Question

有很好的教程可以映射休斯顿犯罪数据，但没有简单的例子来说明如何清理HPD提供的原始数据。 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"))

这将返回Lon和Lat：

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

选择完成数据的示例。

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"))

以下是地理编码的功能：

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，用[address，long，lat]的输出再添加另外3列？

即..

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

这是问题的阻碍点：

#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

您可以通过分割范围和平均来计算平均块范围：

e.g。

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

要进行扩展，我们可以使用separate包中的tidyr。这会做一些很酷的事情，比如自动将blockrange的最小值/最大值放入一个新列，将类型从字符串转换为数字（convert=T, type.convert=as.numeric）。我先filter输出“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（虽然我会考虑如何限制最大数量的请求等）。

关于一次添加输出lat和lon列，我不认为dplyr会这样做（参见this feature request）。

如果你真的想在这里使用dplyr语法，最好的办法就是改变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, .)

但我真的认为你应该跳过这个特定步骤的dplyr糖并做一个手动循环并cbind结果，这样可以更好地控制请求限制。

创建一个向量函数来清除Houston Crime Data的地址数据

1 个答案: