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What is an efficient way from inverting a list of character vectors as shown below?
lov <- list(v1=c("a", "b"), v2=c("a", "c"), v3=c("a"))
list(a=c("v1", "v2", "v3"), b=c("v1"), c=c("v2"))
Similar to Revert list structure, but involving vectors:
903
R – Reverse a List. To reverse a list in R programming, call rev() function and pass given list as argument to it. rev() function returns returns a new list with the contents of given list in reversed order.
The rev() method in R is used to return the reversed order of the R object, be it dataframe or a vector. It computes the reverse columns by default. The resultant dataframe returns the last column first followed by the previous columns. The ordering of the rows remains unmodified.
We can either convert the list to a data.frame (using stack or melt from library(reshape2)) and then split the 'ind' column by the 'values' in 'd1'.
d1 <- stack(lov)
split(as.character(d1$ind), d1$values)
Or if the above method is slow, we can replicate (rep) the names of 'lov' by the length of each list element (lengths gives a vector output of the length of each element) and split it by unlisting the 'lov'.
split(rep(names(lov), lengths(lov)), unlist(lov))
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