One common reason for editing histograms is to add frequency count cells without greatly increasing the number of steps. The changes you will need to make to histograms vary, depending on whether the values represent a dense or sparse frequency count.
To add a frequency cell for a given column value, check the column value just less than the value for the new cell. If the next-lesser value is as close as possible to the value to be added, then the frequency count determined simply.
If the next lesser column value to the step to be changed is as close as possible to the frequency count value, then the frequency count cell can be extracted simply.
For example, if a column contains at least one 19 and many 20s, and the histogram uses a single cell to represent all the values greater than 17 and less than or equal to 22, optdiag output shows the following information for the cell:
Step Weight Value ... 4 0.100000000 <= 17 5 0.400000000 <= 22 ...
Altering this histogram to place the value 20 on its own step requires adding two steps, as shown here:
... 4 0.100000000 <= 17 5 0.050000000 <= 19 6 0.300000000 <= 20 7 0.050000000 <= 22 ...
In the altered histogram above, step 5 represents all values greater than 17 and less than or equal to 19. The sum of the weights of steps 5, 6, and 7 in the modified histogram equals the original weight value for step 5.
If the column has no values greater than 17 and less than 20, the representation for a sparse frequency count must be used instead. Here are the original histogram steps:
Step Weight Value ... 4 0.100000000 <= 17 5 0.400000000 <= 22 ...
The following example shows the zero-weight step, step 5, required for a sparse frequency count:
... 4 0.100000000 <= 17 5 0.000000000 < 20 6 0.350000000 = 20 7 0.050000000 <= 22 ...
The operator for step 5 must be <. Step 6 must specify the weight for the value 20, and its operator must be =.
