来自 R 中数据 Table 的词云

Question

我有一个数据 table 由正面和负面的单词联想组成。我想创建两个词云，一个用于正面词，一个用于负面词。

sentiment_wordstable的例子：

          element_id    sentence_id   negative     positive
1115:          1        1115          limits       agree,available
1116:          1        1116          slow         strongly,agree
1117:          1        1117                       management
1118:          1        1118                                      
1119:          1        1119          concerns     strongly,agree,better,

我正在使用 library(wordcloud) 和 library(sentimentr)

例如，如何只从 "positive" 列中提取单词来创建词云？我不确定如何解决每行关联多个单词的事实（例如，"agree, available" 应视为两个条目）

我对 wordcloud() 函数做了不同的尝试，例如 wordcloud(words = sentiment_words$positive, freq = 3, min.freq = 1, max.words = 200, random.order = FALSE, rot.per=0.35, colors=brewer.pal(8, "Dark2")) 但这只是 returns 第一个词条中的云

编辑：我尝试了下面的 tidyverse 答案，得到的结果是： words n <chr> <int> 1 " \"ability\"" 3 2 " \"ability\")" 1 3 " \"acceptable\")" 1 4 " \"accomplish\"" 1 5 " \"accomplished\")" 1 6 " \"accountability\"" 1 7 " \"accountability\")" 1 8 " \"accountable\"" 2 9 " \"accountable\")" 1

我已经尝试了 gsub() 和 apply 的乘法变体来删除额外的 ) 和 c( 但还没有找到任何有效的方法。结果是应该一起计数的词被单独计数（例如，"acceptable" 和 "acceptable)" 在词云中是两个不同的词）

编辑：为了让它正常工作，我必须先按照下面的建议清理我的 sentiment_words。

for (j in seq(sentiment_words)) {
  sentiment_words[[j]] <- gsub("character(0)", "", sentiment_words[[j]])
  sentiment_words[[j]] <- gsub('"', "", sentiment_words[[j]])
  sentiment_words[[j]] <- gsub("c\(", "", sentiment_words[[j]])
  sentiment_words[[j]] <- gsub(" ", "", sentiment_words[[j]])
  sentiment_words[[j]] <- gsub("\)", "", sentiment_words[[j]])  
}

而且我还必须过滤掉 count_words 函数中剩余的 "character(0" 字符串。请注意，它过滤 "character(0" 而不是 "character(0)" 因为我删除了上面的右括号

filter(!!var != "character(0") %>%

执行上述操作给出了基于文本极性的最干净的词云

Answer 1

这是一个基于 tidyverse 的方法，应该可以帮助您入门。我同意 Mr_Z，因为我不完全清楚问题出在哪里。

让我们定义一个函数，该函数根据源数据 df 的特定列 var 中以逗号分隔的单词生成带有单词计数的 data.frame .

library(tidyverse)
count_words <- function(df, var) {
    var <- enquo(var)
    df %>%
        separate_rows(!!var, sep = ",") %>%
        filter(!!var != "") %>%
        group_by(!!var) %>%
        summarise(n = n()) %>%
        rename(words = !!var)
}

然后我们可以为 positive 和 negative 列生成字数统计

df.pos <- count_words(df, positive)
df.neg <- count_words(df, negative)

让我们检查 data.frames

df.pos
# A tibble: 5 x 2
  words          n
  <chr>      <int>
1 agree          3
2 available      1
3 better         1
4 management     1
5 strongly       2

df.neg
# A tibble: 3 x 2
  words        n
  <chr>    <int>
1 concerns     1
2 limits       1
3 slow         1

让我们绘制词云

library(wordcloud)
wordcloud(words = df.pos$words, freq = df.pos$n, min.freq = 1,
          max.words = 200, random.order = FALSE, rot.per = 0.35,
          colors = brewer.pal(8, "Dark2"))

wordcloud(words = df.neg$words, freq = df.neg$n, min.freq = 1,
          max.words = 200, random.order = FALSE, rot.per = 0.35,
          colors = brewer.pal(8, "Dark2"))

Answer 2

我强烈建议不要在此处使用已接受的答案，因为它忽略了 sentimentr 已经 returns 为您计算的计数（通过 attributes(sentiment_words)$counts）。 documentation for extract_sentiment_terms shows examples that makes this more clear (there's was room for improving the documentation about what is returned and has been added in the dev version: https://github.com/trinker/sentimentr/blob/master/R/extract_sentiment_terms.R)。下面我展示了如何提取用于词云和一些潜在布局的计数：

library(sentimentr)
library(wordcloud)
library(data.table)

set.seed(10)
x <- get_sentences(sample(hu_liu_cannon_reviews[[2]], 1000, TRUE))
sentiment_words <- extract_sentiment_terms(x)

sentiment_counts <- attributes(sentiment_words)$counts
sentiment_counts[polarity > 0,]

par(mfrow = c(1, 3), mar = c(0, 0, 0, 0))
## Positive Words
with(
    sentiment_counts[polarity > 0,],
    wordcloud(words = words, freq = n, min.freq = 1,
          max.words = 200, random.order = FALSE, rot.per = 0.35,
          colors = brewer.pal(8, "Dark2"), scale = c(4.5, .75)
    )
)
mtext("Positive Words", side = 3, padj = 5)

## Negative Words
with(
    sentiment_counts[polarity < 0,],
    wordcloud(words = words, freq = n, min.freq = 1,
          max.words = 200, random.order = FALSE, rot.per = 0.35,
          colors = brewer.pal(8, "Dark2"), scale = c(4.5, 1)
    )
)
mtext("Negative Words", side = 3, padj = 5)

sentiment_counts[, 
    color := ifelse(polarity > 0, 'red', 
        ifelse(polarity < 0, 'blue', 'gray70')
    )]

## Together
with(
    sentiment_counts[polarity != 0,],
    wordcloud(words = words, freq = n, min.freq = 1,
          max.words = 200, random.order = FALSE, rot.per = 0.35,
          colors = color, ordered.colors = TRUE, scale = c(5, .75)
    )
)
mtext("Positive (red) & Negative (blue) Words", side = 3, padj = 5)

来自 R 中数据 Table 的词云

Wordcloud from Data Table in R

r

data-visualization

sentiment-analysis