My first post was removed (not sure why), so here is try number 2…
I’m new to Julia and am stuck on a problem.
I have a dataframe defined as follows:
df = DataFrame((name=["S1","S2","S3","S4","S5"],Z1=[41,12,19,17,11],Z2=[13,28,29,99,41],Z3=[7,86,23,12,71],Z4=[4,13,23,11,19],Z5=[41,12,13,19,22],Z6=[11,18,22,46,5]))
For each row, I want to return the column index of the lowest n values (better yet, return the column name). So for instance, if I wanted the 3 lowest values across Row “S1”, it would be columns 5 (Z4), 4 (Z3), and 7 (Z6).
I can get the column indexes with the following code:
But this returns a vector, which isn’t really ideal. (To clarify, since the input is a df, the expected output is also a df… The indexes themselves aren’t really of value, the value is the column name represented by the index, presented in a package of a dataframe that’s easily read by a human that’s used to looking at spreadsheets.)
I am wondering if there is a better solution here? Is there an available function that allows me to return the indexes or column names for the n lowest values and return the result as a dataframe? If not, what might be the suggested solution? I can imagine perhaps running the above code through a eachrow() loop, building a matrix of the results, converting the matrix to a df, then joining that back to a match table that has the column index/names to get a final result… All of that seems a little overly complex for what I imagine is a simple solution.
Appreciate your help.