目前,我尝试在分组数据中找到群集的中心。通过使用示例数据集和问题定义,我可以创建包含每个组的kmeans
集群。但是,当谈到给定组的每个群集中心时,我不知道如何获取它们。 https://rdrr.io/cran/broom/man/kmeans_tidiers.html
示例数据为from(添加gr
列几乎没有修改)
样本数据
library(dplyr)
library(broom)
library(ggplot2)
set.seed(2015)
sizes_1 <- c(20, 100, 500)
sizes_2 <- c(10, 50, 100)
centers_1 <- data_frame(x = c(1, 4, 6),
y = c(5, 0, 6),
n = sizes_1,
cluster = factor(1:3))
centers_2 <- data_frame(x = c(1, 4, 6),
y = c(5, 0, 6),
n = sizes_2,
cluster = factor(1:3))
points1 <- centers_1 %>%
group_by(cluster) %>%
do(data_frame(x = rnorm(.$n, .$x),
y = rnorm(.$n, .$y),
gr="1"))
points2 <- centers_2 %>%
group_by(cluster) %>%
do(data_frame(x = rnorm(.$n, .$x),
y = rnorm(.$n, .$y),
gr="2"))
combined_points <- rbind(points1, points2)
> combined_points
# A tibble: 780 x 4
# Groups: cluster [3]
cluster x y gr
<fctr> <dbl> <dbl> <chr>
1 1 3.66473833 4.285771 1
2 1 0.51540619 5.565826 1
3 1 0.11556319 5.592178 1
4 1 1.60513712 5.360013 1
5 1 2.18001557 4.955883 1
6 1 1.53998887 4.530316 1
7 1 -1.44165622 4.561338 1
8 1 2.35076259 5.408538 1
9 1 -0.03060973 4.980363 1
10 1 2.22165205 5.125556 1
# ... with 770 more rows
ggplot(combined_points, aes(x, y)) +
facet_wrap(~gr) +
geom_point(aes(color = cluster))
好的,直到这里我的一切都很棒。当我想在每个组中提取每个集群中心时
clust <- combined_points %>%
group_by(gr) %>%
dplyr::select(x, y) %>%
kmeans(3)
> clust
K-means clustering with 3 clusters of sizes 594, 150, 36
Cluster means:
gr x y
1 1.166667 6.080832 6.0074885
2 1.333333 4.055645 0.0654158
3 1.305556 1.507862 5.2417670
我们可以看到gr
号码已更改,我不知道这些中心属于哪个群组。
我们向前迈出一步,看tidy
clust
格式
> tidy(clust)
x1 x2 x3 size withinss cluster
1 1.166667 6.080832 6.0074885 594 1095.3047 1
2 1.333333 4.055645 0.0654158 150 312.4182 2
3 1.305556 1.507862 5.2417670 36 115.2484 3
我仍然看不到gr 2
中心信息。
我希望问题解释清楚。如果您有任何遗漏,请告诉我!提前谢谢!
答案 0 :(得分:4)
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并不了解dplyr分组,所以它只是找到三个整体中心而不是每个群组。此时优选的习惯用法是输入数据的列表列,例如
kmeans
请注意,library(tidyverse)
points_and_models <- combined_points %>%
ungroup() %>% select(-cluster) %>% # cleanup, remove cluster name so data will collapse
nest(x, y) %>% # collapse input data into list column
mutate(model = map(data, kmeans, 3), # iterate model over list column of input data
centers = map(model, broom::tidy)) # extract data from models
points_and_models
#> # A tibble: 2 x 4
#> gr data model centers
#> <chr> <list> <list> <list>
#> 1 1 <tibble [620 × 2]> <S3: kmeans> <data.frame [3 × 5]>
#> 2 2 <tibble [160 × 2]> <S3: kmeans> <data.frame [3 × 5]>
points_and_models %>% unnest(centers)
#> # A tibble: 6 x 6
#> gr x1 x2 size withinss cluster
#> <chr> <dbl> <dbl> <int> <dbl> <fct>
#> 1 1 4.29 5.71 158 441. 1
#> 2 1 3.79 0.121 102 213. 2
#> 3 1 6.39 6.06 360 534. 3
#> 4 2 5.94 5.88 100 194. 1
#> 5 2 4.01 -0.127 50 97.4 2
#> 6 2 1.07 4.57 10 15.7 3
列来自模型结果,而不是输入数据。
你也可以用cluster
做同样的事情,例如
do
但是combined_points %>%
group_by(gr) %>%
do(model = kmeans(.[c('x', 'y')], 3)) %>%
ungroup() %>% group_by(gr) %>%
do(map_df(.$model, broom::tidy)) %>% ungroup()
和rowwise的分组在这一点上有点软弃用,代码变得有点笨拙,正如你可以看到显式do
的需要那么多。