如何在 tibble 的每一行和数据框的嵌套列表上应用 function

Question

I have the following tibble and a nested list of data frames:

>source

# A tibble: 6 × 2
    lon   lat
  <dbl> <dbl>
1  6.02  55.1
2  6.02  55.0
3  6.02  54.9


>dest

[[1]][[1]]
         lon      lat
1   54.98908 6.900084
2   54.92777 6.772623
3   55.09501 6.911837

[[1]][[2]]
         lon      lat
1   54.98908 6.900084
2   54.92777 6.772623
3   55.09501 6.911837

[[1]][[3]]
         lon      lat
1   54.98908 6.900084
2   54.92777 6.772623
3   55.09501 6.911837

[[2]][[1]]
         lon      lat
1   54.98908 6.900084
2   54.92777 6.772623
3   55.09501 6.911837

[[2]][[2]]
         lon      lat
1   54.98908 6.900084
2   54.92777 6.772623
3   55.09501 6.911837

[[2]][[3]]
         lon      lat
1   54.98908 6.900084
2   54.92777 6.772623
3   55.09501 6.911837

I would like to apply a function on a row from a tible source and to each "block" from dest.

Example:

row 1 from source should by applied to each row from dest[[1]][[1]] and dest[[2]][[1]]

row 2 from source should by applied to each row from dest[[1]][[2]] and dest[[2]][[2]]

row 3 from source should by applied to each row from dest[[1]][[3]] and dest[[2]][[3]]

and so on.

How could I make this happen? I got tangled up with apply,lappl and maply and would appreciate any help.

source<-structure(list(lon = c(6.02125801226333, 6.02125801226333, 6.02125801226333, 
6.02125801226333, 6.02125801226333, 6.02125801226333), lat = c(55.0579432585625, 
54.9681151832365, 54.8782857724705, 54.7884550247254, 54.6986229384757, 
54.6087895122085)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame"))

dest<-list(list(structure(list(lon = c(55.0446726604773, 55.0911992769466, 
55.1399831259253), lat = c(6.11070373013145, 5.93718385855719, 
6.05909963519238)), class = "data.frame", row.names = c(NA, -3L
)), structure(list(lon = c(54.963042116042, 54.9238652445021, 
54.9948148730435), lat = c(6.11154210955708, 6.10009257140253, 
5.93487232950475)), class = "data.frame", row.names = c(NA, -3L
)), structure(list(lon = c(54.9181540526, 54.9628448755405, 54.8174082489187
), lat = c(5.94011737583315, 5.98947008604159, 6.08806491235748
)), class = "data.frame", row.names = c(NA, -3L)), structure(list(
    lon = c(54.7263291045393, 54.8728552727446, 54.8675223815364
    ), lat = c(5.95561986508533, 6.0534792303467, 5.97754320721106
    )), class = "data.frame", row.names = c(NA, -3L)), structure(list(
    lon = c(54.7185472365059, 54.7069293987346, 54.78280968399
    ), lat = c(5.93305860952388, 5.93121414118021, 5.9884946645099
    )), class = "data.frame", row.names = c(NA, -3L)), structure(list(
    lon = c(54.560413160877, 54.5853088068835, 54.5185005363673
    ), lat = c(6.0976246910947, 5.93394019791707, 6.02387338808233
    )), class = "data.frame", row.names = c(NA, -3L))), list(
    structure(list(lon = c(55.050226235055, 55.0240838617402, 
    54.9636263846607), lat = c(5.90235917535441, 5.90965086672992, 
    5.97880750058409)), class = "data.frame", row.names = c(NA, 
    -3L)), structure(list(lon = c(55.0746706563331, 55.0478637437921, 
    54.8541974469044), lat = c(5.98859383669152, 5.92618888252071, 
    6.04742105597978)), class = "data.frame", row.names = c(NA, 
    -3L)), structure(list(lon = c(54.7575000883344, 54.7676512681177, 
    54.9427732774055), lat = c(6.06061526193956, 6.09764527834345, 
    5.90903632630959)), class = "data.frame", row.names = c(NA, 
    -3L)), structure(list(lon = c(54.7776555082601, 54.8462348683655, 
    54.7620026570004), lat = c(6.1346781687426, 6.12031707754559, 
    5.91627897917598)), class = "data.frame", row.names = c(NA, 
    -3L)), structure(list(lon = c(54.6176186034159, 54.7833923796146, 
    54.6922873458308), lat = c(6.10088997672983, 6.09177636538747, 
    6.14915348430183)), class = "data.frame", row.names = c(NA, 
    -3L)), structure(list(lon = c(54.5680535136696, 54.5386600427152, 
    54.5879440622283), lat = c(6.13919150641202, 5.91144136237118, 
    5.89113937054887)), class = "data.frame", row.names = c(NA, 
    -3L))))

Answer 1

We could split the source into a list by rows, and then use mapply with lapply :

Example using dplyr::bind_cols as the function to be applied.

lapply(dest,
       \(x) mapply(dplyr::bind_cols, split(source, seq(nrow(source))), x, SIMPLIFY = FALSE))

Output :

[[1]]
[[1]]$`1`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    55.1    55.0    6.11
2    6.02    55.1    55.1    5.94
3    6.02    55.1    55.1    6.06

[[1]]$`2`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    55.0    55.0    6.11
2    6.02    55.0    54.9    6.10
3    6.02    55.0    55.0    5.93

[[1]]$`3`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.9    54.9    5.94
2    6.02    54.9    55.0    5.99
3    6.02    54.9    54.8    6.09

[[1]]$`4`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.8    54.7    5.96
2    6.02    54.8    54.9    6.05
3    6.02    54.8    54.9    5.98

[[1]]$`5`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.7    54.7    5.93
2    6.02    54.7    54.7    5.93
3    6.02    54.7    54.8    5.99

[[1]]$`6`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.6    54.6    6.10
2    6.02    54.6    54.6    5.93
3    6.02    54.6    54.5    6.02


[[2]]
[[2]]$`1`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    55.1    55.1    5.90
2    6.02    55.1    55.0    5.91
3    6.02    55.1    55.0    5.98

[[2]]$`2`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    55.0    55.1    5.99
2    6.02    55.0    55.0    5.93
3    6.02    55.0    54.9    6.05

[[2]]$`3`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.9    54.8    6.06
2    6.02    54.9    54.8    6.10
3    6.02    54.9    54.9    5.91

[[2]]$`4`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.8    54.8    6.13
2    6.02    54.8    54.8    6.12
3    6.02    54.8    54.8    5.92

[[2]]$`5`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.7    54.6    6.10
2    6.02    54.7    54.8    6.09
3    6.02    54.7    54.7    6.15

[[2]]$`6`
# A tibble: 3 × 4
  lon...1 lat...2 lon...3 lat...4
    <dbl>   <dbl>   <dbl>   <dbl>
1    6.02    54.6    54.6    6.14
2    6.02    54.6    54.5    5.91
3    6.02    54.6    54.6    5.89

Answer 2

A loop could do it (the function here is a simple addition):

for(each_row in 1:nrow(source)) {
  for(each_list in 1:length(dest)) {
    dest[[each_list]][[each_row]][["lon"]] <- dest[[each_list]][[each_row]][["lon"]]+source[[each_row, "lon"]]
    dest[[each_list]][[each_row]][["lat"]] <- dest[[each_list]][[each_row]][["lat"]]+source[[each_row, "lat"]]
  }
}

Output:

[[1]]
[[1]][[1]]
       lon      lat
1 61.06593 61.16865
2 61.11246 60.99513
3 61.16124 61.11704

[[1]][[2]]
       lon      lat
1 60.98430 61.07966
2 60.94512 61.06821
3 61.01607 60.90299

[[1]][[3]]
       lon      lat
1 60.93941 60.81840
2 60.98410 60.86776
3 60.83867 60.96635

[[1]][[4]]
       lon      lat
1 60.74759 60.74407
2 60.89411 60.84193
3 60.88878 60.76600

[[1]][[5]]
       lon      lat
1 60.73981 60.63168
2 60.72819 60.62984
3 60.80407 60.68712

[[1]][[6]]
       lon      lat
1 60.58167 60.70641
2 60.60657 60.54273
3 60.53976 60.63266


[[2]]
[[2]][[1]]
       lon      lat
1 61.07148 60.96030
2 61.04534 60.96759
3 60.98488 61.03675

[[2]][[2]]
       lon      lat
1 61.09593 60.95671
2 61.06912 60.89430
3 60.87546 61.01554

[[2]][[3]]
       lon      lat
1 60.77876 60.93890
2 60.78891 60.97593
3 60.96403 60.78732

[[2]][[4]]
       lon      lat
1 60.79891 60.92313
2 60.86749 60.90877
3 60.78326 60.70473

[[2]][[5]]
       lon      lat
1 60.63888 60.79951
2 60.80465 60.79040
3 60.71355 60.84778

[[2]][[6]]
       lon      lat
1 60.58931 60.74798
2 60.55992 60.52023
3 60.60920 60.49993

Answer 3

If I follow, each destination is someplace near the equator and each source is someplace in the north. For each destination block you want to add the source lat and long so you can do something like compute the distance between the two.

So the result should look something like:

> dest2[[1]][[1]]
       lon      lat  lon_src  lat_src
1 55.04467 6.110704 6.021258 55.05794
2 55.09120 5.937184 6.021258 55.05794
3 55.13998 6.059100 6.021258 55.05794

This code will accomplish this. The code could be more efficient if you are working with a large data set.

dest2 <- dest

addStart <- function(startRow, destElements, group) {
  start <- source[startRow, ]
  
  for (i in destElements) {
    rows007 <- nrow(dest[[i]][[group]])
    toadd = data.frame( matrix(rep(start, each = rows007), ncol = 2) )
    names(toadd) = c("lon_src","lat_src")
    dest2[[i]][[group]] <- cbind(dest[[i]][[group]],toadd)
  
  }
  return(dest2)
}

dest2 <- addStart(1, 1:2, 1)
dest2[[1]][[1]]
dest2[[2]][[1]]

dest2 <- addStart(2, 1:2, 2)
dest2[[1]][[2]]
dest2[[2]][[2]]

dest2 <- addStart(3, 1:2, 3)
dest2[[1]][[3]]
dest2[[2]][[3]]