r - 将字符串拆分为每个字符一列的数据框

Question

在以下数据集中:

df <- data.frame(barcode=c("B1","B2", "B3"), 
                 sequence= sapply(1:3, function(x) paste(sample(c("A","C","T","G"), 8, replace=T), collapse=""))

我要拆分 df$sequence分成 8 个附加列，其中包含以正确顺序排列的每个字符串。
我知道如何分割字符向量，但这最终会出现在一个列表中:

library(stringr)
list1 <- str_extract_all(df$sequence,boundary("character"))
[[1]]
[1] "A" "T" "C" "G" "T" "G" "A" "A"

[[2]]
[1] "T" "C" "C" "T" "A" "T" "A" "T"

[[3]]
[1] "C" "G" "T" "T" "A" "A" "G" "G"

str(list1)
List of 3
 $ : chr [1:8] "A" "T" "C" "G" ...
 $ : chr [1:8] "T" "C" "C" "T" ...
 $ : chr [1:8] "C" "G" "T" "T" ...

如何将此列表转换为数据框或有更简单的方法？
编辑:
我可以去:

df$pos1 <- sapply(list1, function(x) x[1])
df$pos2 <- sapply(list1, function(x) x[2])

但我想有更好的解决方案。

Answer 1

使用 R 基础:

> data.frame(do.call(rbind, strsplit(df$sequence, "")))
  X1 X2 X3 X4 X5 X6 X7 X8
1  T  A  A  T  C  A  A  A
2  T  T  A  A  A  T  G  G
3  C  G  A  A  T  C  C  T