Question

我正在尝试对两个数据框中的值进行评估，并使用结果创建一个新的数据框。我是R的新手，我正在努力避免旧的编码习惯。换句话说，我拼命想避免使用循环，但在这种情况下无法弄清楚plyr等。

在样本中，我创建了机场，飞行员以及以公里为单位的距离。我的问题是试图确定每个飞行员最接近哪个主要机场以及每个机场的距离。

#Build Airports
code <- c("IAH", "DFW", "Denver", "STL")
lat <- c(29.97, 32.90, 39.75, 38.75)
long <- c(95.35, 97.03, 104.87, 90.37)
airports <- data.frame(code, lat, long)

#Build Pilots
names <- c("James", "Fiona", "Seamus")
lat <- c(32.335131, 44.913223, 28.849631)
long <- c(-84.989067, -97.151334, -96.917240)
pilots <- data.frame(names, lat, long)

#Create distance function
distInKm <- function(lat1, long1, lat2, long2) {
    dlat = (lat2 * 0.01745329) - (lat1 * 0.01745329) #pi/180 convert to radians
    dlong = (long2 * 0.01745329) - (long1 * 0.01745329)
    step1 = (sin(dlat / 2)) ^ 2 + cos(lat1 * 0.01745329) * cos(long2 * 0.01745329) * (sin(dlong / 2)) ^ 2
    step2 = 2 * atan2(sqrt(step1), sqrt(1 - step1))
    dist = 6372.798 * step2 #R is the radius of earth (40041.47 / (2 * pi))
    dist
}

感谢您的时间。

Answer 1

首先，你的机场经度是积极的，因为它们应该是负面的，这会导致结果。让我们解决它们，以便结果更有意义：

airports$long <- -airports$long

现在，您可以使用apply评估每个机场的所有飞行员。 geosphere包有几个计算直线距离的函数，包括distGeo和distHaversine。

library(geosphere)

pilots$closest_airport <- apply(pilots[, 3:2], 1, function(x){
    airports[which.min(distGeo(x, airports[, 3:2])), 'code']
})

pilots$airport_distance <- apply(pilots[, 3:2], 1, function(x){
    min(distGeo(x, airports[, 3:2])) / 1000    # /1000 to convert m to km
})

pilots
##    names      lat      long closest_airport airport_distance
## 1  James 32.33513 -84.98907             STL         862.5394
## 2  Fiona 44.91322 -97.15133          Denver         855.8088
## 3 Seamus 28.84963 -96.91724             IAH         196.3559

或者如果您想要所有距离而不是最小距离，cbind由apply产生的矩阵：

pilots <- cbind(pilots, t(apply(pilots[, 3:2], 1, function(x){
    setNames(distGeo(x, airports[, 3:2]) / 1000, airports$code)
})))

pilots
##    names      lat      long closest_airport       IAH       DFW    Denver       STL
## 1  James 32.33513 -84.98907             STL 1021.6523 1131.2129 1965.6586  862.5394
## 2  Fiona 44.91322 -97.15133          Denver 1666.0359 1333.6842  855.8088  885.8480
## 3 Seamus 28.84963 -96.91724             IAH  196.3559  449.1838 1412.0664 1253.4874

转换为dplyr，plyr的继承者，

library(dplyr)

pilots %>% rowwise() %>% 
        mutate(closest_airport = airports[which.min(distGeo(c(long, lat), airports[, 3:2])), 'code'], 
               airport_distance = min(distGeo(c(long, lat), airports[, 3:2])) / 1000)

## Source: local data frame [3 x 5]
## Groups: <by row>
## 
## # A tibble: 3 × 5
##    names      lat      long closest_airport airport_distance
##   <fctr>    <dbl>     <dbl>          <fctr>            <dbl>
## 1  James 32.33513 -84.98907             STL         862.5394
## 2  Fiona 44.91322 -97.15133          Denver         855.8088
## 3 Seamus 28.84963 -96.91724             IAH         196.3559

或对于所有距离，使用bind_cols上述方法，或unnest列表列并重新塑造：

library(tidyverse)

pilots %>% rowwise() %>% 
    mutate(closest_airport = airports[which.min(distGeo(c(long, lat), airports[, 3:2])), 'code'], 
           data = list(data_frame(airport = airports$code, 
                                  distance = distGeo(c(long, lat), airports[, 3:2]) / 1000))) %>% 
    unnest() %>% 
    spread(airport, distance)

## # A tibble: 3 × 8
##    names      lat      long closest_airport    Denver       DFW       IAH       STL
## * <fctr>    <dbl>     <dbl>          <fctr>     <dbl>     <dbl>     <dbl>     <dbl>
## 1  Fiona 44.91322 -97.15133          Denver  855.8088 1333.6842 1666.0359  885.8480
## 2  James 32.33513 -84.98907             STL 1965.6586 1131.2129 1021.6523  862.5394
## 3 Seamus 28.84963 -96.91724             IAH 1412.0664  449.1838  196.3559 1253.4874

或更直接，但不太清晰，

pilots %>% rowwise() %>% 
    mutate(closest_airport = airports[which.min(distGeo(c(long, lat), airports[, 3:2])), 'code'], 
           data = (distGeo(c(long, lat), airports[, 3:2]) / 1000) %>% 
                   setNames(airports$code) %>% t() %>% as_data_frame() %>% list()) %>% 
    unnest()

## # A tibble: 3 × 8
##    names      lat      long closest_airport       IAH       DFW    Denver       STL
##   <fctr>    <dbl>     <dbl>          <fctr>     <dbl>     <dbl>     <dbl>     <dbl>
## 1  James 32.33513 -84.98907             STL 1021.6523 1131.2129 1965.6586  862.5394
## 2  Fiona 44.91322 -97.15133          Denver 1666.0359 1333.6842  855.8088  885.8480
## 3 Seamus 28.84963 -96.91724             IAH  196.3559  449.1838 1412.0664 1253.4874

在R中，评估两个数据帧之间的功能

1 个答案: