Create a column based on the name of the element list that contain the data frame in R
An alternate approach is to collapse your list into a single data frame and use the name of the list as an additional column.
dplyr::bind_rows(list_df, .id = "meta_information")# # A tibble: 20 x 4# meta_information id x y# <chr> <int> <dbl> <dbl># 1 jan_2013 1 -1.09 0.877 # 2 jan_2013 2 0.136 0.828 # 3 jan_2013 3 -0.376 0.0376# 4 jan_2013 4 -0.793 0.780 # 5 jan_2013 5 0.259 0.179 # 6 jan_2013 6 0.971 0.556 # 7 jan_2013 7 -0.787 0.579 # 8 jan_2013 8 -0.294 0.563 # 9 jan_2013 9 0.331 0.896 # 10 jan_2013 10 -0.392 0.577 # 11 feb_2013 1 0.0139 0.0381# 12 feb_2013 2 0.640 0.0744# 13 feb_2013 3 0.813 0.270 # 14 feb_2013 4 -0.748 0.305 # 15 feb_2013 5 0.528 0.380 # 16 feb_2013 6 -0.627 0.832 # 17 feb_2013 7 -1.21 0.0529# 18 feb_2013 8 1.45 0.494 # 19 feb_2013 9 0.490 0.402 # 20 feb_2013 10 -0.765 0.531
If it is really necessary to keep the lists separate, we can use an indexed map from purrr
purrr::imap(list_df, ~mutate(.x, meta_information = .y))# $jan_2013# id x y meta_information# 1 1 -1.0867168 0.87674573 jan_2013# 2 2 0.1357794 0.82798892 jan_2013# 3 3 -0.3763973 0.03761698 jan_2013# 4 4 -0.7934503 0.77968454 jan_2013# 5 5 0.2586395 0.17917052 jan_2013# 6 6 0.9707220 0.55617247 jan_2013# 7 7 -0.7871748 0.57870521 jan_2013# 8 8 -0.2939041 0.56255010 jan_2013# 9 9 0.3307507 0.89646137 jan_2013# 10 10 -0.3917830 0.57723403 jan_2013# # $feb_2013# id x y meta_information# 1 1 0.01386418 0.03814336 feb_2013# 2 2 0.64030914 0.07435783 feb_2013# 3 3 0.81281978 0.26987216 feb_2013# 4 4 -0.74768467 0.30482967 feb_2013# 5 5 0.52820991 0.38045027 feb_2013# 6 6 -0.62720336 0.83191998 feb_2013# 7 7 -1.20532079 0.05291640 feb_2013# 8 8 1.45277032 0.49355127 feb_2013# 9 9 0.48985425 0.40229656 feb_2013# 10 10 -0.76508432 0.53114667 feb_2013
I found a way to do the task with purrr::map2
iterating over two arguments in parallel: list_df
and the names(list_df)
. Then an anonymous function used these two arguments, taking a data frame (df
) and creating a constant column based on the name of the element (name_elem_contain_df
) that contain the data frame (df
)
purrr::map2(list_df, names(list_df), function(df, name_elem_contain_df) mutate(df, meta_information = name_elem_contain_df))