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Generate tree structure from csv

I have scratched my head over this problem for a while now. I am basically trying to generate a tree hierarchy from a set of CSV data. The CSV data is not necessarily ordered. This is like something as follows:

Header: Record1,Record2,Value1,Value2
Row: A,XX,22,33
Row: A,XX,777,888
Row: A,YY,33,11
Row: B,XX,12,0
Row: A,YY,13,23
Row: B,YY,44,98

I am trying to make the way the grouping is performed as flexible as possible. The simplest for of grouping would to do it for Record1 and Record2 with the Value1 and Value2 stored under Record2 so that we get the following output:

Record1
    Record2
        Value1 Value2

Which would be:

A
    XX
        22,33
        777,888
    YY
        33,11
        13,23
B
    XX
        12,0
    YY
        44,98 

I am storing my group settings in a List at present - which I don't know if this is hindering my thoughts. This list contains a hierarchy of the groups for example:

Record1 (SchemaGroup)
    .column = Record1
    .columns = null
    .childGroups =
        Record2 (SchemaGroup)
            .column = Record1
            .columns = Value1 (CSVColumnInformation), Value2 (CSVColumnInformation)
            .childGroups = null

The code for this looks like as follows:

private class SchemaGroup {
    private SchemaGroupType type = SchemaGroupType.StaticText;  // default to text
    private String text;
    private CSVColumnInformation column = null;
    private List<SchemaGroup> childGroups = new ArrayList<SchemaGroup>();
    private List<CSVColumnInformation> columns = new ArrayList<CSVColumnInformation>();
}


private enum SchemaGroupType {
    /** Allow fixed text groups to be added */
    StaticText,
    /** Related to a column with common value */
    ColumnGroup
}

I am stuggling producing an algorithm for this, trying to think of the underlying structure to use. At present I am parsing the CSV top to bottom, using my own wrapper class:

CSVParser csv = new CSVParser(content);
String[] line;
while((line = csv.readLine()) != null ) {
    ...
}

I am just trying to kick start my coding brain.

Any thoughts?

like image 441
Andez Avatar asked Oct 05 '11 13:10

Andez


2 Answers

The basic idea isn't difficult: group by the first record, then by the second record, etc. until you get something like this:

(A,XX,22,33)
(A,XX,777,888)
-------------------------
(A,YY,33,11)
(A,YY,13,23)
=============
(B,XX,12,0)
-------------------------
(B,YY,44,98)

and then work backwards to build the trees.

However, there is a recursive component that makes it somewhat hard to reason about this problem, or show it step by step, so it's actually easier to write pseudocode.

I'll assume that every row in your csv is represented like a tuple. Each tuple has "records" and "values", using the same terms you use in your question. "Records" are the things that must be put into a hierarchic structure. "Values" will be the leaves of the tree. I'll use quotations when I use these terms with these specific meanings.

I also assume that all "records" come before all "values".

Without further ado, the code:

// builds tree and returns a list of root nodes
// list_of_tuples: a list of tuples read from your csv
// curr_position: used to keep track of recursive calls
// number_of_records: assuming each csv row has n records and then m values, number_of_records equals n
function build_tree(list_of_tuples, curr_position, number_of_records) {
    // check if we have already reached the "values" (which shouldn't get converted into trees)
    if (curr_position == number_of_records) {
        return list of nodes, each containing a "value" (i.e. everything from position number_of_records on)
    }

    grouped = group tuples in list_of_tuples that have the same value in position curr_position, and store these groups indexed by such common value
    unique_values = get unique values in curr_position

    list_of_nodes = empty list

   // create the nodes and (recursively) their children
    for each val in unique_values {
        the_node = create tree node containing val
        the_children = build_tree(grouped[val], curr_position+1, number_of_records)
        the_node.set_children(the_children)

        list_of_nodes.append(the_node)
    }

    return list_of_nodes
}

// in your example, this returns a node with "A" and a node with "B"
// third parameter is 2 because you have 2 "records"
build_tree(list_parsed_from_csv, 0, 2)

Now you'd have to think about the specific data structures to use, but hopefully this shouldn't be too difficult if you understand the algorithm (as you mention, I think deciding on a data structure early on may have been hindering your thoughts).

like image 98
Jong Bor Lee Avatar answered Nov 14 '22 05:11

Jong Bor Lee


Here is the basic working solution in the form of junit (no assertions though) simplified by using google-guava collections. The code is self-explanatory and instead of file io you use csv libraries for reading the csv. This should give you the basic idea.

import java.io.File;
import java.io.IOException;
import java.util.Collection;
import java.util.Collections;
import java.util.List;
import java.util.Set;

import org.junit.Test;

import com.google.common.base.Charsets;
import com.google.common.base.Splitter;
import com.google.common.collect.ArrayListMultimap;
import com.google.common.collect.Iterables;
import com.google.common.collect.Multimap;
import com.google.common.collect.Sets;
import com.google.common.io.Files;

public class MyTest
{
    @Test
    public void test1()
    {
        List<String> rows = getAllDataRows();

        Multimap<Records, Values> table = indexData(rows);

        printTree(table);

    }

    private void printTree(Multimap<Records, Values> table)
    {
        Set<String> alreadyPrintedRecord1s = Sets.newHashSet();

        for (Records r : table.keySet())
        {
            if (!alreadyPrintedRecord1s.contains(r.r1))
            {
                System.err.println(r.r1);
                alreadyPrintedRecord1s.add(r.r1);
            }

            System.err.println("\t" + r.r2);

            Collection<Values> allValues = table.get(r);

            for (Values v : allValues)
            {
                System.err.println("\t\t" + v.v1 + " , " + v.v2);
            }
        }
    }

    private Multimap<Records, Values> indexData(List<String> lines)
    {
        Multimap<Records, Values> table = ArrayListMultimap.create();

        for (String row : lines)
        {
            Iterable<String> split = Splitter.on(",").split(row);
            String[] data = Iterables.toArray(split, String.class);

            table.put(new Records(data[0], data[1]), new Values(data[2], data[3]));
        }
        return table;
    }

    private List<String> getAllDataRows()
    {
        List<String> lines = Collections.emptyList();

        try
        {
            lines = Files.readLines(new File("C:/test.csv"), Charsets.US_ASCII);
        }
        catch (IOException e)
        {
            e.printStackTrace();
        }

        lines.remove(0);// remove header

        return lines;
    }
}



public class Records
{
    public final String r1, r2;

    public Records(final String r1, final String r2)
    {
        this.r1 = r1;
        this.r2 = r2;
    }

    @Override
    public int hashCode()
    {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((r1 == null) ? 0 : r1.hashCode());
        result = prime * result + ((r2 == null) ? 0 : r2.hashCode());
        return result;
    }

    @Override
    public boolean equals(final Object obj)
    {
        if (this == obj)
        {
            return true;
        }
        if (obj == null)
        {
            return false;
        }
        if (!(obj instanceof Records))
        {
            return false;
        }
        Records other = (Records) obj;
        if (r1 == null)
        {
            if (other.r1 != null)
            {
                return false;
            }
        }
        else if (!r1.equals(other.r1))
        {
            return false;
        }
        if (r2 == null)
        {
            if (other.r2 != null)
            {
                return false;
            }
        }
        else if (!r2.equals(other.r2))
        {
            return false;
        }
        return true;
    }

    @Override
    public String toString()
    {
        StringBuilder builder = new StringBuilder();
        builder.append("Records1and2 [r1=").append(r1).append(", r2=").append(r2).append("]");
        return builder.toString();
    }

}


public class Values
{
    public final String v1, v2;

    public Values(final String v1, final String v2)
    {
        this.v1 = v1;
        this.v2 = v2;
    }

    @Override
    public int hashCode()
    {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((v1 == null) ? 0 : v1.hashCode());
        result = prime * result + ((v2 == null) ? 0 : v2.hashCode());
        return result;
    }

    @Override
    public boolean equals(final Object obj)
    {
        if (this == obj)
        {
            return true;
        }
        if (obj == null)
        {
            return false;
        }
        if (!(obj instanceof Values))
        {
            return false;
        }
        Values other = (Values) obj;
        if (v1 == null)
        {
            if (other.v1 != null)
            {
                return false;
            }
        }
        else if (!v1.equals(other.v1))
        {
            return false;
        }
        if (v2 == null)
        {
            if (other.v2 != null)
            {
                return false;
            }
        }
        else if (!v2.equals(other.v2))
        {
            return false;
        }
        return true;
    }

    @Override
    public String toString()
    {
        StringBuilder builder = new StringBuilder();
        builder.append("Values1and2 [v1=").append(v1).append(", v2=").append(v2).append("]");
        return builder.toString();
    }

}
like image 42
Aravind Yarram Avatar answered Nov 14 '22 07:11

Aravind Yarram