R 如何在 tibble 的每一行和数据框的嵌套列表上应用一个函数

发布于11月30日

我有以下Tibble和嵌套的数据框列表:

>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

我想对来自可用源的行和来自DEST的每个"块"应用一个函数.

示例:

从来源开始的row 1应从dest[[1]][[1]]和dest[[2]][[1]]应用到each row

从来源开始的row 2应从dest[[1]][[2]]和dest[[2]][[2]]应用到each row

从来源开始的row 3应从dest[[1]][[3]]和dest[[2]][[3]]应用到each row

诸若此类.

我怎么才能做到这一点呢？我被应用、LAPPL和MAPLY搞得焦头烂额，如果有任何帮助，我将不胜感激.

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))))

[[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

R 如何在 tibble 的每一行和数据框的嵌套列表上应用一个函数

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