R 按降序计算字母频率

我有一个数据看起来像这样

df<-structure(list(col = structure(c(9L, 2L, 13L, 11L, 5L, 7L, 10L, 
6L, 8L, 3L, 12L, 4L, 1L), .Label = c("HHRGGVCTS", "MGSSN", "MVKTTYYDVG", 
"RRHYNGAYDD", "RTSTN", "S", "SNCWC", "sp|P31689|DNJA1_HUMAN DnaJ homolog GN=DNAJA1 PE=1 SV=2  ", 
"sp|Q9H9K5|MER34_HUMAN Endogenous PE=1 SV=1", "THYDT", "TVHAV", 
"VCMCVVDDNR", "YATTA"), class = "factor")), class = "data.frame", row.names = c(NA, 
-13L))

也就是说，有2个

S

，1，

M

，1，

G

，1，

N

，其他字母为空

字符频率按降序排列

最终输出如下所示

output<-structure(list(col = structure(c(9L, 2L, 13L, 11L, 5L, 7L, 10L, 
6L, 8L, 3L, 12L, 4L, 1L), .Label = c("HHRGGVCTS", "MGSSN", "MVKTTYYDVG", 
"RRHYNGAYDD", "RTSTN", "S", "SNCWC", "sp|P31689|DNJA1_HUMAN DnaJ homolog GN=DNAJA1 PE=1 SV=2  ", 
"sp|Q9H9K5|MER34_HUMAN Endogenous PE=1 SV=1", "THYDT", "TVHAV", 
"VCMCVVDDNR", "YATTA"), class = "factor"), Col2 = structure(c(8L, 
2L, 3L, 2L, 2L, 2L, 2L, 1L, 7L, 5L, 6L, 5L, 4L), .Label = c("1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0", 
"2,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0", "2,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0", 
"2,2,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0", "2,2,2,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0", 
"3,2,2,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0", "sp|P31689|DNJA1_HUMAN DnaJ homolog GN=DNAJA1 PE=1 SV=2  ", 
"sp|Q9H9K5|MER34_HUMAN Endogenous PE=1 SV=1"), class = "factor")), class = "data.frame", row.names = c(NA, 
-13L))

---

output我们可以使用
stru计数

library(stringr)
i1 <- !grepl("^sp", df$col)
df$col2[i1] <- sapply(as.character(df$col[i1]), function(x)
     paste(sort(str_count(x, LETTERS), decreasing = TRUE), collapse=", "))
df$col2[!i1] <- df$col[!i1]

现在还不清楚你用的是什么标准。你能详细说明一下吗？试试
df%>%mutate（col=case_当（！str_detect（col，“^sp”）~str_count（col，LETTERS）%%>%str_c（collapse=“，”，TRUE~as.character（col））
第一种解决方案是否可能，我在第二列得到的是
sp |
，而不是数字？@Learner你是指第二列
sp |
后面的字符吗是的，那一行
sp |
@Learner你能更新你帖子中的预期输出吗。为了检查你想要的是什么，我已经展示了上面的输出。你看到了吗？我只想在我们创建的下一个专栏中完整介绍一下
sp |
library(stringr)
i1 <- !grepl("^sp", df$col)
df$col2[i1] <- sapply(as.character(df$col[i1]), function(x)
     paste(sort(str_count(x, LETTERS), decreasing = TRUE), collapse=", "))
df$col2[!i1] <- df$col[!i1]

library(tidyverse)
df %>%
    mutate(col = as.character(col),
            col2 = map(col, ~ if(str_detect(.x, "^sp")) .x 
               else str_count(.x, LETTERS) %>% 
             sort(decreasing = TRUE)))