如何迭代制作图表？
How to iterate making a graph?
我正在处理西雅图的犯罪数据。下面是一个示例数据集。我有两个问题。
library(ggplot2)
library(ggmap)
SPD_2015 <- structure(list(summarized.offense.description = c("OTHER PROPERTY",
"CAR PROWL", "ASSAULT", "SHOPLIFTING", "VEHICLE THEFT", "OTHER PROPERTY",
"OTHER PROPERTY", "PROSTITUTION", "CAR PROWL", "PROPERTY DAMAGE",
"CAR PROWL", "ASSAULT", "FRAUD", "SHOPLIFTING", "ROBBERY", "WARRANT ARREST",
"VEHICLE THEFT", "CAR PROWL", "PROPERTY DAMAGE", "ASSAULT", "VEHICLE THEFT",
"OTHER PROPERTY", "CAR PROWL", "FRAUD", "CAR PROWL", "CAR PROWL",
"CAR PROWL", "THREATS", "CAR PROWL", "DISTURBANCE", "CAR PROWL",
"CAR PROWL", "EMBEZZLE", "THREATS", "CAR PROWL", "PROPERTY DAMAGE",
"STOLEN PROPERTY", "ASSAULT", "LOST PROPERTY", "BURGLARY-SECURE PARKING-RES",
"THREATS", "PROPERTY DAMAGE", "FRAUD", "CAR PROWL", "BURGLARY",
"ASSAULT", "THEFT OF SERVICES", "OTHER PROPERTY", "DISTURBANCE",
"BIKE THEFT", "BURGLARY", "CAR PROWL", "FRAUD", "CAR PROWL",
"VEHICLE THEFT", "DISTURBANCE", "BURGLARY", "BURGLARY", "BURGLARY",
"OTHER PROPERTY", "CAR PROWL", "CAR PROWL", "BURGLARY", "BURGLARY",
"OTHER PROPERTY", "FRAUD", "CAR PROWL", "BURGLARY", "NARCOTICS",
"THREATS", "PROPERTY DAMAGE", "TRESPASS", "ASSAULT", "FRAUD",
"CAR PROWL", "BURGLARY", "CAR PROWL", "BURGLARY-SECURE PARKING-RES",
"FRAUD", "CAR PROWL", "FRAUD", "THREATS", "CAR PROWL", "BURGLARY",
"TRESPASS", "TRESPASS", "OTHER PROPERTY", "STOLEN PROPERTY",
"STOLEN PROPERTY", "WARRANT ARREST", "WARRANT ARREST", "FRAUD",
"CAR PROWL", "OTHER PROPERTY", "PROPERTY DAMAGE", "BURGLARY",
"FRAUD", "OTHER PROPERTY", "FRAUD", "CAR PROWL"), longitude = c(-122.300109863,
-122.385444641, -122.269958496, -122.341133118, -122.311935425,
-122.256233215, -122.344665527, -122.302001953, -122.344993591,
-122.311782837, -122.325790405, -122.337394714, -122.317298889,
-122.365219116, -122.33140564, -122.343269348, -122.300140381,
-122.280647278, -122.349700928, -122.340240479, -122.354415894,
-122.345626831, -122.317359924, -122.378921509, -122.390213013,
-122.354415894, -122.337089539, -122.280601501, -122.359313965,
-122.337791443, -122.330421448, -122.343261719, -122.396110535,
-122.311546326, -122.316917419, -122.262084961, -122.340454102,
-122.320770264, -122.315254211, -122.344444275, -122.304519653,
-122.319442749, -122.36756134, -122.330039978, -122.337348938,
-122.330810547, -122.303710938, -122.327880859, -122.382667542,
-122.322769165, -122.313537598, -122.301094055, -122.4034729,
-122.333267212, -122.32888031, -122.382377625, -122.310951233,
-122.318778992, -122.326576233, -122.354827881, -122.382377625,
-122.378768921, -122.315391541, -122.311248779, -122.311393738,
-122.32408905, -122.367424011, -122.338768005, -122.297531128,
-122.374198914, -122.348678589, -122.326385498, -122.33303833,
-122.381492615, -122.338088989, -122.282745361, -122.316902161,
-122.355461121, -122.389198303, -122.32635498, -122.404212952,
-122.313087463, -122.343833923, -122.304168701, -122.3854599,
-122.296226501, -122.318733215, -122.332801819, -122.316726685,
-122.323440552, -122.332260132, -122.290527344, -122.337585449,
-122.344940186, -122.31678009, -122.376319885, -122.31816864,
-122.335906982, -122.355148315, -122.355621338), latitude = c(47.595077515,
47.556591034, 47.670768738, 47.610042572, 47.664890289, 47.497062683,
47.702514648, 47.583400726, 47.725036621, 47.526573181, 47.700252533,
47.612663269, 47.564403534, 47.521022797, 47.602767944, 47.608207703,
47.610794067, 47.535404205, 47.57101059, 47.612014771, 47.634437561,
47.660072327, 47.669715881, 47.680427551, 47.521442413, 47.66809082,
47.607299805, 47.724998474, 47.687664032, 47.60974884, 47.620243073,
47.61145401, 47.549030304, 47.60710907, 47.619354248, 47.509685516,
47.686210632, 47.613517761, 47.664012909, 47.608901978, 47.589576721,
47.717647552, 47.642562866, 47.606300354, 47.52047348, 47.600463867,
47.609523773, 47.623706818, 47.665958405, 47.649650574, 47.593112946,
47.602165222, 47.573997498, 47.58398056, 47.630302429, 47.591312408,
47.595115662, 47.660381317, 47.626041412, 47.549259186, 47.591312408,
47.567428589, 47.662197113, 47.629676819, 47.62008667, 47.602870941,
47.673809052, 47.606601715, 47.610782623, 47.56407547, 47.613479614,
47.607337952, 47.604553223, 47.666133881, 47.712303162, 47.727027893,
47.618183136, 47.705989838, 47.652839661, 47.600868225, 47.665912628,
47.66399765, 47.688751221, 47.691646576, 47.561988831, 47.707542419,
47.670059204, 47.611839294, 47.624809265, 47.604129791, 47.605373383,
47.632568359, 47.726856232, 47.71957016, 47.605884552, 47.551052094,
47.615837097, 47.600463867, 47.632316589, 47.635715485)), .Names = c("summarized.offense.description",
"longitude", "latitude"), row.names = c(NA, -100L), class = c("tbl_df",
"tbl", "data.frame"))
我总结了我的数据，看看我在做什么：
group_by(SPD_2015, summarized.offense.description) %>%
summarize(count = n()) %>%
arrange(desc(count))
# A tibble: 21 × 2
summarized.offense.description count
<chr> <int>
1 CAR PROWL 24
2 BURGLARY 11
3 FRAUD 11
4 OTHER PROPERTY 10
5 ASSAULT 6
6 PROPERTY DAMAGE 6
7 THREATS 5
8 VEHICLE THEFT 4
9 DISTURBANCE 3
10 STOLEN PROPERTY 3
# ... with 11 more rows
我目前一直在创建我感兴趣的 "summarized.offense.description" 的新过滤数据集，例如 CAR PROWL：
car.prowl <- SPD_2015 %>%
filter(summarized.offense.description == "CAR PROWL")
然后映射所述数据：
ggmap(seattle.map) +
geom_point(data = car.prowl, aes(car.prowl$longitude, car.prowl$latitude),
alpha = 0.2, color = "tomato4", size = 0.7) +
theme(axis.text = element_blank()) +
theme(axis.title = element_blank())
问题 1：与其为我感兴趣的每个 summarized.offense.description 创建一个新的过滤数据集，不如如何为我总结的前四个 summarized.offense.description 创建并保存一个新地图数据集（在本例中为 CAR PROWL、BURGLARY、FRAUD、OTHER 属性）？
问题 2：如何制作前四位的分面图 summarized.offense.description？
如果能深入了解我的问题，我们将不胜感激。
谢谢！
对于第一个请求，我将执行以下操作：
# Get Seattle Map
seattle.map <- qmap("seattle", zoom = 11, source="stamen", maptype="toner",darken = c(.3,"#BBBBBB"))
# Get ordered summarized data
ordered <- group_by(SPD_2015, summarized.offense.description) %>%
summarize(count = n()) %>%
arrange(desc(count))
# Make the ordered summarized data a dataframe and then use only the first column
# to be a character vector with the top 4
ordered <- as.data.frame(ordered[1:4,1])
ordered <- ordered$summarized.offense.description
# Make a list to comprehend the 4 ggmap plots
ggmap_list <- list()
# Plot the 4 ggmaps
for(i in 1:4){
subst <- SPD_2015[SPD_2015$summarized.offense.description == ordered[i],]
ggmap_list[[i]] <- seattle.map +
geom_point(data = subst, aes(longitude, latitude),
alpha = 0.2, color = "red", size = 0.7) +
theme(axis.text = element_blank()) +
theme(axis.title = element_blank())
}
关于第二个请求，我会尝试：
# Make a subset of the data with the top 4
subst <- SPD_2015[SPD_2015$summarized.offense.description %in% ordered,]
# Plot all 4 ggmaps on the same plotting region
seattle.map +
geom_point(data = subst, aes(longitude, latitude),
alpha = 0.2, color = "red", size = 0.7) +
facet_wrap(~summarized.offense.description) +
theme(axis.text = element_blank()) +
theme(axis.title = element_blank())
请注意，我 更改了颜色 以获得更好的视觉效果，并且我试图重现您的 西雅图地图 。
这是你想要的吗？
我正在处理西雅图的犯罪数据。下面是一个示例数据集。我有两个问题。
library(ggplot2)
library(ggmap)
SPD_2015 <- structure(list(summarized.offense.description = c("OTHER PROPERTY",
"CAR PROWL", "ASSAULT", "SHOPLIFTING", "VEHICLE THEFT", "OTHER PROPERTY",
"OTHER PROPERTY", "PROSTITUTION", "CAR PROWL", "PROPERTY DAMAGE",
"CAR PROWL", "ASSAULT", "FRAUD", "SHOPLIFTING", "ROBBERY", "WARRANT ARREST",
"VEHICLE THEFT", "CAR PROWL", "PROPERTY DAMAGE", "ASSAULT", "VEHICLE THEFT",
"OTHER PROPERTY", "CAR PROWL", "FRAUD", "CAR PROWL", "CAR PROWL",
"CAR PROWL", "THREATS", "CAR PROWL", "DISTURBANCE", "CAR PROWL",
"CAR PROWL", "EMBEZZLE", "THREATS", "CAR PROWL", "PROPERTY DAMAGE",
"STOLEN PROPERTY", "ASSAULT", "LOST PROPERTY", "BURGLARY-SECURE PARKING-RES",
"THREATS", "PROPERTY DAMAGE", "FRAUD", "CAR PROWL", "BURGLARY",
"ASSAULT", "THEFT OF SERVICES", "OTHER PROPERTY", "DISTURBANCE",
"BIKE THEFT", "BURGLARY", "CAR PROWL", "FRAUD", "CAR PROWL",
"VEHICLE THEFT", "DISTURBANCE", "BURGLARY", "BURGLARY", "BURGLARY",
"OTHER PROPERTY", "CAR PROWL", "CAR PROWL", "BURGLARY", "BURGLARY",
"OTHER PROPERTY", "FRAUD", "CAR PROWL", "BURGLARY", "NARCOTICS",
"THREATS", "PROPERTY DAMAGE", "TRESPASS", "ASSAULT", "FRAUD",
"CAR PROWL", "BURGLARY", "CAR PROWL", "BURGLARY-SECURE PARKING-RES",
"FRAUD", "CAR PROWL", "FRAUD", "THREATS", "CAR PROWL", "BURGLARY",
"TRESPASS", "TRESPASS", "OTHER PROPERTY", "STOLEN PROPERTY",
"STOLEN PROPERTY", "WARRANT ARREST", "WARRANT ARREST", "FRAUD",
"CAR PROWL", "OTHER PROPERTY", "PROPERTY DAMAGE", "BURGLARY",
"FRAUD", "OTHER PROPERTY", "FRAUD", "CAR PROWL"), longitude = c(-122.300109863,
-122.385444641, -122.269958496, -122.341133118, -122.311935425,
-122.256233215, -122.344665527, -122.302001953, -122.344993591,
-122.311782837, -122.325790405, -122.337394714, -122.317298889,
-122.365219116, -122.33140564, -122.343269348, -122.300140381,
-122.280647278, -122.349700928, -122.340240479, -122.354415894,
-122.345626831, -122.317359924, -122.378921509, -122.390213013,
-122.354415894, -122.337089539, -122.280601501, -122.359313965,
-122.337791443, -122.330421448, -122.343261719, -122.396110535,
-122.311546326, -122.316917419, -122.262084961, -122.340454102,
-122.320770264, -122.315254211, -122.344444275, -122.304519653,
-122.319442749, -122.36756134, -122.330039978, -122.337348938,
-122.330810547, -122.303710938, -122.327880859, -122.382667542,
-122.322769165, -122.313537598, -122.301094055, -122.4034729,
-122.333267212, -122.32888031, -122.382377625, -122.310951233,
-122.318778992, -122.326576233, -122.354827881, -122.382377625,
-122.378768921, -122.315391541, -122.311248779, -122.311393738,
-122.32408905, -122.367424011, -122.338768005, -122.297531128,
-122.374198914, -122.348678589, -122.326385498, -122.33303833,
-122.381492615, -122.338088989, -122.282745361, -122.316902161,
-122.355461121, -122.389198303, -122.32635498, -122.404212952,
-122.313087463, -122.343833923, -122.304168701, -122.3854599,
-122.296226501, -122.318733215, -122.332801819, -122.316726685,
-122.323440552, -122.332260132, -122.290527344, -122.337585449,
-122.344940186, -122.31678009, -122.376319885, -122.31816864,
-122.335906982, -122.355148315, -122.355621338), latitude = c(47.595077515,
47.556591034, 47.670768738, 47.610042572, 47.664890289, 47.497062683,
47.702514648, 47.583400726, 47.725036621, 47.526573181, 47.700252533,
47.612663269, 47.564403534, 47.521022797, 47.602767944, 47.608207703,
47.610794067, 47.535404205, 47.57101059, 47.612014771, 47.634437561,
47.660072327, 47.669715881, 47.680427551, 47.521442413, 47.66809082,
47.607299805, 47.724998474, 47.687664032, 47.60974884, 47.620243073,
47.61145401, 47.549030304, 47.60710907, 47.619354248, 47.509685516,
47.686210632, 47.613517761, 47.664012909, 47.608901978, 47.589576721,
47.717647552, 47.642562866, 47.606300354, 47.52047348, 47.600463867,
47.609523773, 47.623706818, 47.665958405, 47.649650574, 47.593112946,
47.602165222, 47.573997498, 47.58398056, 47.630302429, 47.591312408,
47.595115662, 47.660381317, 47.626041412, 47.549259186, 47.591312408,
47.567428589, 47.662197113, 47.629676819, 47.62008667, 47.602870941,
47.673809052, 47.606601715, 47.610782623, 47.56407547, 47.613479614,
47.607337952, 47.604553223, 47.666133881, 47.712303162, 47.727027893,
47.618183136, 47.705989838, 47.652839661, 47.600868225, 47.665912628,
47.66399765, 47.688751221, 47.691646576, 47.561988831, 47.707542419,
47.670059204, 47.611839294, 47.624809265, 47.604129791, 47.605373383,
47.632568359, 47.726856232, 47.71957016, 47.605884552, 47.551052094,
47.615837097, 47.600463867, 47.632316589, 47.635715485)), .Names = c("summarized.offense.description",
"longitude", "latitude"), row.names = c(NA, -100L), class = c("tbl_df",
"tbl", "data.frame"))
我总结了我的数据，看看我在做什么：
group_by(SPD_2015, summarized.offense.description) %>%
summarize(count = n()) %>%
arrange(desc(count))
# A tibble: 21 × 2
summarized.offense.description count
<chr> <int>
1 CAR PROWL 24
2 BURGLARY 11
3 FRAUD 11
4 OTHER PROPERTY 10
5 ASSAULT 6
6 PROPERTY DAMAGE 6
7 THREATS 5
8 VEHICLE THEFT 4
9 DISTURBANCE 3
10 STOLEN PROPERTY 3
# ... with 11 more rows
我目前一直在创建我感兴趣的 "summarized.offense.description" 的新过滤数据集，例如 CAR PROWL：
car.prowl <- SPD_2015 %>%
filter(summarized.offense.description == "CAR PROWL")
然后映射所述数据：
ggmap(seattle.map) +
geom_point(data = car.prowl, aes(car.prowl$longitude, car.prowl$latitude),
alpha = 0.2, color = "tomato4", size = 0.7) +
theme(axis.text = element_blank()) +
theme(axis.title = element_blank())
问题 1：与其为我感兴趣的每个 summarized.offense.description 创建一个新的过滤数据集，不如如何为我总结的前四个 summarized.offense.description 创建并保存一个新地图数据集（在本例中为 CAR PROWL、BURGLARY、FRAUD、OTHER 属性）？
问题 2：如何制作前四位的分面图 summarized.offense.description？
如果能深入了解我的问题，我们将不胜感激。
谢谢！
对于第一个请求，我将执行以下操作：
# Get Seattle Map
seattle.map <- qmap("seattle", zoom = 11, source="stamen", maptype="toner",darken = c(.3,"#BBBBBB"))
# Get ordered summarized data
ordered <- group_by(SPD_2015, summarized.offense.description) %>%
summarize(count = n()) %>%
arrange(desc(count))
# Make the ordered summarized data a dataframe and then use only the first column
# to be a character vector with the top 4
ordered <- as.data.frame(ordered[1:4,1])
ordered <- ordered$summarized.offense.description
# Make a list to comprehend the 4 ggmap plots
ggmap_list <- list()
# Plot the 4 ggmaps
for(i in 1:4){
subst <- SPD_2015[SPD_2015$summarized.offense.description == ordered[i],]
ggmap_list[[i]] <- seattle.map +
geom_point(data = subst, aes(longitude, latitude),
alpha = 0.2, color = "red", size = 0.7) +
theme(axis.text = element_blank()) +
theme(axis.title = element_blank())
}
关于第二个请求，我会尝试：
# Make a subset of the data with the top 4
subst <- SPD_2015[SPD_2015$summarized.offense.description %in% ordered,]
# Plot all 4 ggmaps on the same plotting region
seattle.map +
geom_point(data = subst, aes(longitude, latitude),
alpha = 0.2, color = "red", size = 0.7) +
facet_wrap(~summarized.offense.description) +
theme(axis.text = element_blank()) +
theme(axis.title = element_blank())
请注意，我 更改了颜色 以获得更好的视觉效果，并且我试图重现您的 西雅图地图 。
这是你想要的吗？