将坐标从非常规格式的度转换为十进制度 [英] Converting coordinates from degree with unconventional format to decimal degree
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问题描述
我正在尝试转换我的数据,以便可以将其绘制在地图上。例如,数据看起来像:
I am trying to convert my data so that it can be plotting on a map. For example the data looks like:
# A tibble: 2 x 2
Latitud Longitud
<chr> <chr>
1 10º 35' 28.98'' N 3º 41' 33.91'' O
2 10º 35' 12.63'' N 3º 45' 46.22'' O
我正在尝试使用以下方法对其进行突变:
I am trying to mutate it using the following:
df %>%
mutate(
Latitud = str_replace_all(Latitud, "''", ""),
lat_edit = sp::char2dms(Latitud), "°")
哪个返回和错误:
Error in if (any(abs(object@deg) > 90)) return("abs(degree) > 90") :
missing value where TRUE/FALSE needed
In addition: Warning message:
In asMethod(object) : NAs introduced by coercion
我想在ggplot(或其他空间包)中的地图上绘制这两个点
I would like to plot these two points on a map in ggplot (or another spatial package)
数据:
structure(list(Latitud = c("40º 25' 25.98'' N", "40º 25' 17.63'' N"
), Longitud = c("3º 42' 43.91'' O", "3º 40' 56.22'' O")), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -2L))
推荐答案
您可以使用以下自定义函数(我假设 N
, S
, W
, E
。不确定 O
在经度中是什么意思):
You can use the following custom function (I am assuming N
, S
, W
, E
. Not sure what O
means in longitude):
angle2dec <- function(angle) {
angle <- as.character(angle)
angle <- ifelse(grepl("S|W", angle), paste0("-", angle), angle)
angle <- trimws(gsub("[^- +.0-9]", "", angle))
x <- do.call(rbind, strsplit(angle, split=' '))
x <- apply(x, 1L, function(y) {
y <- as.numeric(y)
(abs(y[1]) + y[2]/60 + y[3]/3600) * sign(y[1])
})
return(x)
}
应用数据:
Applying on the data:
df1[] <- lapply(df1, angle2dec)
df1
#> Latitud Longitud
#> 1 -40.42388 3.712197
#> 2 40.42156 -3.682283
绘图:
Plotting:
library(ggplot2)
ggplot(df1, aes(x = Longitud, y = Latitud)) +
geom_point()
稍加修改的数据以显示不同的半球:
Slightly Modified Data to Show for Different Hemispheres:
df1 <- structure(list(Latitud = c("40<U+623C><U+3E61> 25' 25.98'' S",
"40<U+623C><U+3E61> 25' 17.63'' N"),
Longitud = c("3<U+623C><U+3E61> 42' 43.91'' E",
"3<U+623C><U+3E61> 40' 56.22'' W")),
class = c("tbl_df", "tbl", "data.frame"),
row.names = c(NA, -2L))
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