How to avoid function name conflict without using the package identifier every time

General Usage
1

Hello everyone,

I was wondering if there was a way to avoid conflicts with a function name between packages without using the package identifier every time. To explain what I mean, when I write the following:

using DataFramesMeta
using ShiftedArrays

rev_df = DataFrame(firm_id = ["A", "A", "A", "A", "B", "B", "B", "B"] , 
	rev = [100, 200, 300, 400, 50, 67, 75, 90],
	yr = [2001,2002,2003,2004,2001,2002,2003,2004])

using IterTools

new_rev_df = @chain rev_df begin 
	groupby(:firm_id)
	@transform :rev_lag = lag(:rev)
end

I get the output:

julia> new_rev_df = @chain rev_df begin 
           groupby(:firm_id)
           @transform :rev_lag = lag(:rev)
       end
WARNING: both IterTools and DataFramesMeta export "groupby"; uses of it in module Main must be qualified
ERROR: UndefVarError: groupby not defined
Stacktrace:
 [1] top-level scope
   @ REPL[5]:2

Now this can be resolved by the following:

new_rev_lag = @chain rev_df begin 
	DataFramesMeta.groupby(:firm_id)
	@transform :rev_lag = lag(:rev)
end

This works because I’ve specified before groupby that I’m calling the function from DataFramesMeta.

My issue is, in my code I have to use IterTools just once, whereas I have to use DataFramesMeta’s groupby() function many, many times. So having to write DataFramesMeta.groupby() every time I want to do a grouping actually ends up being quite tedious.

Is there a way around this? The simplest I could think of was to load IterTools when I need it, then “unload” the package so that there’s no conflict when the DataFramesMeta's groupby() function needs to be used. As far as I can tell, Julia doesn’t let the user unload a package like this, although I might be missing something.

I’d appreciate some sort of workaround that anyone is aware of. Thanks!

2

You could also import only the one method from IterTools that you need, with something like

using DataFramesMeta
using IterTools: chain

See Modules · The Julia Language for why that works

10 Likes
3

Putting a line like this should help:

const groupby = DataFramesMeta.groupby

Here is a completely reproducible example which does not require 3rd-party packages:

julia> module Foo
       export groupby
       groupby() = "Foo.groupby"
       end
Main.Foo

julia> module Bar
       export groupby
       groupby() = "Bar.groupby"
       end
Main.Bar

julia> using .Foo

julia> using .Bar

julia> groupby()
WARNING: both Bar and Foo export "groupby"; uses of it in module Main must be qualified
ERROR: UndefVarError: groupby not defined
Stacktrace:
 [1] top-level scope
   @ REPL[5]:1

julia> const groupby = Bar.groupby
groupby (generic function with 1 method)

julia> groupby()
"Bar.groupby"
7 Likes
4

This works, thanks.

EDITED: To be accurate, I need to do this:

using IterTools: product
product(rev_df.rev, rev_df.yr)

If instead I did this:

using IterTools: product 
IterTools.product(rev_df.rev, rev_df.yr)

then I get the following error:

julia> IterTools.product(rev_df.rev, rev_df.yr)
ERROR: UndefVarError: IterTools not defined
Stacktrace:
 [1] top-level scope
   @ REPL[6]:1

In both of the above cases the conflict with DataFramesMeta.groupby goes away, but in the former case I can use IterTools.product() as well.

5

This works as well, thanks!

6

There is the import X as Y syntax, personally I think this is a great but underused feature. You could replace using IterTools by import IterTools as IT and then use IT.product etc.

3 Likes
7

Which version of Julia introduced this? I believe 1.5 and below do not support it.

8

The using X: y has imo some unintuitive behavior. using DataFrames: groupby brings groupby into Main but not DataFrames. This is a useful stopgap is module names conflict with other names.

julia> using DataFrames: groupby;

julia> groupby
groupby (generic function with 1 method)

julia> DataFrames
ERROR: UndefVarError: DataFrames not defined
2 Likes
9

I think so. For older julia versions I recommend the pattern import IterTools; const IT=IterTools.

1 Like
10

Ah okay makes sense. That’s why IterTools.product() was throwing an error about how IterTools wasn’t defined, but using product() worked fine.

1 Like
12

This should also work

using DataFrameMeta
groupby
using IterTools

Now groupby is the one from DataFramesMeta.

1 Like