Query.@orderby destroys DataFrame

General Usage
,
1

This code does not work:

using DataFrames, Query
df = DataFrame(A=[1,2,3], B=[4,5,6]);
fromdata = @from i in df begin
           @where i.B > 3
           @select {i.A}
           @orderby i.A
           @collect DataFrame
        end 
println(typeof(fromdata))
println(nrow(fromdata))
println(typeof(fromdata.A))

However, when you put a comment # in front of @orderBy, it works. Is this intended - why? (is an ordered DataFrame no longer a DataFrame?) … or is it a bug?

// Edit: Creating an explicit DataFrame again makes it work:

...
fromData = DataFrame(@from ... @orderby ... end)
...

// Edit: Rewritten question so that a naive copy/paste of the code should show the error.

2

where are these macros coming from?

3

Query (I added the using in my question).

4

Odd, it works for me. (DataFrames v0.19.1, Query v0.12.0.)

julia-1.3> using DataFrames, Query

julia-1.3> df = DataFrame(A=[1,2,3], B=[4,5,6]);

julia-1.3> fromdata = @from i in df begin
                      @where i.B > 3
                      @select {i.A}
                      @orderby i.A
                      @collect DataFrame
                   end
3x1 query result
A
─
1
2
3

First suggestion - a restart. I see many world age errors when I make mistakes in Query… :slight_smile:

I’m not sure, but is this alternative syntax equivalent?

df |> @filter(_.B > 3) |> @select(:A) |> @orderby(_.A) |> collect |> DataFrame
5

Thanks - I went down some other track (using sort!) … maybe it’s a problem with versions - I was using 1.1 (and am now on 1.2), you tried it in 1.3. I also did not check which package version I use(d) …

6

I just tried your code in 1.2, and it works.

7

Yeah, I also can’t replicate that error.

8

… when commenting in the @orderby? I get it “reliably” (in my Julia 1.1.1 installation - I have not yet upgraded … //edit: but let me upgrade - maybe I’m happy then!) - see the image below. More readably the output for the first two println()s is:

julia> println(typeof(fromdata))
QueryOperators.EnumerableMap{NamedTuple{(:A,),Tuple{Int64}},QueryOperators.EnumerableFilter{NamedTuple{(:A, :B),Tuple{Int64,Int64}},QueryOperators.EnumerableIterable{NamedTuple{(:A, :B),Tuple{Int64,Int64}},Tables.DataValueRowIterator{NamedTuple{(:A, :B),Tuple{Int64,Int64}},Tables.RowIterator{NamedTuple{(:A, :B),Tuple{Array{Int64,1},Array{Int64,1}}}}}},getfield(Main, Symbol("##2384#2386"))},getfield(Main, Symbol("##2385#2387"))}

julia> println(nrow(fromdata))
ERROR: MethodError: no method matching nrow(::QueryOperators.EnumerableMap{NamedTuple{(:A,),Tuple{Int64}},QueryOperators.EnumerableFilter{NamedTuple{(:A, :B),Tuple{Int64,Int64}},QueryOperators.EnumerableIterable{NamedTuple{(:A, :B),Tuple{Int64,Int64}},Tables.DataValueRowIterator{NamedTuple{(:A, :B),Tuple{Int64,Int64}},Tables.RowIterator{NamedTuple{(:A, :B),Tuple{Array{Int64,1},Array{Int64,1}}}}}},getfield(Main, Symbol("##2384#2386"))},getfield(Main, Symbol("##2385#2387"))})
Closest candidates are:
  nrow(::DataFrame) at C:\Users\h.mueller\.julia\packages\DataFrames\VrZOl\src\dataframe\dataframe.jl:278
  nrow(::SubDataFrame) at C:\Users\h.mueller\.julia\packages\DataFrames\VrZOl\src\subdataframe\subdataframe.jl:114
Stacktrace:
 [1] top-level scope at none:0

.

orderbyproblem
orderbyproblem1078×980 29.6 KB

9

No - also does not work with 1.2.0 (DataFrames is 0.19.2, Query is 0.12.1).

(Edit: I have now rewritten the original question so that the code shown shows the error - just to make sure that people who copy it without reading the earlier instruction to remove the # will “hopefully” get the error).