Question

我有一个带有横断面GPS位置的data.frame，这里只是其中的一小部分：

structure(list(X = c(432532.3435, 432533.3316, 432534.3198, 432535.3068, 
432536.339, 432528.3127, 432529.2956, 432530.2271, 432531.2019, 
432532.1574, 432533.1353, 432534.0987, 432703.2786, 432702.2761, 
432701.4092, 432700.3743, 432699.4523), Y = c(179892.6113, 179892.7918, 
179892.9953, 179893.2271, 179893.3646, 179931.3134, 179931.5124, 
179931.7763, 179932.0264, 179932.256, 179932.5104, 179932.7853, 
179432.1222, 179432.2754, 179432.5235, 179432.7024, 179432.9146
), plot_raai = c("F1", "F1", "F1", "F1", "F1", "F6", "F6", "F6", 
"F6", "F6", "F6", "F6", "A3", "A3", "A3", "A3", "A3")), .Names = c("X", 
"Y", "plot_raai"), row.names = c(1L, 2L, 3L, 4L, 5L, 200L, 201L, 
202L, 203L, 204L, 205L, 206L, 1039L, 1040L, 1041L, 1042L, 1043L
), class = "data.frame")

我想添加一个列，其中每行（位置）的距离到它的横断面的第一行。所以预期的结果将是

            X        Y plot_raai     dist
1    432532.3 179892.6        F1 0.000000
2    432533.3 179892.8        F1 1.004451
3    432534.3 179893.0        F1 2.013260
4    432535.3 179893.2        F1 3.026608
5    432536.3 179893.4        F1 4.065892
200  432528.3 179931.3        F6 0.000000
201  432529.3 179931.5        F6 1.002843
202  432530.2 179931.8        F6 1.969569
203  432531.2 179932.0        F6 2.975877
204  432532.2 179932.3        F6 3.958562
205  432533.1 179932.5        F6 4.968931
206  432534.1 179932.8        F6 5.970284
1039 432703.3 179432.1        A3 0.000000
1040 432702.3 179432.3        A3 1.014138
1041 432701.4 179432.5        A3 1.911988
1042 432700.4 179432.7        A3 2.961687
1043 432699.5 179432.9        A3 3.907489

以下是我的尝试：

 #created distance function (Pythagoras)
 distance <- function(x1,y1,x2,y2) {sqrt((x2-x1)^2+(y2-y1)^2)}

 #applied that to the rows with sapply (however, no grouping yet)
 sapply(2:nrow(mydf), function(x) distance(mydf$X[x],mydf$Y[x],mydf$X[1], mydf$Y[1]))

#then tried grouping using dplyr 
library(dplyr)
test1 <- mydf %>%
      group_by(., plot_raai) %>%
      mutate(dist = c(0,sapply(2:nrow(.), function(x) 
        distance(X[x],Y[x],X[1],Y[1]) )))

但是，这会计算到de数据帧中第一行的距离，而不是组中的第一行：

          X        Y plot_raai       dist
1  432532.3 179892.6        F1   0.000000
2  432533.3 179892.8        F1   1.004451
3  432534.3 179893.0        F1   2.013260
4  432535.3 179893.2        F1   3.026608
5  432536.3 179893.4        F1   4.065892
6  432528.3 179931.3        F6  38.911437
7  432529.3 179931.5        F6  39.020319
8  432530.2 179931.8        F6  39.222141
9  432531.2 179932.0        F6  39.431629
10 432532.2 179932.3        F6  39.645137
11 432533.1 179932.5        F6  39.906956
12 432534.1 179932.8        F6  40.212324
13 432703.3 179432.1        A3 491.191429
14 432702.3 179432.3        A3 490.699734
15 432701.4 179432.5        A3 490.167313
16 432700.4 179432.7        A3 489.643284
17 432699.5 179432.9        A3 489.128211

我知道这一定很简单，但我已经尝试了其他几种方式，而且我已经挣扎了一个多小时。任何人都可以帮我这个吗？

Answer 1

试试这个：

mydf%>%
  group_by(plot_raai)%>%
  mutate(dist=distance(first(X),first(Y),X,Y))

它给了我：

          X        Y plot_raai     dist
      <dbl>    <dbl>     <chr>    <dbl>
1  432532.3 179892.6        F1 0.000000
2  432533.3 179892.8        F1 1.004451
3  432534.3 179893.0        F1 2.013260
4  432535.3 179893.2        F1 3.026608
5  432536.3 179893.4        F1 4.065892
6  432528.3 179931.3        F6 0.000000
7  432529.3 179931.5        F6 1.002843
8  432530.2 179931.8        F6 1.969569
9  432531.2 179932.0        F6 2.975877
10 432532.2 179932.3        F6 3.958562
11 432533.1 179932.5        F6 4.968931
12 432534.1 179932.8        F6 5.970284
13 432703.3 179432.1        A3 0.000000
14 432702.3 179432.3        A3 1.014138
15 432701.4 179432.5        A3 1.911988
16 432700.4 179432.7        A3 2.961687
17 432699.5 179432.9        A3 3.907489

计算分组位置之间的距离（横断面）

1 个答案: