如何根据分层字典名称结构在行名和列名上聚合data.frame？

Question

(道歉,我不确定这篇文章的最佳标题是什么,随意编辑).

可以说我在单词和它们的类型(即字典)之间有以下关系结构:

dictionary <- data.frame(level1=c(rep("Positive", 3), rep("Negative", 3)), level2 = c("happy", "fantastic", "great", "sad", "rubbish", "awful"))

#     level1    level2
# 1 Positive     happy
# 2 Positive fantastic
# 3 Positive     great
# 4 Negative       sad
# 5 Negative   rubbish
# 6 Negative     awful

我们已经计算了七个文件(即术语 - 文档矩阵)的出现次数:

set.seed(42)
range = 0:3
df <- data.frame(row.names = c("happy", "fantastic", "great", "sad", "rubbish", "awful"), doc1 = sample(x=range, size=6, replace=TRUE), doc2 = sample(x=range, size=6, replace=TRUE), doc3 = sample(x=range, size=6, replace=TRUE), doc4 = sample(x=range, size=6, replace=TRUE), doc5 = sample(x=range, size=6, replace=TRUE), doc6 = sample(x=range, size=6, replace=TRUE), doc7 = sample(x=range, size=6, replace=TRUE))

#           doc1 doc2 doc3 doc4 doc5 doc6 doc7
# happy        3    2    3    1    0    2    0
# fantastic    3    0    1    2    2    3    0
# great        1    2    1    3    1    1    3
# sad          3    2    3    0    3    2    2
# rubbish      2    1    3    3    1    0    1
# awful        2    2    0    3    3    3    1    

然后我可以很容易地计算出两个单词出现在同一个文档中的频率(即共现或邻接矩阵):

# binary to indicate a co-occurrence
df[df > 0] <- 1
# sum co-occurrences
m <- as.matrix(df) %*% t(as.matrix(df))

#           happy fantastic great sad rubbish awful
# happy         5         4     5   4       4     4
# fantastic     4         5     5   4       4     4
# great         5         5     7   6       6     6
# sad           4         4     6   6       5     5
# rubbish       4         4     6   5       6     5
# awful         4         4     6   5       5     6

问题:如何重构我的共生矩阵,以便我查看字典中的单词类型(level1)而不是单词本身(level2)？

即我想:

data.frame(row.names = c("Positive", "Negative"), Positive = c(5+4+5+4+5+5+5+5+7, 4+4+6+4+4+6+4+4+6), Negative = c(4+4+4+4+4+4+6+6+6, 6+5+5+5+6+5+5+5+6))

#          Positive Negative
# Positive       45       42
# Negative       42       48

---

到目前为止我所做的事情:以前我曾希望能够从这个问题中推断出基于名称类型的数据列的总和.

然而,虽然我可以减少行:

require(data.table)
dt <- data.table(m)
dt[, level1:=c(rep("Positive", 3), rep("Negative", 3))]
dt[, lapply(.SD, sum), by = "level1"]

#      level1 happy fantastic great sad rubbish awful
# 1: Positive    14        14    17  14      14    14
# 2: Negative    12        12    18  16      16    16

我无法弄清楚如何根据需要减少列.

Answer 1

继续 df[df > 0] <- 1

library(reshape)
library(reshape2)
library(data.table)

# incorporating @RicardoSaporta's suggestion of using data.table(keep.rownames = TRUE)
dt <- data.table(as.matrix(df) %*% t(as.matrix(df)), keep.rownames = TRUE)

#reducing matrix format to plain data format, look at dt to see the change
dt <- melt(dt, "rn")

#getting positive/negative for word1 and word2
dt <- merge(dt,dictionary, all.x = TRUE, by.y = "level2", by.x = "rn")
dt <- merge(dt,dictionary, all.x = TRUE, by.y = "level2", by.x = "variable", suffixes = c("_1","_2"))


#getting counts for each positive/negative - positive/negative combination
dt <- data.table(dt)
dt[,list(value = sum(value)), by = c("level1_1","level1_2")]

#structuring
cast(dt,level1_1~level1_2, fun.aggregate=sum)

产量

> cast(dt,level1_1~level1_2, fun.aggregate=sum)
  level1_1 Negative Positive
1 Negative       48       42
2 Positive       42       45

Answer 2

到目前为止基本上与其他两个解决方案相同,只是更紧凑,可能更快一点:

library(reshape2)
library(data.table)

mdt = data.table(melt(m), key = 'Var1')
dic = data.table(dictionary, key = 'level2')

dcast(dic[setkey(dic[mdt], Var2)], level1 ~ level1.1, fun.aggregate = sum)
#    level1 Negative Positive
#1 Negative       48       42
#2 Positive       42       45