[英]Calculate nearest point coordinate and distance by differing years between two data frames using sf
[英]Distance to nearest point by group using sf
我有一个看起来与下面的示例相似的数据集。 对于每个code
我想计算到与它属于同一area
的下一个最近代码的距离。 因此,在我的示例中,对于属于区域A001的每个代码,我将在数据集中的另一列之后,该列包含与属于区域A001的其他点之一的最小距离。 我认为应该有一种使用st_distance
来实现这一目标?
require("data.table")
require("sf")
dt1 <- data.table(
code=c("A00111", "A00112","A00113","A00211","A00212","A00213","A00214","A00311","A00312"),
area=c("A001", "A001","A001","A002","A002","A002","A002","A003","A003"),
x=c(325147,323095,596020,257409,241206,248371,261076,595218,596678),
y=c(286151,284740,335814,079727,084266,078283,062045,333889,337836))
sf1 <- st_as_sf(dt1, coords = c("x","y"), crs=27700, na.fail=FALSE)
到达这里可能会有一种“更清洁”的方法,但这可以为您提供正确的值。
library(tidyverse)
# intermediate fun to help later in apply()
smallest_non_zero <- function(x) {
min_val <- min(x[x != 0])
x[match(min_val, x)]
}
closest_grp_distances <- sf1 %>%
group_split(area) %>%
map(~st_distance(., .) %>% # returns matrix
apply(1, smallest_non_zero)) %>%
unlist()
sf1$closest_grp_distances <- closest_grp_distances
我想使用baseR split
但是它没有用于sf
对象的方法。
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