创建一个向量函数来清理 Houston Crime Data 的地址数据

Question

有很好的映射休斯顿犯罪数据的教程，但没有关于如何清理 HPD 提供的原始数据的简单示例。 https://github.com/hadley/ggplot2/wiki/Crime-in-Downtown-Houston,-Texas-:-Combining-ggplot2-and-Google-Maps

d <- structure(list(BlockRange = c("5400-5499", "3700-3799", "2200-2299", 
    "1000-1099", "1200-1299", "UNK", "1900-1999", "500-599", "1200-1299"
    ), StreetName = c("BELL", "BELL", "BELL", "BELL", "BELL", "BELL", 
    "BELL", "BELL", "BELL"), Date = c("4/28/2015", "4/11/2015", "4/26/2015", 
    "4/9/2015", "4/9/2015", "4/21/2015", "4/26/2015", "4/26/2015", 
    "4/17/2015")), row.names = c(60L, 75L, 88L, 4972L, 4990L, 5096L, 
    5098L, 5099L, 5155L), class = "data.frame", .Names = c("BlockRange", 
    "StreetName", "Date"))

这将 return 经度和纬度：

x <- gGeoCode("1950 Bell St, Houston, TX")
#[1]  29.74800 -95.35926

但是，它需要一个函数来对整个数据库进行地理编码并为 Lon 和 Lat 添加列

完成数据的选择示例。

structure(list(address = c("9650 marlive ln", "4750 telephone rd", 
"5050 wickview ln", "1050 ashland st", "8350 canyon", "9350 rowan ln", 
"2550 southmore blvd", "6350 rupley cir", "5050 georgi ln", "10750 briar forest dr"
), lon = c(-95.4373883, -95.2988769, -95.455864, -95.4033373, 
-95.3779081, -95.5483009, -95.3733977, -95.3156032, -95.4665841, 
-95.565934), lat = c(29.6779015, 29.6917121, 29.5992174, 29.7902425, 
29.6706341, 29.7022336, 29.7198936, 29.6902746, 29.8297359, 29.747596
)), row.names = 82729:82738, class = "data.frame", .Names = c("address", 
"lon", "lat"))

以下是地理编码函数：

library(RCurl)
library(RJSONIO)
library(dplyr)
library(gdata)


construct.geocode.url <- function(address, return.call = "json", sensor = "false") {
  root <- "http://maps.google.com/maps/api/geocode/"
  u <- paste(root, return.call, "?address=", address, "&sensor=", sensor, sep = "")
  return(URLencode(u))
}

gGeoCode <- function(address,verbose=FALSE) {
  if(verbose) cat(address,"\n")
  u <- construct.geocode.url(address)
  doc <- getURL(u)
  x <- fromJSON(doc,simplify = FALSE)
  if(x$status=="OK") {
    lat <- x$results[[1]]$geometry$location$lat
    lng <- x$results[[1]]$geometry$location$lng
    return(c(lat, lng))
  } else {
    return(c(NA,NA))
  }
}

我们如何使用 dplyr 或另一种方法编写一个函数，将 [address, long, lat] 的输出再添加 3 列？

即..

data.frame <- mutate(d, address = ConvertBlockRange(BlockRange) + StreetName, "Houston, TX"), Lon = geocode(address)[0] , lat = geocode(address)[1])

这是问题的阻塞点：

#function to convert - "2200-2299" to integer 2250.. i.e find the middle of the block.    


    library(stringr)             
    ConvertBlockRange <- function(blockRange){
    m <-   unlist(str_split(d$BlockRange, "-"))
  m2 <- mean(c(as.numeric(m[1]),as.numeric(m[2]))) + .5
  m2
}

Answer 1

您可以通过拆分范围和平均来计算平均块范围：

例如

x <- '5400-5499'
mean(as.numeric(strsplit(x, '-')[[1]])) # 5449.5

要扩大规模，我们可以使用 tidyr 包中的 separate。这会做一些很酷的事情，比如自动将 blockrange 的 min/max 放入新列，将类型从字符串转换为数字 (convert=T, type.convert=as.numeric)。我先 filter 列出了 "UNK" 地址 - 你必须单独处理它们。

library(dplyr)
library(tidyr) 
d %>%
  filter(BlockRange != "UNK") %>%
  # this is a df with blockmin & blockmax
  separate(BlockRange, c("blockmin", "blockmax"), sep = "-", 
           convert=T, type.convert=as.numeric, remove=FALSE) %>%
  # calc average (round down) and address
  mutate(block=floor((blockmin + blockmax)/2),
         address=paste(block, StreetName))

#   BlockRange blockmin blockmax StreetName      Date block   address
# 1  5400-5499     5400     5499       BELL 4/28/2015  5449 5449 BELL
# 2  3700-3799     3700     3799       BELL 4/11/2015  3749 3749 BELL
# 3  2200-2299     2200     2299       BELL 4/26/2015  2249 2249 BELL
# 4  1000-1099     1000     1099       BELL  4/9/2015  1049 1049 BELL
# 5  1200-1299     1200     1299       BELL  4/9/2015  1249 1249 BELL
# 6  1900-1999     1900     1999       BELL 4/26/2015  1949 1949 BELL
# 7    500-599      500      599       BELL 4/26/2015   549  549 BELL
# 8  1200-1299     1200     1299       BELL 4/17/2015  1249 1249 BELL

然后您可以 %>% group_by(address) 获取唯一地址和地理编码（尽管我会考虑如何在这里限制最大请求数等）。

关于同时添加输出纬度和经度列，我认为 dplyr 还没有这样做（参见 this feature request）。

如果你真的想在这里使用 dplyr 语法，最好的办法是更改 gGeoCode 以便它被矢量化，例如

gGeoCode2 <- function (addresses) {
    x <- data.frame(t(sapply(addresses[[1]], gGeoCode)), row.names=NULL)
    names(x) <- c('lat', 'lng')
    x
}

d2 %>%
   select(address) %>%
   gGeoCode2 %>%
   bind_cols(d2, .)

但我真的真的认为您应该跳过此特定步骤的 dplyr 糖并执行手动循环和 cbind 结果，这使您可以更好地控制请求限制。

创建一个向量函数来清理 Houston Crime Data 的地址数据

Create a vector function to clean address data for Houston Crime Data

r

ggplot2

ggmap

dplyr