Question

我有一个由正词和负词关联组成的数据表。我想创建两个词云，一个用于肯定词，一个用于否定词。

sentiment_words表的示例：

          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，如何仅从“正”列中拉出单词以创建wordcloud？我不确定如何解决每行有多个单词的事实（例如，“同意，可用”应视为两个条目）

我对wordcloud()函数进行了不同的尝试，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"))，但这只会返回术语在第一个条目中的云

编辑：我已经尝试过下面的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

我已经尝试将gsub()和apply的变体相乘以删除多余的)和c(，但是还没有找到任何可行的方法。结果是应该一起计算的单词被分别计算（例如，“可接受的”和“可接受的”是单词云中的两个不同的单词）

编辑：为了使其能够正常工作，我必须按照以下建议先清理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函数中其余的“ character（0”）字符串，请注意，它过滤了“ character（0”而不是“ character（0）”，因为我删除了右括号以上

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

实施上述操作可以根据文本的极性提供最干净的文字云

Answer 1

这是一种基于tidyverse的方法，可以帮助您入门。我同意Mr_Z的意见，因为我不清楚问题出在哪里。

让我们定义一个函数，该函数根据源数据data.frame的特定列var中逗号分隔的单词来生成一个单词计数的df。

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列生成单词计数

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

让我们检查data.frame

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

让我们画出云这个词

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

我强烈建议您不要在此处使用已接受的答案，因为它忽略了 sentimentr 已经为您返回了计算的计数（通过attributes(sentiment_words)$counts）。 documentation for extract_sentiment_terms shows examples使这一点更加清楚（存在改进返回的文档的空间，并且已在开发版本https://github.com/trinker/sentimentr/blob/master/R/extract_sentiment_terms.R中添加了该文档）。下面我展示了如何提取用于wordcloud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)

R中的数据表中的Wordcloud

2 个答案: