Min-Max Scaling

Apply Scaling to a New Row

A new row uses the fitted min and max table; the endpoints are not recomputed from that row.

highlighted = computed this step

Keep the fit table fixed

The fit rows keep the same endpoints: small min 0, small max 4, large min 0, and large max 100.

fit endpoints: [0,4][0,100]\text{fit endpoints: }[0,4]\quad [0,100]
Apply Fitted Scaling to a New RowExact min-max fractions for one new displayed row.apply fitted min-max scale to a new rowfeaturefit minfit maxnew value(new-min)/(max-min)small0411/4large01001255/4new row uses the fitted min/max table5/4 is outside 0..1 because 125 is beyond the fit max 100fit rows define min and max; new row is apply onlyoutside 0..1 can happenNOT refit; NOT data leakageNOT training; NOT learning; NOT generalizationNOT accuracy; NOT calibration; NOT a guarantee of model qualityNOT probability truth

Apply those endpoints to N

New row N has small value 1 and large value 125. Using the fitted endpoints gives scaled values 1/4 and 5/4.

Nscaled=(1/4,5/4)N_{\text{scaled}}=(1/4,5/4)
Apply Fitted Scaling to a New RowExact min-max fractions for one new displayed row.apply fitted min-max scale to a new rowfeaturefit minfit maxnew value(new-min)/(max-min)small0411/4large01001255/4new row uses the fitted min/max table5/4 is outside 0..1 because 125 is beyond the fit max 100fit rows define min and max; new row is apply onlyoutside 0..1 can happenNOT refit; NOT data leakageNOT training; NOT learning; NOT generalizationNOT accuracy; NOT calibration; NOT a guarantee of model qualityNOT probability truth

Do not refit on the new row

The value 5/4 is outside the zero-to-one interval because 125 is beyond the fitted large max 100. That is an apply-only preprocessing step, not a refit and not data leakage.

5/4  >  15/4\;>\;1
Apply Fitted Scaling to a New RowExact min-max fractions for one new displayed row.apply fitted min-max scale to a new rowfeaturefit minfit maxnew value(new-min)/(max-min)small0411/4large01001255/4new row uses the fitted min/max table5/4 is outside 0..1 because 125 is beyond the fit max 100fit rows define min and max; new row is apply onlyoutside 0..1 can happenNOT refit; NOT data leakageNOT training; NOT learning; NOT generalizationNOT accuracy; NOT calibration; NOT a guarantee of model qualityNOT probability truth