Question

在下面的数据（包含在dput中）中，我对三个人（IndIDII）进行了重复观察（经纬度）。请注意，每个人都有不同数量的位置。

> Dat
  IndIDII      IndYear  WintLat  WintLong
1 BHS_265 BHS_265-2015 47.61025 -112.7210
2 BHS_265 BHS_265-2016 47.59884 -112.7089
3 BHS_770 BHS_770-2016 42.97379 -109.0400
4 BHS_770 BHS_770-2017 42.97129 -109.0367
5 BHS_770 BHS_770-2018 42.97244 -109.0509
6 BHS_377 BHS_377-2015 43.34744 -109.4821
7 BHS_377 BHS_377-2016 43.35559 -109.4445
8 BHS_377 BHS_377-2017 43.35195 -109.4566
9 BHS_377 BHS_377-2018 43.34765 -109.4892

我想为每个人计算连续点之间的欧式距离。我最初的工作是使用dplyr在lead()中工作，如下所示。 distm函数需要一个矩阵，我无法在dplyr中创建矩阵。可以生成矩阵并将其用作distm的参数吗？

Dat %>% 
  group_by(IndIDII) %>% 
  mutate(WitnGeoDist = distm(as.matrix(c("WintLong", "WintLat")), lead(as.matrix(c("WintLong", "WintLat"))), fun = distVincentyEllipsoid))

或者，还有其他可能性吗？提前谢谢了。

数据：

Dat <- structure(list(IndIDII = c("BHS_265", "BHS_265", "BHS_770", "BHS_770", 
"BHS_770", "BHS_377", "BHS_377", "BHS_377", "BHS_377"), IndYear = c("BHS_265-2015", 
"BHS_265-2016", "BHS_770-2016", "BHS_770-2017", "BHS_770-2018", 
"BHS_377-2015", "BHS_377-2016", "BHS_377-2017", "BHS_377-2018"
), WintLat = c(47.6102519805014, 47.5988417247191, 42.9737859090909, 
42.9712914772727, 42.9724390816327, 43.3474354347826, 43.3555934579439, 
43.3519543396226, 43.3476466990291), WintLong = c(-112.720994832869, 
-112.708887595506, -109.039964727273, -109.036693522727, -109.050923061224, 
-109.482114456522, -109.444522149533, -109.45659254717, -109.489241553398
)), class = "data.frame", row.names = c(NA, -9L))

Answer 1

这是一种sf和tidyverse的方法，尽管我认为这不是最干净的方法。我无法让geosphere::distm来优雅地处理缺失值（这会让我们使用group_by），所以我只能将split与st_distance一起使用。

这些步骤基本上是将坐标转换为点几何，在分组列上拆分以创建数据框列表，使用添加距离列的函数在此列表上进行映射，然后rbind将数据框放回原处

Dat <- structure(list(IndIDII = c("BHS_265", "BHS_265", "BHS_770", "BHS_770", "BHS_770", "BHS_377", "BHS_377", "BHS_377", "BHS_377"), IndYear = c("BHS_265-2015", "BHS_265-2016", "BHS_770-2016", "BHS_770-2017", "BHS_770-2018", "BHS_377-2015", "BHS_377-2016", "BHS_377-2017", "BHS_377-2018"), WintLat = c(47.6102519805014, 47.5988417247191, 42.9737859090909, 42.9712914772727, 42.9724390816327, 43.3474354347826, 43.3555934579439, 43.3519543396226, 43.3476466990291), WintLong = c(-112.720994832869, -112.708887595506, -109.039964727273, -109.036693522727, -109.050923061224, -109.482114456522, -109.444522149533, -109.45659254717, -109.489241553398)), class = "data.frame", row.names = c(NA, -9L))
library(tidyverse)
library(sf)

Dat %>%
  st_as_sf(coords = c("WintLong", "WintLat"), crs = 4326, remove = FALSE) %>%
  split(.$IndIDII) %>%
  map(function(df){
    dist <- st_distance(df[2:nrow(df), ], df[1:(nrow(df)- 1), ], by_element = TRUE)
    df %>% mutate(WitnGeoDist = c(NA, dist))
  }) %>%
  invoke(rbind, .x = .)
#> Simple feature collection with 9 features and 5 fields
#> geometry type:  POINT
#> dimension:      XY
#> bbox:           xmin: -112.721 ymin: 42.97129 xmax: -109.0367 ymax: 47.61025
#> epsg (SRID):    4326
#> proj4string:    +proj=longlat +datum=WGS84 +no_defs
#>           IndIDII      IndYear  WintLat  WintLong WitnGeoDist
#> BHS_265.1 BHS_265 BHS_265-2015 47.61025 -112.7210          NA
#> BHS_265.2 BHS_265 BHS_265-2016 47.59884 -112.7089   1561.4776
#> BHS_377.1 BHS_377 BHS_377-2015 43.34744 -109.4821          NA
#> BHS_377.2 BHS_377 BHS_377-2016 43.35559 -109.4445   3179.6929
#> BHS_377.3 BHS_377 BHS_377-2017 43.35195 -109.4566   1058.7986
#> BHS_377.4 BHS_377 BHS_377-2018 43.34765 -109.4892   2689.9938
#> BHS_770.1 BHS_770 BHS_770-2016 42.97379 -109.0400          NA
#> BHS_770.2 BHS_770 BHS_770-2017 42.97129 -109.0367    384.7117
#> BHS_770.3 BHS_770 BHS_770-2018 42.97244 -109.0509   1167.7996
#>                             geometry
#> BHS_265.1  POINT (-112.721 47.61025)
#> BHS_265.2 POINT (-112.7089 47.59884)
#> BHS_377.1 POINT (-109.4821 43.34744)
#> BHS_377.2 POINT (-109.4445 43.35559)
#> BHS_377.3 POINT (-109.4566 43.35195)
#> BHS_377.4 POINT (-109.4892 43.34765)
#> BHS_770.1   POINT (-109.04 42.97379)
#> BHS_770.2 POINT (-109.0367 42.97129)
#> BHS_770.3 POINT (-109.0509 42.97244)

由reprex package（v0.2.0）于2018-09-19创建。

Answer 2

这是另一种方法，可以更好地利用group_by并通过使用geosphere::distm使purrr::possibly正常工作。这样，我们就可以为NA填写距离没有意义的行，因为没有以前的值可以使用。

Dat <- structure(list(IndIDII = c("BHS_265", "BHS_265", "BHS_770", "BHS_770", "BHS_770", "BHS_377", "BHS_377", "BHS_377", "BHS_377"), IndYear = c("BHS_265-2015", "BHS_265-2016", "BHS_770-2016", "BHS_770-2017", "BHS_770-2018", "BHS_377-2015", "BHS_377-2016", "BHS_377-2017", "BHS_377-2018"), WintLat = c(47.6102519805014, 47.5988417247191, 42.9737859090909, 42.9712914772727, 42.9724390816327, 43.3474354347826, 43.3555934579439, 43.3519543396226, 43.3476466990291), WintLong = c(-112.720994832869, -112.708887595506, -109.039964727273, -109.036693522727, -109.050923061224, -109.482114456522, -109.444522149533, -109.45659254717, -109.489241553398)), class = "data.frame", row.names = c(NA, -9L))
library(tidyverse)
poss_dist <- possibly(geosphere::distm, otherwise = NA)
Dat %>%
  nest(WintLong, WintLat, .key = "coords") %>%
  group_by(IndIDII) %>%
  mutate(prev_coords = lag(coords)) %>%
  ungroup() %>%
  mutate(WitnGeoDist = map2_dbl(coords, prev_coords, poss_dist))
#> # A tibble: 9 x 5
#>   IndIDII IndYear      coords              prev_coords         WitnGeoDist
#>   <chr>   <chr>        <list>              <list>                    <dbl>
#> 1 BHS_265 BHS_265-2015 <data.frame [1 x 2~ <lgl [1]>                   NA 
#> 2 BHS_265 BHS_265-2016 <data.frame [1 x 2~ <data.frame [1 x 2~       1561.
#> 3 BHS_770 BHS_770-2016 <data.frame [1 x 2~ <lgl [1]>                   NA 
#> 4 BHS_770 BHS_770-2017 <data.frame [1 x 2~ <data.frame [1 x 2~        385.
#> 5 BHS_770 BHS_770-2018 <data.frame [1 x 2~ <data.frame [1 x 2~       1168.
#> 6 BHS_377 BHS_377-2015 <data.frame [1 x 2~ <lgl [1]>                   NA 
#> 7 BHS_377 BHS_377-2016 <data.frame [1 x 2~ <data.frame [1 x 2~       3180.
#> 8 BHS_377 BHS_377-2017 <data.frame [1 x 2~ <data.frame [1 x 2~       1059.
#> 9 BHS_377 BHS_377-2018 <data.frame [1 x 2~ <data.frame [1 x 2~       2690.

由reprex package（v0.2.0）于2018-09-19创建。

计算分组数据中点之间的欧式距离

2 个答案: