无法使用 geom_contour() 在 R 中绘制决策边界
Unable to plot Decision Boundary in R with geom_contour()
我知道有几个类似的问题，但我仍在努力理解它背后的概念。
我根据 knn 预测 类（“真”和“假”）：
hype_knn_prediction = knn(hype_quant_train[,2:3], hype_quant_test[,2:3], cl = hype_quant_train$Selected, k = k)
而且我还可以绘制测试数据：
ggplot(hype_quant_test, aes(x= Comments, y= Votings, color = Selected, shape = Selected)) +
geom_point(size = 3) +
labs(y="Votes", x = "Comments")+
ggtitle("Testdata")+
theme(plot.title = element_text(hjust = 0.5))+
theme(legend.position="bottom")
Plot of test data
现在我想将 类 的决策边界添加到测试数据的图中。因此，我将 hype_knn_prediction 作为一列添加到 hype_quant_test 数据框中。但是当我将 geom_contour(data=hype_quant_test, aes(x=Comments, y=Votings, z= as.numberic(Selected)), breaks=c(0,.5)) 添加到情节时，我收到以下消息：
stat_contour() 中的计算失败：x 坐标的数量必须与密度矩阵中的列数匹配。
我该如何解决这个问题？我假设我必须转换一些数据但不知道如何
编辑
训练数据：
Selected Votings Comments
1 true 0.2348563517 0.162454874
2 false 0.0027691243 0.001805054
3 false 0.0136725511 0.027075812
4 false 0.1128418138 0.077617329
5 false 0.0529595016 0.016245487
6 false 0.0190377293 0.012635379
7 false 0.0231914157 0.001805054
8 false 0.3367947387 0.019855596
9 false 0.0036344756 0.005415162
10 false 0.0051921080 0.005415162
11 false 0.0202492212 0.014440433
12 false 0.0178262375 0.007220217
13 false 0.0029421945 0.010830325
14 false 0.0680166147 0.036101083
15 false 0.0053651783 0.003610108
16 false 0.2397023191 0.034296029
17 false 0.0001730703 0.000000000
18 false 0.0228452752 0.023465704
19 false 0.0129802700 0.000000000
20 false 0.0192107996 0.018050542
21 false 0.0010384216 0.000000000
22 false 0.0129802700 0.005415162
23 false 0.0000000000 0.000000000
24 false 0.0134994808 0.003610108
25 false 0.0742471443 0.039711191
26 false 0.0256143994 0.009025271
27 false 0.0039806161 0.001805054
28 true 0.4110418830 0.050541516
29 false 0.0114226376 0.063176895
30 false 0.0185185185 0.016245487
31 false 0.0051921080 0.003610108
32 false 0.1952232606 0.021660650
33 false 0.1138802354 0.012635379
34 false 0.0048459675 0.016245487
35 false 0.0242298373 0.009025271
36 false 0.0167878159 0.001805054
37 false 0.0039806161 0.001805054
38 true 0.7727587400 0.146209386
39 false 0.0154032537 0.000000000
40 false 0.0057113188 0.007220217
41 false 0.0038075459 0.000000000
42 false 0.0046728972 0.001805054
43 false 0.0152301835 0.003610108
44 false 0.0408445829 0.025270758
45 false 0.0131533403 0.007220217
46 false 0.0578054690 0.037906137
47 false 0.0046728972 0.005415162
48 false 0.0001730703 0.001805054
49 false 0.1169955002 0.122743682
50 false 0.0044998269 0.003610108
51 false 0.0000000000 0.000000000
52 false 0.1439944618 0.036101083
53 false 0.0072689512 0.005415162
54 false 0.0064035999 0.009025271
55 false 0.0614399446 0.027075812
56 false 0.0719972309 0.005415162
57 true 0.3418137764 0.018050542
58 false 0.0117687781 0.012635379
59 false 0.0072689512 0.014440433
60 true 0.0313257182 0.018050542
61 false 0.1021114573 0.019855596
62 false 0.0024229837 0.003610108
63 false 0.0072689512 0.000000000
64 false 0.0169608861 0.003610108
65 false 0.0340948425 0.014440433
66 true 0.7069920388 0.332129964
67 true 0.7377985462 0.175090253
68 false 0.0919003115 0.007220217
69 false 0.0065766701 0.001805054
70 false 0.0401523018 0.027075812
71 false 0.0223260644 0.005415162
72 false 0.0635167878 0.018050542
73 false 0.0013845621 0.000000000
74 false 0.0060574593 0.000000000
75 true 0.6102457598 0.909747292
76 false 0.0022499135 0.001805054
77 false 0.0316718588 0.007220217
78 false 0.0019037729 0.000000000
79 true 1.0000000000 1.000000000
80 false 0.0240567670 0.016245487
添加预测列后的测试数据：
Selected Votings Comments Prediction
1 false 0.329525787 0.023465704 false
2 false 0.299930772 0.075812274 false
3 true 0.962443752 0.178700361 true
4 false 0.032191070 0.001805054 false
5 false 0.036863967 0.025270758 false
6 false 0.014884043 0.005415162 false
7 false 0.034787124 0.005415162 false
8 false 0.007615092 0.000000000 false
9 false 0.005538249 0.000000000 false
10 false 0.006403600 0.005415162 false
11 false 0.006749740 0.005415162 false
12 false 0.048286604 0.072202166 false
13 false 0.057286258 0.021660650 false
14 false 0.067324334 0.012635379 false
15 false 0.004153686 0.001805054 false
16 false 0.004845967 0.003610108 false
17 false 0.089131187 0.055956679 false
18 false 0.010384216 0.001805054 false
19 false 0.040671513 0.021660650 false
20 false 0.001903773 0.001805054 false
编辑
我尝试使用默认的 Iris 数据测试绘图，但消息仍然相同：
library(datasets)
library(class)
library(ggplot2)
library(caret)
iris_df = as.data.frame(iris)
normalize = function(x) {
return ((x - min(x)) / (max(x) - min(x)))
}
iris_df$Sepal.Length = normalize(iris_df$Sepal.Length)
iris_df$Sepal.Width = normalize(iris_df$Sepal.Width)
iris_df$Petal.Length = normalize(iris_df$Petal.Length)
iris_df$Petal.Width = normalize(iris_df$Petal.Width)
set.seed(1234)
#sampling
sample = sample(nrow(iris_df), nrow(iris_df)*0.8, replace = FALSE)
#Training data
iris_train=iris_df[sample,]
#Testdata
iris_test=iris_df[-sample,]
ggplot(iris_train, aes(x= Sepal.Length, y= Sepal.Width, color = Species, shape = Species)) +
geom_point(size = 3) +
theme(legend.position="bottom")
ggplot(iris_test, aes(x= Sepal.Length, y= Sepal.Width, color = Species, shape = Species)) +
geom_point(size = 3) +
theme(legend.position="bottom")
k = round(sqrt(nrow(iris_train)))
knn_predict = knn(iris_train[,1:4], iris_test[,1:4], cl = iris_train$Species, k = k)
iris_test$Prediction = knn_predict
confusionMatrix(iris_test$Prediction, iris_test$Species)
ggplot(iris_test, aes(x= Sepal.Length, y= Sepal.Width, color = Species, shape = Species)) +
geom_point(size = 3) +
geom_contour(data = iris_test, aes(x= Sepal.Length, y= Sepal.Width, z = as.numeric(Prediction)),breaks = c(0,.5))
stat_contour()计算失败：
x 坐标的数量必须与密度矩阵中的列数匹配。
我认为 geom_contour 方法的一个关键部分是对变量矩阵执行预测，而我没有在您的 iris 代码中看到这一点。这是你如何制作一个。我没有对预处理做任何花哨的事情。
library(class)
library(ggplot2)
train <- sample(150, 75)
train_dat <- iris[train, -5]
test_dat <- iris[-train, -5]
vars <- c("Sepal.Width", "Sepal.Length")
# First make a grid
n <- 40
pred.mat <- expand.grid(
Sepal.Width = with(iris, seq(min(Sepal.Width), max(Sepal.Width), length.out = n)),
Sepal.Length = with(iris, seq(min(Sepal.Length), max(Sepal.Length), length.out = n))
)
# Then ask for prediction on the grid
pred.mat$pred <- knn(train_dat[, vars], pred.mat, cl = iris$Species[train], k = 3)
# Use grid as input for geom_contour
ggplot(pred.mat, aes(Sepal.Width, Sepal.Length)) +
geom_point(data = iris, aes(color = Species)) +
geom_contour(aes(z = as.numeric(pred == "setosa"),
colour = "setosa"),
breaks = 0.5) +
geom_contour(aes(z = as.numeric(pred == "virginica"),
colour = "virginica"),
breaks = 0.5) +
geom_contour(aes(z = as.numeric(pred == "versicolor"),
color = "versicolor"),
breaks = 0.5)
由 reprex package (v0.3.0)
于 2020-04-18 创建
我知道有几个类似的问题，但我仍在努力理解它背后的概念。
我根据 knn 预测 类（“真”和“假”）：
hype_knn_prediction = knn(hype_quant_train[,2:3], hype_quant_test[,2:3], cl = hype_quant_train$Selected, k = k)
而且我还可以绘制测试数据：
ggplot(hype_quant_test, aes(x= Comments, y= Votings, color = Selected, shape = Selected)) +
geom_point(size = 3) +
labs(y="Votes", x = "Comments")+
ggtitle("Testdata")+
theme(plot.title = element_text(hjust = 0.5))+
theme(legend.position="bottom")
Plot of test data
现在我想将 类 的决策边界添加到测试数据的图中。因此，我将 hype_knn_prediction 作为一列添加到 hype_quant_test 数据框中。但是当我将 geom_contour(data=hype_quant_test, aes(x=Comments, y=Votings, z= as.numberic(Selected)), breaks=c(0,.5)) 添加到情节时，我收到以下消息：
stat_contour() 中的计算失败：x 坐标的数量必须与密度矩阵中的列数匹配。
我该如何解决这个问题？我假设我必须转换一些数据但不知道如何
编辑
训练数据：
Selected Votings Comments
1 true 0.2348563517 0.162454874
2 false 0.0027691243 0.001805054
3 false 0.0136725511 0.027075812
4 false 0.1128418138 0.077617329
5 false 0.0529595016 0.016245487
6 false 0.0190377293 0.012635379
7 false 0.0231914157 0.001805054
8 false 0.3367947387 0.019855596
9 false 0.0036344756 0.005415162
10 false 0.0051921080 0.005415162
11 false 0.0202492212 0.014440433
12 false 0.0178262375 0.007220217
13 false 0.0029421945 0.010830325
14 false 0.0680166147 0.036101083
15 false 0.0053651783 0.003610108
16 false 0.2397023191 0.034296029
17 false 0.0001730703 0.000000000
18 false 0.0228452752 0.023465704
19 false 0.0129802700 0.000000000
20 false 0.0192107996 0.018050542
21 false 0.0010384216 0.000000000
22 false 0.0129802700 0.005415162
23 false 0.0000000000 0.000000000
24 false 0.0134994808 0.003610108
25 false 0.0742471443 0.039711191
26 false 0.0256143994 0.009025271
27 false 0.0039806161 0.001805054
28 true 0.4110418830 0.050541516
29 false 0.0114226376 0.063176895
30 false 0.0185185185 0.016245487
31 false 0.0051921080 0.003610108
32 false 0.1952232606 0.021660650
33 false 0.1138802354 0.012635379
34 false 0.0048459675 0.016245487
35 false 0.0242298373 0.009025271
36 false 0.0167878159 0.001805054
37 false 0.0039806161 0.001805054
38 true 0.7727587400 0.146209386
39 false 0.0154032537 0.000000000
40 false 0.0057113188 0.007220217
41 false 0.0038075459 0.000000000
42 false 0.0046728972 0.001805054
43 false 0.0152301835 0.003610108
44 false 0.0408445829 0.025270758
45 false 0.0131533403 0.007220217
46 false 0.0578054690 0.037906137
47 false 0.0046728972 0.005415162
48 false 0.0001730703 0.001805054
49 false 0.1169955002 0.122743682
50 false 0.0044998269 0.003610108
51 false 0.0000000000 0.000000000
52 false 0.1439944618 0.036101083
53 false 0.0072689512 0.005415162
54 false 0.0064035999 0.009025271
55 false 0.0614399446 0.027075812
56 false 0.0719972309 0.005415162
57 true 0.3418137764 0.018050542
58 false 0.0117687781 0.012635379
59 false 0.0072689512 0.014440433
60 true 0.0313257182 0.018050542
61 false 0.1021114573 0.019855596
62 false 0.0024229837 0.003610108
63 false 0.0072689512 0.000000000
64 false 0.0169608861 0.003610108
65 false 0.0340948425 0.014440433
66 true 0.7069920388 0.332129964
67 true 0.7377985462 0.175090253
68 false 0.0919003115 0.007220217
69 false 0.0065766701 0.001805054
70 false 0.0401523018 0.027075812
71 false 0.0223260644 0.005415162
72 false 0.0635167878 0.018050542
73 false 0.0013845621 0.000000000
74 false 0.0060574593 0.000000000
75 true 0.6102457598 0.909747292
76 false 0.0022499135 0.001805054
77 false 0.0316718588 0.007220217
78 false 0.0019037729 0.000000000
79 true 1.0000000000 1.000000000
80 false 0.0240567670 0.016245487
添加预测列后的测试数据：
Selected Votings Comments Prediction
1 false 0.329525787 0.023465704 false
2 false 0.299930772 0.075812274 false
3 true 0.962443752 0.178700361 true
4 false 0.032191070 0.001805054 false
5 false 0.036863967 0.025270758 false
6 false 0.014884043 0.005415162 false
7 false 0.034787124 0.005415162 false
8 false 0.007615092 0.000000000 false
9 false 0.005538249 0.000000000 false
10 false 0.006403600 0.005415162 false
11 false 0.006749740 0.005415162 false
12 false 0.048286604 0.072202166 false
13 false 0.057286258 0.021660650 false
14 false 0.067324334 0.012635379 false
15 false 0.004153686 0.001805054 false
16 false 0.004845967 0.003610108 false
17 false 0.089131187 0.055956679 false
18 false 0.010384216 0.001805054 false
19 false 0.040671513 0.021660650 false
20 false 0.001903773 0.001805054 false
编辑
我尝试使用默认的 Iris 数据测试绘图，但消息仍然相同：
library(datasets)
library(class)
library(ggplot2)
library(caret)
iris_df = as.data.frame(iris)
normalize = function(x) {
return ((x - min(x)) / (max(x) - min(x)))
}
iris_df$Sepal.Length = normalize(iris_df$Sepal.Length)
iris_df$Sepal.Width = normalize(iris_df$Sepal.Width)
iris_df$Petal.Length = normalize(iris_df$Petal.Length)
iris_df$Petal.Width = normalize(iris_df$Petal.Width)
set.seed(1234)
#sampling
sample = sample(nrow(iris_df), nrow(iris_df)*0.8, replace = FALSE)
#Training data
iris_train=iris_df[sample,]
#Testdata
iris_test=iris_df[-sample,]
ggplot(iris_train, aes(x= Sepal.Length, y= Sepal.Width, color = Species, shape = Species)) +
geom_point(size = 3) +
theme(legend.position="bottom")
ggplot(iris_test, aes(x= Sepal.Length, y= Sepal.Width, color = Species, shape = Species)) +
geom_point(size = 3) +
theme(legend.position="bottom")
k = round(sqrt(nrow(iris_train)))
knn_predict = knn(iris_train[,1:4], iris_test[,1:4], cl = iris_train$Species, k = k)
iris_test$Prediction = knn_predict
confusionMatrix(iris_test$Prediction, iris_test$Species)
ggplot(iris_test, aes(x= Sepal.Length, y= Sepal.Width, color = Species, shape = Species)) +
geom_point(size = 3) +
geom_contour(data = iris_test, aes(x= Sepal.Length, y= Sepal.Width, z = as.numeric(Prediction)),breaks = c(0,.5))
stat_contour()计算失败：
x 坐标的数量必须与密度矩阵中的列数匹配。
我认为 geom_contour 方法的一个关键部分是对变量矩阵执行预测，而我没有在您的 iris 代码中看到这一点。这是你如何制作一个。我没有对预处理做任何花哨的事情。
library(class)
library(ggplot2)
train <- sample(150, 75)
train_dat <- iris[train, -5]
test_dat <- iris[-train, -5]
vars <- c("Sepal.Width", "Sepal.Length")
# First make a grid
n <- 40
pred.mat <- expand.grid(
Sepal.Width = with(iris, seq(min(Sepal.Width), max(Sepal.Width), length.out = n)),
Sepal.Length = with(iris, seq(min(Sepal.Length), max(Sepal.Length), length.out = n))
)
# Then ask for prediction on the grid
pred.mat$pred <- knn(train_dat[, vars], pred.mat, cl = iris$Species[train], k = 3)
# Use grid as input for geom_contour
ggplot(pred.mat, aes(Sepal.Width, Sepal.Length)) +
geom_point(data = iris, aes(color = Species)) +
geom_contour(aes(z = as.numeric(pred == "setosa"),
colour = "setosa"),
breaks = 0.5) +
geom_contour(aes(z = as.numeric(pred == "virginica"),
colour = "virginica"),
breaks = 0.5) +
geom_contour(aes(z = as.numeric(pred == "versicolor"),
color = "versicolor"),
breaks = 0.5)
由 reprex package (v0.3.0)
于 2020-04-18 创建