How many ways can it be reshaped?

I have found a few ways to solve the following problem, but I’m sure there are many others.


###sol. 1#####
cols=[c.Y for c in grp]
dict=[n=>v for (n,v) in zip(names,cols)]

###sol. 2#####
D=[c.X[1]=>c.Y for c in grp]

I would like to see solutions that make use of named-tuples, or the unstack function, or any other function suitable for the purpose.

Here’s a few :slight_smile: (some of these produce additional columns but it’s easy to get rid of them if necessary):

vcat((permutedims(df[i:i+3, :], :X) for i in 1:4:nrow(df)-1)...)
vcat((permutedims(combine(grp, g->g[i,:]), :X) for i in 1:3)...)

# Using named tuples
DataFrame((; (Symbol.(df[i:i+3, :X]) .=> df[i:i+3, :Y])...) for i in 1:4:9)
DataFrame((; (Symbol(g[i, :X]) => g[i, :Y] for g in grp)...) for i in 1:3)

DataFrame([df[j+4i,2] for i in 0:2, j in 1:4], df[1:4, 1])

df.Z = repeat(1:3, inner=4)
unstack(df, :X, :Y)
1 Like

Just what I was looking for. Thank you very much @sudete.
Could you explain me the rationale of the named tuple syntax?
That is, where does it (; (keys. => values) …) come from

You asked for named tuples so I looked for a way to use them :slight_smile:

The DataFrame() constructor accepts an iterator of named tuples (one per row). You can construct a named tuple with (a=1, b=2) or (; a=1, b=2) (names are literal) or with (; :a => 1, :b => 2) (names can be any expression that returns a symbol).

In your problem the names are not literal, they are computed, so I have to use that last version with pairs.