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Challenge #300: Where is the party?

ChrisWaspe
8 - Asteroid
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ChrisWaspe_0-1641208038814.png

 

aantoni
5 - Atom

I want to post my solution. 

Luke_C
15 - Aurora
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Luke_C_0-1641216743463.png

 

MatthewBr
Alteryx
Alteryx
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Probably could have done it a different way - but happy to complete my 1st challenge of 2022.
alexnajm
8 - Asteroid

Happy Holidays everyone!

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Challenge 300.PNG
cgoodman3
13 - Pulsar
13 - Pulsar

Slightly different answers to the provided solution as having a filter only containing New Year instead of = New Year's Day gives results such as Russia like @AngelosPachis identified, but also finds countries such as Bolivia and China who have other celebrations during the year to celebrate New Year.

 

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Chris
Check out my collaboration with fellow ACE Joshua Burkhow at AlterTricks.com
patrick_digan
16 - Nebula
16 - Nebula
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patrick_digan_0-1641223677272.png

 

Ladarthure
13 - Pulsar
13 - Pulsar

my solution!

PhilipMannering
14 - Magnetar
14 - Magnetar

 

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PhilipMannering_0-1641230084342.png

 

 

 

TH
7 - Meteor

The question is somewhat ambiguous. Having two days off for New Years is not specific. Personally, I would think that having New Years' Eve off also counts as a day off for New Years, but that's not what the challenge expected. Unfortunately, the challenge text did not address this, so I had to adjust my workflow to get to the expected answer.

 

It is important to notice that, as has been pointed out, the data in certain fields is not consistent. The "english name" is, mostly, "New Year's Day". However, for Russia it is "New Year holiday". There are a lot of other distinct "english name"s too. This project actually should have some rather in-depth cultural research. Does the new year celebration always have a title with "New Year" in it? Is it always on 01 January of every calendar?

 

Also inconvenient is Alteryx's lack of a set minus tool (or a single tool that could be configured to perform a set minus function). I'd find it very useful to have a function that would do something like the following -

take as input a base set list and a subtract set list

base = [A,B,C,D,E,F,G]

subtract = [B,D,G]

and then would do a set-subtract - that is, return all the elements in the first list that are not in the second list

base - subtract = [A,C,E,F]

 

All that being said, here's my solution to challenge 300.

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The option to insert a photo is broken, but you can look at my workflow.