R如何从理货制作词云？

Question

这是我的计数的重复：

word_tally <- data.frame(scarred  = c(1,1,0,0,0,0,0,0,0,0,0,0,0), 
                         happy    = c(0,0,1,0,0,0,0,0,0,0,0,0,0),
                         cheerful = c(0,0,0,1,0,0,0,0,0,0,0,0,0),
                         mad      = c(0,0,0,0,1,1,1,1,1,0,0,0,0),
                         curious  = c(0,0,0,0,0,0,0,0,0,1,1,1,1))

要制作词云，我似乎需要 1 个包含所有词的列。我怎样才能转换上述数据框来为词云制作这种类型的结构？

Answer 1

您可以获取长格式的数据并删除 value = 0 中的行。

library(dplyr)

tidyr::pivot_longer(word_tally, cols = everything(), names_to = "word") %>%
  filter(value != 0) %>%
  select(word)  


# A tibble: 13 x 1
#   word    
#   <chr>   
# 1 scarred 
# 2 scarred 
# 3 happy   
# 4 cheerful
# 5 mad     
# 6 mad     
# 7 mad     
# 8 mad     
# 9 mad     
#10 curious 
#11 curious 
#12 curious 
#13 curious

这将给出一列中的所有单词，可以用作词云的输入。

在 base R 中，另一种方式可能是：

names(word_tally)[which(word_tally != 0, arr.ind = TRUE)[,2]]

Answer 2

使用 rep 和 colSums:

words <- rep(names(word_tally), colSums(word_tally))
words
 [1] "scarred"  "scarred"  "happy"    "cheerful" "mad"     
 [6] "mad"      "mad"      "mad"      "mad"      "curious" 
[11] "curious"  "curious"  "curious"

或者因为频率是列总和，所以只使用数据。

wordcloud(names(word_tally), freq=colSums(word_tally), min.freq = 1)

R如何从理货制作词云？

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