如何计算 R 单元格中由逗号分隔的唯一 2 个单词短语？

Question

我有一个包含不同位置 (Location) 以及在每个位置发现的动物种类 (Spp) 的数据框。动物种类使用其独特的属种名称进行编码。我想知道每个独特的属物种在数据框中的频率。

示例数据

df1 <- data.frame(matrix(ncol = 2, nrow = 3))
x <- c("Location","Spp")
colnames(df1) <- x
df1$Location <- seq(1,3,1)
df1[1,2] <- c("Genus1 species1")
df1[2,2] <- c("Genus1 species1, Genus1 species2")
df1[3,2] <- c("Genus1 species1, Genus1 species2, Genus2 species1")

输出应如下所示

            Spp Freq
Genus1 species1    3
Genus1 species2    2
Genus2 species1    1

我已经尝试使用 corpus 包来解决这个问题，但只能让它计算独特的单词而不是独特的属种短语。

library(tm)
library(corpus)
library(dplyr)

text <- df1[,2]
docs <- Corpus(VectorSource(text))
docs <- docs %>%
  tm_map(removePunctuation)
dtm <- TermDocumentMatrix(docs)
matrix <- as.matrix(dtm)
words <- sort(rowSums(matrix), decreasing = TRUE)
words ### only provides count of unique individual Genus and species words. I want similar but need to keep Genus and species together.

Answer 1

这是一个快速的解决方案：

df1 <- data.frame(matrix(ncol = 2, nrow = 3))
x <- c("Location","Spp")
colnames(df1) <- x
df1$Location <- seq(1,3,1)
df1[1,2] <- c("Genus1 species1")
df1[2,2] <- c("Genus1 species1, Genus1 species2")
df1[3,2] <- c("Genus1 species1, Genus1 species2, Genus2 species1")

table(unlist(strsplit(df1$Spp,', ')))
#> 
#> Genus1 species1 Genus1 species2 Genus2 species1 
#>               3               2               1

^{由 reprex package (v2.0.1)}

于 2021-10-04 创建

Answer 2

我们可以使用 separate_rows 和 count

library(dplyr)
library(tidyr)
df1 %>% 
   separate_rows(Spp, sep = ",\s+") %>%
   count(Spp, name = 'Freq')
# A tibble: 3 × 2
  Spp              Freq
  <chr>           <int>
1 Genus1 species1     3
2 Genus1 species2     2
3 Genus2 species1     1

如何计算 R 单元格中由逗号分隔的唯一 2 个单词短语？

How can I count unique 2 word phrases that are seperated by a comma within a cell in R?

r

corpus