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我正在尝试执行一个非常简单的聚类分析，但无法得到正确的结果。我对大型数据集的问题是“哪些疾病经常一起报告？”。下面的简化数据样本应导致 2 个集群：1) 头痛/头晕 2) 恶心/腹痛。但是，我无法正确获取代码。我正在使用pamanddaisy函数。对于这个例子，我手动分配了 2 个集群 (k=2)，因为我知道所需的结果，但实际上我探索了几个 k 值。

有谁知道我在这里做错了什么？

library(cluster)
library(dplyr)

dat <- data.frame(ID = c("id1","id1","id2","id2","id3","id3","id4","id4","id5","id5"),
                  PTName = c("headache","dizziness","nausea","abd pain","dizziness","headache","abd pain","nausea","headache","dizziness"))


gower_dist <- daisy(dat, metric = "gower")
k <- 2
pam_fit <- pam(gower_dist, diss = TRUE, k)  # performs cluster analysis
pam_results <- dat %>%
  mutate(cluster = pam_fit$clustering) %>%
  group_by(cluster) %>%
  do(the_summary = summary(.))
head(pam_results$the_summary)
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1 回答 1

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将数据集提供给聚类算法的格式不适合您的目标。事实上，如果您想将报告的疾病分组在一起，但同时在相异矩阵中包含 ID，它们将在矩阵构造中占有一席之地，而您不希望这样，因为您的目标仅涉及疾病。

因此，我们需要建立一个数据集，其中每一行是一个患有他/她报告的所有疾病的患者，然后仅在数字特征上构建相异矩阵。presence对于这个任务，如果患者报告了疾病，我将添加一个值为 1 的列，否则为 0；pivot_wider函数（链接）将自动填充零。

这是我使用的代码，我想我达到了你想要的，如果是这样，请告诉我。

library(cluster)
library(dplyr)
library(tidyr)

dat <- data.frame(ID = c("id1","id1","id2","id2","id3","id3","id4","id4","id5","id5"),
                  PTName = c("headache","dizziness","nausea","abd pain","dizziness","headache","abd pain","nausea","headache","dizziness"),
                  presence = 1)
# build the wider dataset: each row is a patient
dat_wider <- pivot_wider(
    dat,
    id_cols = ID,
    names_from = PTName,
    values_from = presence,
    values_fill = list(presence = 0)
)

# in the dissimalirity matrix construction, we leave out the column ID
gower_dist <- daisy(dat_wider %>% select(-ID), metric = "gower")
k <- 2

set.seed(123)
pam_fit <- pam(gower_dist, diss = TRUE, k) 
pam_results <- dat_wider %>%
    mutate(cluster = pam_fit$clustering) %>%
    group_by(cluster) %>%
    do(the_summary = summary(.))
head(pam_results$the_summary)

此外，由于您只使用二进制数据，如果它们更适合您的数据，您可以考虑使用简单匹配或Jaccard距离而不是 Gower 距离。在 R 中，您可以使用它们

sm_dist <- dist(dat_wider %>% select(-ID), method = "manhattan")/p
j_dist <- dist(dat_wider %>% select(-ID), method = "binary")

分别p是您要考虑的二进制变量的数量。

于 2020-02-28T08:54:44.013 回答