在 R 中列出每 n 个具有序列的列

Question

Sorry for my probably easy to solve question. I have a dataframe of 400 columns and 17532 rows and I want to unlist this to create another dataframe of 8 columns and 876600 rows. Basically unlist from 1 to 50, 51 to 100, 101 to 150 etc... However I'm running into some problems:

flw<-data.frame(matrix(NA, ncol=8, nrow=876600)) ## create an empty dataframe

seq(1,length(fl),by=50) ## sequence of columns which effectively is  '1  51 101 151 201 251 301 351' with length(fl)=400

for (i in seq(1,length(fl),by=50) ){
  flw[i] <- as.data.frame(unlist(fl[i:(i+49)]))
}

I get the error:

Error in `[<-.data.frame`(`*tmp*`, i, value = list(`unlist(fl[(i):(i + 49)])` = c(13.24512,  : 
  new columns would leave holes after existing columns

I don't understand why as it shouldn't leave any holes. It should unlist from 1 to 50, and then 51 to 100 etc and this will be 8 columns x 876600. What am I missing?

Answer 1

Your problem is the indexing into the flw . You are using the same index that you are using for fl . You end up trying to assigning to row 1, then row 51. Your holes are rows 2:49.

If you're not tidyverse adverse this can be done without the loop using the purrr and dplyr

purrr::map_dfc(
  seq(1,length(fl),by=50),
  ~dplyr::select(fl, .x:(.x+49)) %>% unlist(),
)