LINQ query toarray() - how to access array?

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i have a LINQ query that gives me an average linked to an index. I've turned the query into an array, but i can't figure out how do anything with the array (i need to use the query data). My LINQ is only so so

C#

var vavg = (from z in data
                       where z.Y > 0 && z.Y < 0.1
                       group z by z.X into XGroup
                       select new
                       {
                           X = XGroup.Key,
                           vavg = XGroup.Average(z => z.A),
                       }).ToArray();

i get the correct results but the query isn't returned in a true array, i need to do operations on them so i need a double[][] where [0] is X and [1] is vavg

What I have tried:

vavg.Average(); returns error
double[][] test = new double[2][];
test[0] = vavg[0]; returns error

the errors seem to be related to the structure of the Array which isn't a "real" array and looks more like a data structure to me

Posted 5-Nov-16 17:05pm

Member 12815488

Updated 5-Nov-16 17:51pm

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[no name] 5-Nov-16 23:45pm

ToArray does not return a multi-dimensional array.

1 solution

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Midi_Mick · Accepted Answer · 2016-11-05T17:51:00

Solution 1

you need to change your select so that it creates your jagged array elements.

C#

var vavg = (from z in data
                       where z.Y > 0 && z.Y < 0.1
                       group z by z.X into XGroup
                       select new double []
                       {
                           XGroup.Key,
                           XGroup.Average(z => z.A),
                       }).ToArray();

See the declaration of the new double[] on the select. Also got rid of the temporary assignments in there - they're not needed.

Posted 5-Nov-16 17:51pm

Midi_Mick

Updated 5-Nov-16 17:54pm

v2

Comments

Member 12815488 6-Nov-16 0:49am

thanks! funny how simple some things end up. i literally couldn't find that answer on the web in an hour of googling. appreciate it.

Midi_Mick 6-Nov-16 1:13am

No worries - it was a fun one, and had me scratching up a bald patch for a little while. But once I hit upon that what you needed was an array of arrays, the answer became obvious.

Wendelius 6-Nov-16 2:42am

Nice example, a 5.