Question

我有一堆由同一个人写的文本，我正在尝试估算他们用于每个文本的模板。我的方式是：

为所有文字

TermDocumentMatrix

取每对的原始欧几里德距离
删除任何大于X距离的对（为了争论而为10）
压扁森林
返回每个模板的一个示例，其中包含一些汇总统计信息

我能够达到距离对的程度，但我无法将dist实例转换为可以使用的实例。 底部有一个可重现的例子。

dist实例中的数据如下所示：

行名和列名对应于原始文本列表中的索引，我可以用它来完成第5步。

我一直想要摆脱的是一个带有col name, row name, value的稀疏矩阵。

col, row, value
  1    2  14.966630
  1    3  12.449900
  1    4  13.490738
  1    5  12.688578
  1    6  12.369317
  2    3  12.449900
  2    4  13.564660
  2    5  12.922848
  2    6  12.529964
  3    4   5.385165
  3    5   5.830952
  3    6   5.830952
  4    5   7.416198
  4    6   7.937254
  5    6   7.615773

从这一点开始，我会很乐意切掉比我的截止值更大的所有对并使森林变平，即在此示例中返回3个模板，仅包含文档1的组，仅包含文档2的组和包含文档的第三组3,4,5和6。

我已经尝试了一些东西，从创建一个矩阵，然后尝试使其稀疏，直接使用dist类中的向量，我似乎无法想象它进行。

可重复的例子：

tdm <- matrix(c(1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,3,1,2,2,2,3,2,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,2,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,4,1,1,1,1,1,0,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,2,0,0,1,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,1,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,1,1,1,1,0,1,0,1,0,0,2,0,0,0,0,0,1,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,3,1,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,1,1,1,0,0,1,1,1,1,0,0,0,1,0,0,2,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,3,1,1,1,1,0,1,0,0,0,0,1,2,0,1,1,0,0,0,0,1,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,1,0,1,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,0,0,1,1,1,2,1,1,1,0,0,0,0,1,2,2,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,1,0,2,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,2,0,2,2,3,2,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,2,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,2,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,2,1,1,1,1,1,0,1,0,0,0,0,1,1,0,0,0,0,1,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,1,0,0,1,1,1,0,0,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,2,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,1,0,2,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,3,0,1,1,1,1,0,0,1,0,1,1,1,0,0,0,0,0,1,0,0,0,0,0,4,2,4,6,4,3,1,0,1,2,1,1,0,1,0,0,0,0,2,0,0,0,0,0,0,1,1,1,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,2,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,2,1,2,2,2,2,1,0,1,2,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,2,2,2,2,2,2,3,3,4,5,3,1,2,1,1,1,1,1,1,0,0,0,0,3,3,0,0,1,1,0,1,0,0,0,0), nrow=6)
rownames(tdm) <- 1:6
colnames(tdm) <- paste("term", 1:229, sep="")
tdm.dist <- dist(tdm)
# I'm stuck turning tdm.dist into what I have shown

Answer 1

将类似“矩阵”的对象转换为[row，col，value]“data.frame”的经典方法是as.data.frame(as.table(.))路由。具体来说，我们需要：

subset(as.data.frame(as.table(as.matrix(tdm.dist))), as.numeric(Var1) < as.numeric(Var2))

但这包括太多的强制和更大的对象的创建只是立即成为子集。

由于dist以“lower.tri”角度形式存储其值，我们可以使用combn生成行/列索引，使用“dist”对象生成cbind：< / p>

data.frame(do.call(rbind, combn(attr(tdm.dist, "Size"), 2, simplify = FALSE)), c(tdm.dist))

此外，“Matrix”软件包具有一定的灵活性，可以在此处使用创建对象的内存效率：

library(Matrix)
tmp = combn(attr(tdm.dist, "Size"), 2)
summary(sparseMatrix(i = tmp[2, ], j = tmp[1, ], x = c(tdm.dist), 
                     dims = rep_len(attr(tdm.dist, "Size"), 2), symmetric = TRUE))

此外，在处理“dist”对象的不同函数中，

cutree(hclust(tdm.dist), h = 10)
#1 2 3 4 5 6 
#1 2 3 3 3 3

通过指定切割高度进行分组。

Answer 2

这就是我过去使用dplyr和tidyr包做过类似的事情。您可以逐行运行链式（%>%）脚本，以查看数据集如何逐步更新。

tdm <- matrix(c(1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,3,1,2,2,2,3,2,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,2,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,4,1,1,1,1,1,0,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,2,0,0,1,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,1,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,1,1,1,1,0,1,0,1,0,0,2,0,0,0,0,0,1,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,3,1,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,1,1,1,0,0,1,1,1,1,0,0,0,1,0,0,2,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,3,1,1,1,1,0,1,0,0,0,0,1,2,0,1,1,0,0,0,0,1,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,1,0,1,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,0,0,1,1,1,2,1,1,1,0,0,0,0,1,2,2,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,1,0,2,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,2,0,2,2,3,2,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,2,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,2,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,2,1,1,1,1,1,0,1,0,0,0,0,1,1,0,0,0,0,1,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,1,0,0,1,1,1,0,0,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,2,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,1,0,2,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,3,0,1,1,1,1,0,0,1,0,1,1,1,0,0,0,0,0,1,0,0,0,0,0,4,2,4,6,4,3,1,0,1,2,1,1,0,1,0,0,0,0,2,0,0,0,0,0,0,1,1,1,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,2,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,2,1,2,2,2,2,1,0,1,2,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,2,2,2,2,2,2,3,3,4,5,3,1,2,1,1,1,1,1,1,0,0,0,0,3,3,0,0,1,1,0,1,0,0,0,0), nrow=6)
rownames(tdm) <- 1:6
colnames(tdm) <- paste("term", 1:229, sep="")
tdm.dist <- dist(tdm)

library(dplyr)
library(tidyr)


tdm.dist %>% 
  as.matrix() %>%                      # update dist object to a matrix
  data.frame() %>%                     # update matrix to a data frame
  setNames(nm = 1:ncol(.)) %>%         # update column names
  mutate(names1 = 1:nrow(.)) %>%       # use rownames as a variable
  gather(names2, value , -names1) %>%  # reshape data
  filter(names1 <= names2)             # keep the values only once

#    names1 names2     value
# 1       1      1  0.000000
# 2       1      2 14.966630
# 3       2      2  0.000000
# 4       1      3 12.449900
# 5       2      3 12.449900
# 6       3      3  0.000000
# 7       1      4 13.490738
# 8       2      4 13.564660
# 9       3      4  5.385165
# 10      4      4  0.000000
# 11      1      5 12.688578
# 12      2      5 12.922848
# 13      3      5  5.830952
# 14      4      5  7.416198
# 15      5      5  0.000000
# 16      1      6 12.369317
# 17      2      6 12.529964
# 18      3      6  5.830952
# 19      4      6  7.937254
# 20      5      6  7.615773
# 21      6      6  0.000000

R重新排列数据

2 个答案: