conditional list selection and appending of variable in R - list

Question

Say i have a list mn like this
i<-c(w=5,n="oes")
p<-c(w=9,n="ty",j="ooe")
mn<-list(i,p,i,p,i,p,i)
Now I´d like to select the list elements with the shortest length (the i´s) and append "unknown" to the list before creating a dataframe. How can I do this?
EDIT: In the end I´d like the list to have every i element in mn as w=5,n="oes", and j="unknown" before mn including p is changed into a dataframe:

Answer 1

To find the lenght of each element in your list, use length wrapped in sapply:
len <- sapply(mn, length)
len
[1] 2 3 2 3 2 3 2
Now, to identify only those elements that have lengths equal to the shortest length:
which(len==min(len))
[1] 1 3 5 7
Use subsetting and as.data.frame to create your data.frame. But this data.frame will have somewhat random column names, so I rename the column names:
df <- as.data.frame(mn[which(len==min(len))])
names(df) <- seq_len(ncol(df))
df
1 2 3 4
w 5 5 5 5
n oes oes oes oes
You will have to clarify what you mean with "append unknown" to this data.frame.

Answer 2

Another possibility is:
all.names = unique( unlist( lapply( mn, names ) ) )
do.call( 'rbind', lapply( mn, function( r ) {
data.frame( sapply( all.names, function( v ) r[ v ], simplify=F ) )
} ) )
which gives:
w n j
w 5 oes <NA>
w1 9 ty ooe
w2 5 oes <NA>
w3 9 ty ooe
w4 5 oes <NA>
w5 9 ty ooe
w6 5 oes <NA>
But I get the feeling there's a much neater route to this solution...
edit
If you want unknown rather than <NA>, you can change the inner sapply to:
sapply( all.names, function( v ) if( is.na( r[v] ) ) 'unknown' else r[v], simplify=F )

Answer 3

There is not very elegant, but it might do the trick:
maxlength <- max(sapply(mn,length))
## make a new list, with the "missing" entries replaced with "unknown"
mn2 <- lapply(mn,function(x)c(x,rep('unknown',maxlength - length(x))))
## convert to a data.frame
mn3 <- data.frame(matrix(unlist(mn2),nrow = 3))
Which gives the following
> mn3
X1 X2 X3 X4 X5 X6 X7
1 5 9 5 9 5 9 5
2 oes ty oes ty oes ty oes
3 unknown ooe unknown ooe unknown ooe unknown
However it is better practice to use NA, rather than "unknown"