The second dataset ,named dataset2 contains Market Basket retails for sale data from one of china's markets unknowns. It contains 11040.0 transactions database, it has over 12 items and most of the transactions contain between 7 and 11 items. The third dataset ,named dataset 3 contains the retail list of medications from a pharmacy data from an anonymous china retail store. It contains 13454.0 transactions over 20 items and most of the transactions contain between 9 and 15 items.
The fourth dataset ,named dataset 4 contains the retail List of meals of different food in a restaurant. It contains 56429.0 transactions over 40 items and most of the transactions contain between 22 and 30 items.
In figure 5.8 for performance comparison of our …show more content…
This experiment showed a good results in the memory cost , because when we put Min_Sup= 0.5 % , the old algorithm generated 422 KB of memory size from frequent closed items, while our algorithm generated only 316 KB of memory size from frequent closed items and when we put Min_Sup= 3.0 % , the old algorithm generated 400 KB of memory size from frequent closed items, while our algorithm generated only 300 KB of memory size from frequent closed items and so on , as shown in figure 5.9.
The results in figure 5.10 showed memory cost with different minimum support, the support of the experiment begin from [0.5% , 1.0% ,1.5 % and 2.5%] . This experiment showed a good results in the memory cost with large database , because when we put Min_Sup= 0.5 % , the old algorithm generated 4010 KB of memory size from frequent closed items, while our algorithm generated only 2750 KB of memory size from frequent closed items and when we put Min_Sup= 1.0 % , the old algorithm generated 2220 KB of memory size from frequent closed items, while our algorithm generated only 1550 KB of memory size from frequent closed items and so on. Figure 5.9 . Comparison The Memory cost with different minimum support in dataset 2
The fourth dataset ,named dataset 4 contains the retail List of meals of different food in a restaurant. It contains 56429.0 transactions over 40 items and most of the transactions contain between 22 and 30 items.
In figure 5.8 for performance comparison of our …show more content…
This experiment showed a good results in the memory cost , because when we put Min_Sup= 0.5 % , the old algorithm generated 422 KB of memory size from frequent closed items, while our algorithm generated only 316 KB of memory size from frequent closed items and when we put Min_Sup= 3.0 % , the old algorithm generated 400 KB of memory size from frequent closed items, while our algorithm generated only 300 KB of memory size from frequent closed items and so on , as shown in figure 5.9.
The results in figure 5.10 showed memory cost with different minimum support, the support of the experiment begin from [0.5% , 1.0% ,1.5 % and 2.5%] . This experiment showed a good results in the memory cost with large database , because when we put Min_Sup= 0.5 % , the old algorithm generated 4010 KB of memory size from frequent closed items, while our algorithm generated only 2750 KB of memory size from frequent closed items and when we put Min_Sup= 1.0 % , the old algorithm generated 2220 KB of memory size from frequent closed items, while our algorithm generated only 1550 KB of memory size from frequent closed items and so on. Figure 5.9 . Comparison The Memory cost with different minimum support in dataset 2