution out there.
The problem you can't use rbind is because you have differing colnames (e.g. T1, T2, T3...)

So my idea was to name each df in the list with

  names(table_list1) <- banner
  Map(cbind, table_list1, SampleID = names(table_list1))

then add new names to each dataframe:

  colnames <- c("name","Median","Mean", "N") 
  
  for (i in seq_along(table_list1)){
    colnames(table_list1[[i]]) <- colnames
  }

Now you can use your code:

  t2 <- do.call(rbind,table_list1)
  t2

func1<-function(db,list_var,var_name_list,....){
  table_list1<-list()
  for (d in 1:length(df_list)) {
    
    table_list<-list()
    for (i in 1:length(list_var)) {
      
      
      table_list[[i]]<-sub_fun(db, list_var[i],var_name_list[i])
      
      t1 <- do.call(rbind,table_list)
      
    }
    
    colnames(t1)[1] <- banner[[d]]
    t1 <- t1 %>%
      add_row() %>%
      mutate_all(~replace(., is.na(.), ""))
    
    table_list1[[d]] <- t1
  }
  names(table_list1) <- banner
  Map(cbind, table_list1, SampleID = names(table_list1))
  
  
  colnames <- c("name","Median","Mean", "N") 
  
  for (i in seq_along(table_list1)){
    colnames(table_list1[[i]]) <- colnames
  }
  
  t2 <- do.call(rbind,table_list1)
  t2
}

  name Median       Mean  N
T1.1 klick      6     6.1875 32
T1.2 Nemar  196.3 230.721875 32
T1.3 Wingo    123   146.6875 32
T1.4                           
T2.1 klick      6     6.1875 32
T2.2 Nemar  196.3 230.721875 32
T2.3 Wingo    123   146.6875 32
T2.4                           
T3.1 klick      6     6.1875 32
T3.2 Nemar  196.3 230.721875 32
T3.3 Wingo    123   146.6875 32
T3.4             

Converting a list of list of data.frames to a single data.frame in R

We can use lapply/Map in base R. We can loop through the list with lapply, rbind the nested list elements, then create a new column with Map and rbind the outer list elements

out <- do.call(rbind, Map(cbind, lapply(L, function(x) 
              do.call(rbind, x)), id = seq_along(L)))
row.names(out) <- NULL
out
#   d SD id
#1  1  3  1
#2  2  4  1
#3  2  1  2
#4  3  2  2
#5  7  6  2
#6  8  7  2
#7  5  3  3
#8  6  4  3
#9  8  1  3
#10 9  2  3
#11 4  6  3
#12 5  7  3

---

Based on the comments, if we need to add another column from the names of the inner list

out1 <- do.call(rbind, Map(cbind, lapply(L, function(dat)
   do.call(rbind, Map(cbind, dat, es.type = names(dat)))), id = seq_along(L)))
row.names(out1) <- NULL

out1
#   d SD es.type id
#1  1  3   Short  1
#2  2  4   Short  1
#3  2  1   Short  2
#4  3  2   Short  2
#5  7  6   Long1  2
#6  8  7   Long1  2
#7  5  3   Short  3
#8  6  4   Short  3
#9  8  1   Long1  3
#10 9  2   Long1  3
#11 4  6   Long2  3
#12 5  7   Long2  3

---

If there are ..\\d+ and want to remove

out1 <- do.call(rbind, Map(cbind, lapply(L, function(dat)
   do.call(rbind, Map(cbind, dat, 
     es.type = sub("\\.*\\d+$", "", names(dat))))), id = seq_along(L)))
row.names(out1) <- NULL
out1
#   d SD es.type id
#1  1  3   Short  1
#2  2  4   Short  1
#3  2  1   Short  2
#4  3  2   Short  2
#5  7  6    Long  2
#6  8  7    Long  2
#7  5  3   Short  3
#8  6  4   Short  3
#9  8  1    Long  3
#10 9  2    Long  3
#11 4  6    Long  3
#12 5  7    Long  3

How can I convert a list of data frames into a data frame of lists using purrr or tidyr?

If I understand the output you want correctly, you can do

list %>% map_df(~.x %>% nest(cols=-ID))

Since nest requires a data.frame, you need to map over the data frames in the list to do the mapping. Then map_df will combine everything into one data.frame in the end.

Convert a list to a data frame

Update July 2020:

The default for the parameter stringsAsFactors is now default.stringsAsFactors() which in turn yields FALSE as its default.

---

Assuming your list of lists is called l:

df <- data.frame(matrix(unlist(l), nrow=length(l), byrow=TRUE))

The above will convert all character columns to factors, to avoid this you can add a parameter to the data.frame() call:

df <- data.frame(matrix(unlist(l), nrow=132, byrow=TRUE),stringsAsFactors=FALSE)

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