Converting each list within a dataframe to a normal column

I produce a data frame from several sources from the web which are cleaned beforehand and then selected with ``` cleans <- ls() cleans <- cleans[grepl("Clean_News", cleans)] ``` My first attempt to bind them together was inspired by a solution on Stack Overflow: ``` All_News <- mapply(get, grep("Clean_News", ls(), value=T)) All_News <- data.frame(t(All_News)) All_News <- as.data.frame(All_News) ``` However, this is a problem for me, since the result is a dataframe, where each column is a list of ints or characters. So, my main question is how to convert each list within the dataframe to a normal column within the df. I tried many hand-made functions on Stack Overflow, but none worked for me (due to my inexperience, I guess...). The df has the form ``` All_News <- data.frame(a=I(list(1,1:2,1:3)), b=I(list(4:6,7:9,10:11))) ``` Alternatively, I tried the following, which works: ``` All_News <- do.call(rbind, lapply(cleans, get)) ``` But has the huge disadvantage that I did not succeed in getting the names of the data frames as rownames / or first column into the data frame... So, my second question would be how to attach the names of the single data frames to each row of the huge df, instead of an id like the line of code below. ``` t2 <- rbindlist(lapply(cleans, get), idcol = "id") ``` This does not much good since I need the names of all data frames x -times repeatedly as an identifier, e.g. AND since this is an automated process with thousands of webpages, I do not know beforehand the number of rows in each data frame. The data looks like: ``` news1 data1 data2 news1 data5 data6 news2 data3 data4 and so on. ``` I tried something along these lines ``` nr <- length(cleans) names <- rep(cleans, nr) names <- sort(names) ``` But without much success.