R 如何转换table（）命令以从第三列接收频率？

有这样一个数据帧：

df1 <- data.frame(stock = c("Google, Yahoo", "Google", "Yahoo, Google", "Amazon, Google", "Google, Amazon"), investor = c("Nathalie","George","Nathalie, George", "Melanie, George","Melanie"))

如何可能添加一个过滤器，以获得每只股票的频率，但基于显示投资者选择偏好的第三列。 以下是预期输出的示例：

data.frame(investor = c("Nathalie", "George", "George", "George", "Melanie", "Malanie"), stock = c("Google, Yahoo", "Google", "Google, Yahoo", "Amazon, Google", "Amazon, Google", "Amazon"), frq = c(2,1,1,1,1,1))

再添加一列：

df1 <- data.frame(stock = c("Google, Yahoo", "Google", "Yahoo, Google", "Amazon, Google", "Google, Amazon"), investor = c("Nathalie","George","Nathalie, George", "Melanie, George","Melanie"), year = c("2017", "2018", "2017", "2018", "2017"))

---

df1在我们像上一步一样对值进行排序后，不要直接执行
表
，而是通过赋值来更新列，然后我们可以使用
tidyverse
方法拆分“investor”行，并使用
add\u count

library(tidyverse)
df1$stock <- sapply(strsplit(as.character(df1$stock), ", "),
                function(x) toString(sort(x)))
df1 %>% 
  mutate_if(is.factor, as.character) %>% 
  separate_rows(investor) %>% 
  add_count(stock, investor, year)

库（tidyverse）
df1美元股票%
如果（is.factor，as.character）%>%，则进行变异
单独行（投资者）%>%
添加计数（股票、投资者、年份）
是否拆分“投资者”的名称@akrun是的，我想为此拆分投资者的名称case@akrun或者可能是像meltI这样的东西，我没有得到您显示的准确输出
df1$stock%mutate\u all（as.character）%%>%separate\u rows（investor）%%>%add\u count（stock，investor）
@akrun谢谢。你又对了！
df1 <- data.frame(stock = c("Google, Yahoo", "Google", "Yahoo, Google", "Amazon, Google", "Google, Amazon"), investor = c("Nathalie","George","Nathalie, George", "Melanie, George","Melanie"), year = c("2017", "2018", "2017", "2018", "2017"))

library(tidyverse)
df1$stock <- sapply(strsplit(as.character(df1$stock), ", "),
                function(x) toString(sort(x)))
df1 %>% 
  mutate_if(is.factor, as.character) %>% 
  separate_rows(investor) %>% 
  add_count(stock, investor, year)