【发布时间】:2019-10-28 04:50:23
【问题描述】:
我有一个 sf 数据框,其中包含标记沿许多 单向 街道的交叉点位置的点。除了几何列之外,一列包含街道名称,另一列包含交叉点在单向街道上的相对位置。
下面是一个玩具示例。第一排是拱街的第一个路口,第二排是拱街的第二个路口,以此类推
library(sf)
intersections <- structure(list(street = c("ARCH ST", "ARCH ST", "ARCH ST", "SANSOM ST",
"SANSOM ST", "SANSOM ST"), number = c(1L, 2L, 3L, 1L, 2L, 3L),
geometry = structure(list(structure(c(2699665.2606043, 236074.947200272
), class = c("XY", "POINT", "sfg")), structure(c(2699402.74765515,
236109.729280198), class = c("XY", "POINT", "sfg")), structure(c(2699202.95996668,
236136.613760229), class = c("XY", "POINT", "sfg")), structure(c(2699431.38476158,
234437.663731016), class = c("XY", "POINT", "sfg")), structure(c(2699162.09261096,
234476.514355583), class = c("XY", "POINT", "sfg")), structure(c(2697100.77148795,
234809.605567052), class = c("XY", "POINT", "sfg"))), precision = 0, bbox = structure(c(xmin = 2697100.77148795,
ymin = 234437.663731016, xmax = 2699665.2606043, ymax = 236136.613760229
), class = "bbox"), crs = structure(list(epsg = 2272L, proj4string = "+proj=lcc +lat_1=40.96666666666667 +lat_2=39.93333333333333 +lat_0=39.33333333333334 +lon_0=-77.75 +x_0=600000 +y_0=0 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=us-ft +no_defs"), class = "crs"), n_empty = 0L, class = c("sfc_POINT",
"sfc"))), row.names = c(NA, -6L), class = c("sf", "tbl_df",
"tbl", "data.frame"), sf_column = "geometry", agr = structure(c(street = NA_integer_,
number = NA_integer_), class = "factor", .Label = c("constant",
"aggregate", "identity")))
> intersections
Simple feature collection with 6 features and 2 fields
geometry type: POINT
dimension: XY
bbox: xmin: 2697101 ymin: 234437.7 xmax: 2699665 ymax: 236136.6
epsg (SRID): 2272
proj4string: +proj=lcc +lat_1=40.96666666666667 +lat_2=39.93333333333333 +lat_0=39.33333333333334 +lon_0=-77.75 +x_0=600000 +y_0=0 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=us-ft +no_defs
# A tibble: 6 x 3
street number geometry
<chr> <int> <POINT [US_survey_foot]>
1 ARCH ST 1 (2699665 236074.9)
2 ARCH ST 2 (2699403 236109.7)
3 ARCH ST 3 (2699203 236136.6)
4 SANSOM ST 1 (2699431 234437.7)
5 SANSOM ST 2 (2699162 234476.5)
6 SANSOM ST 3 (2697101 234809.6)
使用mapsapi 包中的mp_matrix() 和mp_get_matrix(),我想添加一列,显示从每个路口到该街道上的下一个路口的行驶时间(最后一个路口除外,它会得到一个不)。
理想情况下,它应该如下所示:
street number travel_time_sec geometry
1 ARCH ST 1 210 POINT (2699665 236074.9)
2 ARCH ST 2 180 POINT (2699403 236109.7)
3 ARCH ST 3 NA POINT (2699203 236136.6)
4 SANSOM ST 1 150 POINT (2699431 234437.7)
5 SANSOM ST 2 175 POINT (2699162 234476.5)
6 SANSOM ST 3 NA POINT (2697101 234809.6)
如何按组(即街道)循环遍历 sf 数据帧中的行,告诉每一行与该组中的下一行执行操作以填充新列,如果没有则返回 NA下一行存在吗?
最后,由于mp_matrix() 调用Google Maps API 需要花钱,请改用sf 中的st_distance() 函数生成以下内容。
street number travel_distance geometry
1 ARCH ST 1 576 POINT (2699665 236074.9)
2 ARCH ST 2 397 POINT (2699403 236109.7)
3 ARCH ST 3 NA POINT (2699203 236136.6)
4 SANSOM ST 1 410 POINT (2699431 234437.7)
5 SANSOM ST 2 440 POINT (2699162 234476.5)
6 SANSOM ST 3 NA POINT (2697101 234809.6)
非常感谢您的帮助。
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