通过共同 ID 从一个城市到其他城市的平均距离
Mean distance from one city to others by a mutual ID
我完全不明白spatial.data。我一直在学习,但我错过了一些东西。
我有: data.frame enterprises
列:id, parent_subsidiary, city_cod.
我需要:parent所在城市到附属城市的平均距离和最大距离
例如:
id | mean_dist | max_dist
1111 | 25km | 50km
232 | 110km | 180km
333 | 0km | 0km
我做了什么 :
library("tidyverse")
library("sf")
# library("brazilmaps") not working anymore
library("geobr")
parent <- enterprises %>% filter(parent_subsidiary==1)
subsidiary <- enterprises %>% filter(parent_subsidiary==2)
# Cities - polygons
m_city_br <- read_municipality(code_muni="all", year=2019)
# or shp_city<- st_read("/BR_Municipios_2019.shp")
# data.frame with the column geom
map_parent <- left_join(parent, m_city_br, by=c("city_cod"="code_muni"))
map_subsidiary <- left_join(subsidiary, m_city_br, by=c("city_cod"="code_muni"))
st_distance(map_parent$geom[1],map_subsidiary$geom[2]) %>% units::set_units(km)
# it took a long time and the result is different from google.maps
# is it ok?!
# To do by ID -- I also stucked here
distance_p_s <- data.frame(id=as.numeric(),subsidiar=as.numeric(),mean_dist=as.numeric(),max_dist=as.numeric())
id_v <- as.vector(parent$id)
for (i in 1:length(id_v)){
test_p <- map_parent %>% filter(id==id_v[i])
test_s <- map_subsidiary %>% filter(id==id_v[i])
total <- 0
value <- 0
max <- 0
l <- 0
l <- nrow(test_s)
for (j in 1:l){
value <- as.numeric(round(st_distance(test_p$geom[1],test_s$geom[j]) %>% units::set_units(km),2))
total <- total + value
ifelse(value>max,max<-value,NA)
}
mean_dist <- total/l
done <- data.frame(id=id[i],subsidiary=l,mean_dist=round(mean_dist,2),max_dist=max)
distance_p_s <- rbind(distance_p_s,done)
rm(done)
}
}
对吗?
我可以计算城市的质心然后计算距离吗?
我知道从code_muni==4111407到code_muni==4110102的距离是0,但是是另一个城市(Imbituva, PR,Brasil - Ivaí, PR,Brasil ).为什么?
数据示例: structure(list(id = c("1111", "1111", "1111", "1111", "232", "232", "232", "232", "3123", "3123", "4455", "4455", "686", "333", "333", "14112", "14112", "14112", "3633", "3633"), parent_subsidiary = c("1","2", "2", "2", "1", "2", "2", "2", "1", "2", "1", "2", "1", "2", "1", "1", "2", "2", "1", "2"), city_cod = c(4305801L,4202404L, 4314803L, 4314902L, 4318705L, 1303403L, 4304507L, 4314100L, 2408102L, 3144409L, 5208707L, 4205407L, 5210000L, 3203908L, 3518800L, 3118601L, 4217303L, 3118601L, 5003702L, 5205109L)), row.names = c(NA, 20L), class = "data.frame")
PS: 这是巴西城市
https://github.com/ipeaGIT/geobr/tree/master/r-package
大问题。我看了一会儿。然后想了想又回来看了一些。没有计算平均值。仅确定母公司与其子公司之间的距离。
数据已绑定 - 城市数据和数据框数据。然后新的 df 被变异以添加表面上每个点的质心数据。
df 按 id 拆分,结果是一个包含 8 个 df 的列表。每个 df 都包含独立的母公司和相关子公司。 (1:4, 1:3, 1:4, 1:2, ....)
一个带有函数的循环清理了 8 个 df,并计算了每个母公司到每个子公司的距离。
我根据与网站的距离值检查了列表中第一个 df 的距离。 df1的距离和网站的距离几乎一致。
输出显示在[link]
我做了类似的事情:
distance_p_s <- data.frame(id=as.character(),
qtd_subsidiary=as.numeric(),
dist_min=as.numeric(),
dist_media=as.numeric(),
dist_max=as.numeric())
id <- as.vector(mparentid$id)
for (i in 1:length(id)){
eval(parse(text=paste0("
print('Filtering id: ",id[i]," (",i," of ",length(id),")')
")))
teste_m <- mparentid %>% filter(id==id[i]) %>% st_as_sf()
teste_f <- msubsidiaryid %>% filter(id==id[i]) %>% st_as_sf()
teste_f <- st_centroid(teste_f)
teste_m <- st_centroid(teste_m)
teste_f = st_transform(teste_f, 4674)
teste_m = st_transform(teste_m, 4674)
total <- 0
value <- 0
min <- 0
max <- 0
l <- 0
l <- nrow(teste_f)
for (j in 1:l){
eval(parse(text=paste0("
print('Tratando id: ",id[i]," (",i," de ",length(id),"), subsidiary: ",j," de ",l,"')
")))
value <- as.numeric(round(st_distance(teste_m$geom[1],teste_f$geom[j]) %>% units::set_units(km),2))
total <- total + value
ifelse(value>max,max<-value,NA)
if(j==1){
min<-value
} else {
ifelse(value<min,min<-value,NA)}
}
dist_med <- total/l
done <- data.frame(id=id[i],qtd_subsidiary=l,dist_min=min,dist_media=round(dist_med,2),dist_max=max)
distance_p_s <- rbind(distance_p_s,done)
eval(parse(text=paste0("
print('Concluido id: ",id[i]," (",i," de ",length(id),"), subsidiary: ",j," de ",l,"')
")))
rm(done)
}
可能这不是最好的方法,但它暂时解决了我的问题。
我完全不明白spatial.data。我一直在学习,但我错过了一些东西。
我有: data.frame enterprises
列:id, parent_subsidiary, city_cod.
我需要:parent所在城市到附属城市的平均距离和最大距离
例如:
id | mean_dist | max_dist
1111 | 25km | 50km
232 | 110km | 180km
333 | 0km | 0km
我做了什么 :
library("tidyverse")
library("sf")
# library("brazilmaps") not working anymore
library("geobr")
parent <- enterprises %>% filter(parent_subsidiary==1)
subsidiary <- enterprises %>% filter(parent_subsidiary==2)
# Cities - polygons
m_city_br <- read_municipality(code_muni="all", year=2019)
# or shp_city<- st_read("/BR_Municipios_2019.shp")
# data.frame with the column geom
map_parent <- left_join(parent, m_city_br, by=c("city_cod"="code_muni"))
map_subsidiary <- left_join(subsidiary, m_city_br, by=c("city_cod"="code_muni"))
st_distance(map_parent$geom[1],map_subsidiary$geom[2]) %>% units::set_units(km)
# it took a long time and the result is different from google.maps
# is it ok?!
# To do by ID -- I also stucked here
distance_p_s <- data.frame(id=as.numeric(),subsidiar=as.numeric(),mean_dist=as.numeric(),max_dist=as.numeric())
id_v <- as.vector(parent$id)
for (i in 1:length(id_v)){
test_p <- map_parent %>% filter(id==id_v[i])
test_s <- map_subsidiary %>% filter(id==id_v[i])
total <- 0
value <- 0
max <- 0
l <- 0
l <- nrow(test_s)
for (j in 1:l){
value <- as.numeric(round(st_distance(test_p$geom[1],test_s$geom[j]) %>% units::set_units(km),2))
total <- total + value
ifelse(value>max,max<-value,NA)
}
mean_dist <- total/l
done <- data.frame(id=id[i],subsidiary=l,mean_dist=round(mean_dist,2),max_dist=max)
distance_p_s <- rbind(distance_p_s,done)
rm(done)
}
}
对吗? 我可以计算城市的质心然后计算距离吗?
我知道从code_muni==4111407到code_muni==4110102的距离是0,但是是另一个城市(Imbituva, PR,Brasil - Ivaí, PR,Brasil ).为什么?
数据示例: structure(list(id = c("1111", "1111", "1111", "1111", "232", "232", "232", "232", "3123", "3123", "4455", "4455", "686", "333", "333", "14112", "14112", "14112", "3633", "3633"), parent_subsidiary = c("1","2", "2", "2", "1", "2", "2", "2", "1", "2", "1", "2", "1", "2", "1", "1", "2", "2", "1", "2"), city_cod = c(4305801L,4202404L, 4314803L, 4314902L, 4318705L, 1303403L, 4304507L, 4314100L, 2408102L, 3144409L, 5208707L, 4205407L, 5210000L, 3203908L, 3518800L, 3118601L, 4217303L, 3118601L, 5003702L, 5205109L)), row.names = c(NA, 20L), class = "data.frame")
PS: 这是巴西城市 https://github.com/ipeaGIT/geobr/tree/master/r-package
大问题。我看了一会儿。然后想了想又回来看了一些。没有计算平均值。仅确定母公司与其子公司之间的距离。
数据已绑定 - 城市数据和数据框数据。然后新的 df 被变异以添加表面上每个点的质心数据。
df 按 id 拆分,结果是一个包含 8 个 df 的列表。每个 df 都包含独立的母公司和相关子公司。 (1:4, 1:3, 1:4, 1:2, ....)
一个带有函数的循环清理了 8 个 df,并计算了每个母公司到每个子公司的距离。
我根据与网站的距离值检查了列表中第一个 df 的距离。 df1的距离和网站的距离几乎一致。
输出显示在[link]
我做了类似的事情:
distance_p_s <- data.frame(id=as.character(),
qtd_subsidiary=as.numeric(),
dist_min=as.numeric(),
dist_media=as.numeric(),
dist_max=as.numeric())
id <- as.vector(mparentid$id)
for (i in 1:length(id)){
eval(parse(text=paste0("
print('Filtering id: ",id[i]," (",i," of ",length(id),")')
")))
teste_m <- mparentid %>% filter(id==id[i]) %>% st_as_sf()
teste_f <- msubsidiaryid %>% filter(id==id[i]) %>% st_as_sf()
teste_f <- st_centroid(teste_f)
teste_m <- st_centroid(teste_m)
teste_f = st_transform(teste_f, 4674)
teste_m = st_transform(teste_m, 4674)
total <- 0
value <- 0
min <- 0
max <- 0
l <- 0
l <- nrow(teste_f)
for (j in 1:l){
eval(parse(text=paste0("
print('Tratando id: ",id[i]," (",i," de ",length(id),"), subsidiary: ",j," de ",l,"')
")))
value <- as.numeric(round(st_distance(teste_m$geom[1],teste_f$geom[j]) %>% units::set_units(km),2))
total <- total + value
ifelse(value>max,max<-value,NA)
if(j==1){
min<-value
} else {
ifelse(value<min,min<-value,NA)}
}
dist_med <- total/l
done <- data.frame(id=id[i],qtd_subsidiary=l,dist_min=min,dist_media=round(dist_med,2),dist_max=max)
distance_p_s <- rbind(distance_p_s,done)
eval(parse(text=paste0("
print('Concluido id: ",id[i]," (",i," de ",length(id),"), subsidiary: ",j," de ",l,"')
")))
rm(done)
}
可能这不是最好的方法,但它暂时解决了我的问题。