使用R中的纬度和经度计算观测值

诺布问题。我想不出这个代码有什么问题。我试图找出在半径为400米的圆圈内发生的观测次数。我对每一次观察都有一个lat和long。我正在尝试创建一个新的专栏，显示半径为400米的圆圈内竞争餐厅的数量。我包括了我正在使用的代码的数据示例，以及数据帧的STR。提前谢谢

for (i in seq(nrow(expandedDataFrame2)))
{
  # circle's centre
  xcentre <- df[i,'latitude']
  ycentre <- df[i,'longitude']

  # checking how many restaurants lie within 400 m of the above centre, noofcloserest   column will contain this value
  expandedDataFrame2[i,'noofcloserest'] <- sum(
(expandedDataFrame2[,'latitude'] - xcentre)^2 + 
  (expandedDataFrame2[,'longitude'] - ycentre)^2 
<= 400^2
) - 1

# logging part for deeper analysis
cat(i,': ')

 cat((expandedDataFrame2[,'latitude'] - xcentre)^2 + 
    (expandedDataFrame2[,'longitude'] - ycentre)^2 
  <= 400^2)

cat('\n')

}

---

这里有一种方法，它使用包

sp

中的函数

spDistsN1（…）

。调用数据帧

df

library(sp)

get.dists <- function(i) {
  ref.pt <- with(df[i,],c(longitude,latitude))
  points <- as.matrix(with(df[-i,],cbind(longitude,latitude)))
  dists  <- spDistsN1(points, ref.pt, longlat=T)
  return(length(which(dists<0.4)))
}
df$count <- sapply(1:nrow(df),get.dists)

库（sp）
get.dists您可以发布一个（最小）可复制的示例吗？我会围绕你的中心画一个400米的圆，并使用
sp
包中的函数（比如
points.in.polygon
）来计算这个圆内的点数。你不会真的期望在lat long degrees上使用毕达哥拉斯距离公式会得到以米为单位的距离，是吗？或者使用一个具有特定地球半径的大圆距离函数，或者如果您的点位于一个小区域，则将其投影到UTM网格系统。太棒了！谢谢你也解释了功能中正在发生的事情。这对帮助像我这样的傻瓜来说是很有帮助的。
str(expandeddataframe2)

'data.frame':   2833 obs. of  28 variables:
 $ business_id   : chr  "--5jkZ3-nUPZxUvtcbr8Uw" "--BlvDO_RG2yElKu9XA1_g" "-_Ke8q969OAwEE_-U0qUjw" "-_npP9XdyzILAjtFfX8UAQ" ...
 $ restaurantType: chr  "Greek" "Sushi Bars" "Beer, Wine & Spirits" "Vietnamese" ...
 $ full_address  : chr  "1336 N Scottsdale Rd\nScottsdale, AZ 85257" "14870 N Northsight Blvd\nSte 103\nScottsdale, AZ 85260" "18555 N 59th Ave\nGlendale, AZ 85308" "6025 N 27th Avenue\nSte 24\nPhoenix, AZ 85073" ...
 $ open          : Factor w/ 2 levels "0","1": 2 2 1 2 2 2 2 2 2 2 ...
 $ city          : chr  "Scottsdale" "Scottsdale" "Glendale" "Phoenix" ...
 $ review_count  : num  11 37 6 15 4 145 255 35 7 7 ...
 $ name          : chr  "George's Gyros Greek Grill" "Asian Island" "Jug 'n Barrel Wine Shop" "Thao's Sandwiches" ...
 $ longitude     : num  -112 -112 -112 -112 -112 ...
 $ state         : chr  "AZ" "AZ" "AZ" "AZ" ...
 $ stars         : num  4.5 4 4.5 3 4 3.5 4.5 4 2.5 4.5 ...
 $ latitude      : num  33.5 33.6 33.7 33.4 33.6 ...
 $ type          : chr  "business" "business" "business" "business" ...
 $ categories1   : chr  "Greek" "Sushi Bars" NA "Vietnamese" ...
 $ categories2   : chr  NA "Hawaiian" "Beer, Wine & Spirits" "Sandwiches" ...
 $ categories3   : chr  NA "Chinese" NA NA ...
 $ categories4   : chr  NA NA NA NA ...
 $ categories5   : chr  NA NA NA NA ...
 $ categories6   : chr  NA NA NA NA ...
 $ categories7   : chr  NA NA NA NA ...
 $ categories8   : chr  NA NA NA NA ...
 $ categories9   : chr  NA NA NA NA ...
 $ categories10  : chr  NA NA NA NA ...
 $ isRestaurant  : logi  TRUE TRUE TRUE TRUE TRUE TRUE ...
 $ Freq          : num  66 58 8 44 166 166 98 35 45 166 ...
 $ avgRev        : num [1:2833(1d)] 31.3 68.6 34.3 63.2 30.8 ...
  ..- attr(*, "dimnames")=List of 1
  .. ..$ : chr  "Greek" "Sushi Bars" "Beer, Wine & Spirits" "Vietnamese" ...
 $ avgStar       : num [1:2833(1d)] 3.69 3.66 3.56 3.58 3.48 ...
  ..- attr(*, "dimnames")=List of 1
  .. ..$ : chr  "Greek" "Sushi Bars" "Beer, Wine & Spirits" "Vietnamese" ...
 $ duration      :Class 'difftime'  atomic [1:2833] 381 690 604 1916 226 ...
  .. ..- attr(*, "units")= chr "days"
 $ delta         : num  0 0 1 0 0 0 0 0 0 0 ...

library(sp)

get.dists <- function(i) {
  ref.pt <- with(df[i,],c(longitude,latitude))
  points <- as.matrix(with(df[-i,],cbind(longitude,latitude)))
  dists  <- spDistsN1(points, ref.pt, longlat=T)
  return(length(which(dists<0.4)))
}
df$count <- sapply(1:nrow(df),get.dists)