My list has 12000 entries. Each entry consists of 16 columns and 8 rows. I would like to create a data frame for every single entry. I'm interested in 3 of the 16 columns (X,Y and Z coordinates)

I already tried this:

data_frame12000 <- as.data.frame(do.call(cbind, list_small_read_laz))

This and other functions only create one big data.frame with all the 16 columns for each entry.

Can anybody help me?

Thank You in advance!

CodePudding user response：

If I am correct, you have a list containing 12000 elements each cintaining a dataframe with 8r*16c. And I suppose the column names are the same for all list elements.

First select X, Y, Z columns from each entry element :

library(tidyverse) 
# assumming your list name is 'list_small_read_laz'
reduced_column <- map(list_small_read_laz,~ select(.,X,Y,Z))

Then combine all entries into a single dataframe:

df_reduced_column <- map_dfr(reduced_column, as.data.frame)

Hope this is what you are looking for.

CodePudding user response：

If you have a list of 12000 dataframes you can generate a list of dataframes with only the desired columns using lapply. Here is an example using mtcars:

cars1 <- mtcars
cars2 <- cars1
cars3 <- cars2
list1 <- list(cars1, cars2, cars3)

df_list <- lapply(list1, function(x) x[, c(2, 4, 6)]) # column numbers are used

final_df <- Reduce(rbind, df_list) # if you want all of the dataframes combined by rows