Considering you want to merge them into two dimension as shown C = {Size, Engine, Colour} on D = {Max-Speed, Acceleration} then re-code the dataset by merging options as:

1) medium_diesel_silver

2) compact_gasoline_white

3) full_diesel_black

4) medium_gasoline_black

5) medium_diesel_silver

6) full_propane_black

7) full_gasoline_white

8) compact_gasoline_white

with D as options

1) m_poor

2) h_excellent

3) h_good

4) m_excellent

5) l_good

6) h_good

7) h_excellent

8) l_good

and then calculate the conditional entropy for only those record which satisfy the conditions

Considering you want to merge them into two dimension as shown C = {Size, Engine, Colour} on D = {Max-Speed, Acceleration} then re-code the dataset by merging options as:

1) medium_diesel_silver

2) compact_gasoline_white

3) full_diesel_black

4) medium_gasoline_black

5) medium_diesel_silver

6) full_propane_black

7) full_gasoline_white

8) compact_gasoline_white

with D as options

1) m_poor

2) h_excellent

3) h_good

4) m_excellent

5) l_good

6) h_good

7) h_excellent

8) l_good

and then calculate the conditional entropy for only those record which satisfy the conditions

Thanks a lot Sir PKs Prakash i will try it

Dear Sir Pks Prakash,

Should i compute the entropy for each row or only for one

for example i worked on the first row and i get the answer 0.796 and when working with the second row i get 0.984

so which on is the true one?

Thanks for your help.

The entropy could be calculated for whose system or for define X. For example, from the above case H(Y | X=medium_diesel_silver) has 2 records thus the marginal distribution would be {0.5, 0, 0, 0, 0.5} for Y responses {m_poor, h_excellent, h_good, m_excellent, l_good}. Thus, H(Y | X=medium_diesel_silver) = -0.5*log2(0.5)-0*log2(0)-0*log2(0)-0*log2(0)-0.5*log2(0.5) = 1 bits

Thanks a lot

really i understand what you are saying but in attached paper page 91 he computed the H(D*|C*) = 0.215

could you help me please understand how did he computed H(D*|C*)

Again Thanks a lot for your help.

Hi Ahmed,

The paper evaluates the overall normalized conditional entropy for the data. See below how its is obtained for the dataset:

-------------------------------------------------

Marginal Distribution Table to get P(Xi):

-------------------------------------------------

m_poor h_excellent h_good m_excellent l_good Prob(Xi)

medium_diesel_silver 0.13 0.00 0.00 0.00 0.13 0.25

compact_gasoline_white 0.00 0.13 0.00 0.00 0.13 0.25

full_diesel_black 0.00 0.00 0.13 0.00 0.00 0.125

medium_gasoline_black 0.00 0.00 0.00 0.13 0.00 0.125

full_propane_black 0.00 0.00 0.13 0.00 0.00 0.125

full_gasoline_white 0.00 0.13 0.00 0.00 0.00 0.125

-------------------------------------------------

Entropy for H(Y|Xi)

-------------------------------------------------

m_poor h_excellent h_good m_excellent l_good H(Y|Xi)

medium_diesel_silver 0.5 0 0 0 0.5 1

compact_gasoline_white 0 0.5 0 0 0.5 1

full_diesel_black 0 0 1 0 0 0

medium_gasoline_black 0 0 0 1 0 0

full_propane_black 0 0 1 0 0 0

full_gasoline_white 0 1 0 0 0 0

-----------------------------------------------

Normalized Entropy (H(Y*|X*)

-----------------------------------------------

X Normalized Entropy (H(Y*|Xi*)

medium_diesel_silver 0.25

compact_gasoline_white 0.25

full_diesel_black 0

medium_gasoline_black 0

full_propane_black 0

full_gasoline_white 0

The total Normalized entropy for H(D* I C * ) = sum(H(Y*|Xi*))/log2(5)=0.215338279 (Reported in Palwak Paper)

Very Special Thanks to you Sir Pks Prakash

You may refer our paper "A Rough Computing based Performance Evaluation Approach for Educational Institutions, International Journal of Software Engineering and Its Applications, Vol. 7 (4), pp. 331-348 (2013)"

Thanks Sir Debi.