我们可以将源代码按行split到列表中，然后将mapply与lapply一起使用：
使用dplyr::bind_cols作为要应用的函数的示例。

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

输出：

[[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

一个循环就可以做到这一点（这里的函数是一个简单的加法）：

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"]]
  }
}

输出量：

[[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

如果我跟着，每个目的地都在赤道附近，每个源都在北方，对于每个目的地，你要把源的纬度和经度相加，这样你就可以计算两者之间的距离。
因此，结果应类似于：

> 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

这段代码将完成这一点。如果你正在处理一个大的数据集，这段代码可能会更有效。

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]]