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Question

在這裏需要幫助。我正在嘗試創建一個新欄目，列出使用經度和緯度在200米餐廳的餐館數量。我找不到任何東西在stackoverflow，我不是R忍者。任何幫助，將不勝感激！

頭（）

business_id      restaurantType            full_address open  city 
1 --5jkZ3-nUPZxUvtcbr8Uw    Greek    1336 N Scottsdale Rd\nScottsdale, AZ 85257 1 Scottsdale 
2 --BlvDO_RG2yElKu9XA1_g   Sushi Bars 14870 N Northsight Blvd\nSte 103\nScottsdale, AZ 85260 1 Scottsdale 
3 -_Ke8q969OAwEE_-U0qUjw Beer, Wine & Spirits     18555 N 59th Ave\nGlendale, AZ 85308 0 Glendale 
4 -_npP9XdyzILAjtFfX8UAQ   Vietnamese   6025 N 27th Avenue\nSte 24\nPhoenix, AZ 85073 1 Phoenix 
5 -2xCV0XGD9NxfWaVwA1-DQ    Pizza      9008 N 99th Ave\nPeoria, AZ 85345 1  Peoria 
6 -3WVw1TNQbPBzaKCaQQ1AQ    Chinese      302 E Flower St\nPhoenix, AZ 85012 1 Phoenix 
review_count      name longitude state stars latitude  type categories1   categories2 
1   11 George's Gyros Greek Grill -111.9269 AZ 4.5 33.46337 business  Greek     <NA> 
2   37    Asian Island -111.8983 AZ 4.0 33.62146 business Sushi Bars    Hawaiian 
3   6 Jug 'n Barrel Wine Shop -112.1863 AZ 4.5 33.65387 business  <NA> Beer, Wine & Spirits 
4   15   Thao's Sandwiches -112.0739 AZ 3.0 33.44990 business Vietnamese   Sandwiches 
5   4   Nino's Pizzeria 2 -112.2766 AZ 4.0 33.56626  business  Pizza     <NA> 
6   145    China Chili -112.0692 AZ 3.5 33.48585 business  Chinese     <NA> 

    avgStar duration delta 
1 3.694030  381  0 
2 3.661017  690  0 
3 3.555556  604  1 
4 3.577778  1916  0 
5 3.482036  226  0 
6 3.535928  2190  0 

STR（）

'data.frame': 2833 obs. of 28 variables: 
$ business_id : Factor w/ 2833 levels "--5jkZ3-nUPZxUvtcbr8Uw",..: 1 2 3 4 5 6 7 8 9 10 ... 
$ restaurantType: Factor w/ 118 levels "Afghan","African",..: 60 106 15 117 89 31 17 7 84 31 ... 
$ full_address : Factor w/ 2586 levels "1 E Jackson St\nPhoenix, AZ 85004",..: 274 371 642 1825 2368 1102 1000 1143 2169 1669 ... 
$ open   : int 1 1 0 1 1 1 1 1 1 1 ... 
$ city   : Factor w/ 44 levels "Ahwatukee","Anthem",..: 34 34 19 31 30 31 34 4 18 31 ... 
$ review_count : int 11 37 6 15 4 145 255 35 7 7 ... 
$ name   : Factor w/ 2652 levels "#1 Brother's Pizza",..: 885 127 1167 2318 1601 453 591 697 1492 1319 ... 
$ longitude  : num -112 -112 -112 -112 -112 ... 
$ state   : Factor w/ 2 levels "AZ","SC": 1 1 1 1 1 1 1 1 1 1 ... 
$ 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   : Factor w/ 1 level "business": 1 1 1 1 1 1 1 1 1 1 ... 
$ categories1 : Factor w/ 103 levels "Afghan","African",..: 50 93 NA 102 78 26 14 7 73 26 ... 
$ Freq   : int 66 58 8 44 166 166 98 35 45 166 ... 
$ avgRev  : num 31.3 68.6 34.3 63.2 30.8 ... 
$ avgStar  : num 3.69 3.66 3.56 3.58 3.48 ... 
$ duration  : int 381 690 604 1916 226 2190 1968 1338 1606 56 ... 

Answer 1

Base R和未經測試的代碼，但你應該明白。

我基本上測試了每個餐廳的圓形方程式x2 + y2 <= R中有多少行，除了該餐廳本身，並將其更新爲列中的值。請注意，我的公式中的半徑是200，但它會有所不同，因爲您的x，y位於經度和緯度，您將不得不將半徑200米縮放至2pi radians/circumference of earth或360 degree/circumference of earth。

df <- data.frame(
    latitude = runif(n=10,min=0,max=1000), 
    longitude = runif(n=10,min=0,max=1000) 
) 

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

    # checking how many restaurants lie within 200 m of the above centre, noofcloserest column will contain this value 
    df[i,'noofcloserest'] <- sum(
    (df[,'latitude'] - xcentre)^2 + 
     (df[,'longitude'] - ycentre)^2 
    <= 200^2 
) - 1 

    # logging part for deeper analysis 
    cat(i,': ') 
    # this prints the true/false vector for which row is within the radius, and which row isn't 
    cat((df[,'latitude'] - xcentre)^2 + 
    (df[,'longitude'] - ycentre)^2 
    <= 200^2) 

    cat('\n') 

} 

輸出 -

1 : TRUE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE FALSE 
2 : FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE 
3 : FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE 
4 : TRUE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE FALSE 
5 : FALSE FALSE TRUE FALSE TRUE FALSE FALSE TRUE FALSE FALSE 
6 : TRUE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE FALSE 
7 : FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE TRUE 
8 : FALSE FALSE FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE 
9 : FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE 
10 : FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE TRUE 
> df 
    latitude longitude noofcloserest 
1 189.38878 270.25004    2 
2 402.36853 879.26657    0 
3 747.46417 581.66627    1 
4 291.64303 157.75450    2 
5 830.10699 736.19586    2 
6 299.06803 157.76147    2 
7 725.68360 58.53049    1 
8 893.31904 772.46217    1 
9 45.47875 701.82201    0 
10 645.44772 226.95042    1 

什麼該輸出機構，是在第1行的座標，三行是200米內。第1行本身以及第4和第6行。

Answer 2

一種方法是計算距離矩陣，然後找出充分靠近的那些（在這裏我演示是在20公里以內，所以數字不全是0）：

# Load the fields library 
library(fields) 

# Create a simple data frame to demonstrate (each row is a restaurant). The rdist.earth function 
# we're about to call takes as input something where the first column is longitude and the second 
# column is latitude. 
df = data.frame(longitude=c(-111.9269, -111.8983, -112.1863, -112.0739, -112.2766, -112.0692), 
       latitude=c(33.46337, 33.62146, 33.65387, 33.44990, 33.56626, 33.48585)) 

# Let's compute the distance between each restaurant. 
distances = rdist.earth(df, miles=F) 
distances 

#   [,1]  [,2]  [,3]   [,4]  [,5]   [,6] 
# [1,] 0.00000 17.79813 32.07533 1.373515e+01 34.41932 1.344867e+01 
# [2,] 17.79813 0.00000 26.93558 2.510519e+01 35.61413 2.189270e+01 
# [3,] 32.07533 26.93558 0.00000 2.498676e+01 12.85352 2.162964e+01 
# [4,] 13.73515 25.10519 24.98676 1.344145e-04 22.84310 4.025824e+00 
# [5,] 34.41932 35.61413 12.85352 2.284310e+01 0.00000 2.122719e+01 
# [6,] 13.44867 21.89270 21.62964 4.025824e+00 21.22719 9.504539e-05 

# Compute the number of restaurants within 20 kilometers of the restaurant in each row. 
df$num.close = colSums(distances <= 20) - 1 
df$num.close 
# [1] 3 1 1 2 1 2