Question

我有11个文本文档的语料库。我使用命令找到了单词关联：

findAssocs(dtm, c("youngster","campaign"), corlimit=0.9) 
findAssocs(dtms, "corruption", corlimit=0.9)

dtm是一个文档术语矩阵。

 dtm <- DocumentTermMatrix(docs)

其中docs是语料库。

dtms是删除10％稀疏术语后的文档术语矩阵。

dtms <- removeSparseTerms(dtm, 0.1)

我想绘制我得到的相关术语（i）2个特定单词和（ii）1个特定单词我尝试过这篇文章：Plot highly correlated words against a specific word of interest

toi <- "corruption" # term of interest
corlimit <- 0.9 #  lower correlation bound limit.
cor_0.9 <- data.frame(corr = findAssocs(dtm, toi, corlimit)[,1],terms=row.names(findAssocs(dtm, toi, corlimit)))

但不幸的是代码：

cor_0.9 <- data.frame(corr = findAssocs(dtm, toi, corlimit)[,1],terms=row.names(findAssocs(dtm, toi, corlimit)))

给了我一个错误：

Error in findAssocs(dtm, toi, corlimit)[, 1]:incorrect number of dimensions

这是文档术语矩阵的结构：

dtm
<<DocumentTermMatrix (documents: 11, terms: 1847)>>
Non-/sparse entries: 8024/12293
Sparsity           : 61%
Maximal term length: 23
Weighting          : term frequency (tf)

并且在environemt中它的形式为：

dtm                 List of 6
i: int [1:8024] 1  1  1  1  1  ...
j: int [1:8024] 17 29 34 43 47 ...
v: num [1:8024] 9  4  9  5  5  ...
nrow : int 11
ncol : int 1847
dimnames: list of 2
...$ Docs : chr [1:11]  "character (0)" "character (0)" "character (0)"
...$ Terms: chr [1:1847] "campaigning"|__truncated__"a"|__"truncated"__ 
attr(*,"class") = chr [1:2] "DocumentTermMatrix" "simple_triplet_matrix"...
attr(*,"weighting") = chr [1:2] "term frequency" "tf"

如何为单个单词和多个单词绘制单词关联？请帮忙。

这是

的输出

findAssocs(dtm, c("youngster","campaign"), corlimit=0.9)

$youngster
   character         colleges      controversi           expect          corrupt             much 
        1.00             1.00             1.00             1.00             0.99             0.99 
        okay          saritha         existing            leads          satisfi           social 
        0.99             0.99             0.98             0.98             0.98             0.98

$campaign
basic      make      lack  internal   general    method satisfied      time 
 0.95      0.95      0.94      0.93      0.92      0.92      0.92      0.92

Answer 1

两个单词需要稍微不同的方法，这是一个快速尝试：

require(tm)
data("crude")
tdm <- TermDocumentMatrix(crude)

# Compute correlations and store in data frame...

toi1 <- "oil" # term of interest
toi2 <- "winter"
corlimit <- 0.7 #  lower correlation bound limit.

corr1 <-  findAssocs(tdm, toi1, corlimit)[[1]]
corr1 <- cbind(read.table(text = names(corr1), stringsAsFactors = FALSE), corr1)
corr2 <- findAssocs(tdm, toi2, corlimit)[[1]]
corr2 <- cbind(read.table(text = names(corr2), stringsAsFactors = FALSE), corr2)

# join them together
library(dplyr)
two_terms_corrs <- full_join(corr1, corr2)

# gather for plotting
library(tidyr)
two_terms_corrs_gathered <- gather(two_terms_corrs, term, correlation, corr1:corr2)

# insert the actual terms of interest so they show up on the legend
two_terms_corrs_gathered$term <- ifelse(two_terms_corrs_gathered$term  == "corr1", toi1, toi2)


# Draw the plot...

require(ggplot2)
ggplot(two_terms_corrs_gathered, aes(x = V1, y = correlation, colour =  term ) ) +
  geom_point(size = 3) +
  ylab(paste0("Correlation with the terms ", "\"", toi1,  "\"", " and ",  "\"", toi2, "\"")) +
  theme_bw() +
  theme(axis.text.x = element_text(angle = 90, hjust = 1, vjust = 0.5))

R：将相关单词与一个或多个单词进行可视化

1 个答案: