这是一种方法，

data.frame(list = na.omit(
  unname(unlist(read.csv(
    text = 'ABCDE,"January 10, 2010",F,,,,GH,"March 9, 2009",,,', 
    check.names = F, header = F)))))
              list
1            ABCDE
2 January 10, 2010
3            FALSE
4               GH
5    March 9, 2009

这里您可能应该使用CSV解析器，但是如果您想使用纯正则表达式方法，您可以尝试：

library(stringr)
library(dplyr)

x <- "ABCDE,\"January 10, 2010\",F,,,,GH,\"March 9, 2009\",,,"
y <- str_match_all(x, "\"(.*?)\"|[^,]+")[[1]]
output <- coalesce(y[,2], y[,1])
output

[1] "ABCDE"            "January 10, 2010" "F"                "GH"
[5] "March 9, 2009"

正则表达式模式使用了一个交替的技巧，并表示要匹配：

• "(.*?)"匹配引号中的日期，但不捕获引号
• |或
• [^,]+匹配单个CSV术语

如果模式如图所示，则regex选项将创建分隔符并使用read.table

read.table(text = gsub('"', '', gsub('("[^,"]+,)(*SKIP)(*FAIL)|,',
   '\n', trimws(gsub(",{2,}", ",", str1), whitespace = ","), perl = TRUE)), 
    header = FALSE, fill = TRUE, sep = "\n")

• 输出

V1
1            ABCDE
2 January 10, 2010
3                F
4               GH
5    March 9, 2009

或者用scan

data.frame(V1 = setdiff(scan(text = str1, sep = ",",
    what = character()), ""))

• 输出

V1
1            ABCDE
2 January 10, 2010
3                F
4               GH
5    March 9, 2009

数据

str1 <- "ABCDE,\"January 10, 2010\",F,,,,GH,\"March 9, 2009\",,,"

另一个选项可以是：
第一个