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Operator for finding hidden / redundant patters

vyronadvyronad Member Posts: 12 Contributor II
Hi All,

I am completely new to Rapid Miner. Which operator can be used to find hidden / redundant patterns?

Thanks and regards,
Veeranna Ronad.

Answers

  • landland RapidMiner Certified Analyst, RapidMiner Certified Expert, Member Posts: 2,531   Unicorn
    Hi,
    this is a very hard question :) In fact you might have to or will be able to utilize each single operator. This heavily depends on your data and what exactly you are going to find...
    You will have to be a little bit more specific than that.

    Greetings,
      Sebastian
  • vyronadvyronad Member Posts: 12 Contributor II
    Thank you very much Sebastian.

    Basically I have a transaction table. Customer 'A' has made some transactions. I want to list out all those customers whose transaction pattern matches customer 'A'.

    Table Transactions
    -----------------------------------------
    Trans_Date | Customer | Amount
    -----------------------------------------
    01-01-2010 | A            |          50
    01-01-2010 | B            |        100
    10-01-2010 | A            |          40
    11-01-2010 | A            |        100
    11-01-2010 | C            |        100
    21-02-2010 | A            |          10
    ...
    ...

    Thanks and Regards,
    Veeranna Ronad.
  • landland RapidMiner Certified Analyst, RapidMiner Certified Expert, Member Posts: 2,531   Unicorn
    Hi,
    I think this should be possible using the FP Growth operator in combination with some filtering operations. You finally might apply the association rules for each other customer's transaction, count the number of fulfilled rules and aggregate them to get a measure for equality.
    The Applier for association rules will be part of the next RapidMiner update.

    Greetings,
      Sebastian
  • vyronadvyronad Member Posts: 12 Contributor II
    Hi Sebastian,

    I used FP-Growth operator, I am getting error "Regular attributes must be of type binominal", inspite of using operator "Nominal to Binominal".

    The xml is:
    -----------------------------------------------
    <?xml version="1.0" encoding="UTF-8" standalone="no"?>
    <process version="5.0">
      <context>
        <input>
          <location/>
        </input>
        <output>
          <location/>
          <location/>
        </output>
        <macros/>
      </context>
      <operator activated="true" class="process" expanded="true" name="Root">
        <description>Using a logistic regression learner for a classification task of numerical data.</description>
        <process expanded="true" height="264" width="656">
          <operator activated="true" class="read_arff" expanded="true" height="60" name="Read ARFF" width="90" x="45" y="30">
            <parameter key="data_file" value="C:\Documents and Settings\36533\Desktop\RM_Data\Outlier.arff"/>
          </operator>
          <operator activated="true" class="nominal_to_binominal" expanded="true" height="94" name="Nominal to Binominal" width="90" x="186" y="29"/>
          <operator activated="true" class="fp_growth" expanded="true" height="76" name="FP-Growth" width="90" x="380" y="30"/>
          <connect from_op="Read ARFF" from_port="output" to_op="Nominal to Binominal" to_port="example set input"/>
          <connect from_op="Nominal to Binominal" from_port="example set output" to_op="FP-Growth" to_port="example set"/>
          <connect from_op="FP-Growth" from_port="example set" to_port="result 1"/>
          <portSpacing port="source_input 1" spacing="0"/>
          <portSpacing port="sink_result 1" spacing="0"/>
          <portSpacing port="sink_result 2" spacing="0"/>
        </process>
      </operator>
    </process>

    -----------------------------------------------
    The data file is:
    -----------------------------------------------
    @RELATION TranData

    @ATTRIBUTE 'AMOUNT' NUMERIC
    @ATTRIBUTE 'Customer' STRING {'A',''B','C','D','E'}
    @ATTRIBUTE 'DATE' DATE "MM-yyyy"

    @DATA
    '4659.545','A','01-2009'
    '4670.889','A','01-2009'
    '7956.187','B','02-2009'
    '1849.152','A','02-2009'
    '6285.257','B','02-2009'
    '28538','C','02-2009'
    '4579.993','C','02-2009'
    '8559.586','A','02-2009'
    '20128','B','02-2009'
    '4416.704','D','02-2009'
    '7479.134','C','02-2009'
    '9500.984','D','07-2009'
    '1223.983','E','07-2009'
    '7098.173','D','07-2009'
    '5047.892','A','07-2009'
    '7350.093','C','08-2009'
    '1896.075','B','08-2009'
    '8184.474','E','09-2009'
    '5401.709','E','09-2009'
    '3193.046','B','09-2009'
    '3228.8','C','09-2009'
    '1754.48','A','09-2009'
    '8326.189','D','09-2009'
    '7713.256','D','09-2009'
    -----------------------------------------------

    Kindly advice me.

    Thanks,
    Best,
    Veeranna Ronad.
  • landland RapidMiner Certified Analyst, RapidMiner Certified Expert, Member Posts: 2,531   Unicorn
    Hi,
    as far as I can see, the data table contains numerical attributes. They won't be transformed by the nominal to binominal operator, since they aren't nominal...You will have to filter them out by using select attributes operator or discretize them first.

    Greetings,
      Sebastian
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