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))))
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