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Corrected formula

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Corrected formula for weighted within-class variance
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vishwesh5 2018-06-14 12:30:06 +05:30
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doc/py_tutorials/py_imgproc/py_thresholding/py_thresholding.markdown
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@ -183,7 +183,7 @@ minimizes the **weighted within-class variance** given by the relation :
where
\f[q_1(t) = \sum_{i=1}^{t} P(i) \quad \& \quad q_1(t) = \sum_{i=t+1}^{I} P(i)\f]\f[\mu_1(t) = \sum_{i=1}^{t} \frac{iP(i)}{q_1(t)} \quad \& \quad \mu_2(t) = \sum_{i=t+1}^{I} \frac{iP(i)}{q_2(t)}\f]\f[\sigma_1^2(t) = \sum_{i=1}^{t} [i-\mu_1(t)]^2 \frac{P(i)}{q_1(t)} \quad \& \quad \sigma_2^2(t) = \sum_{i=t+1}^{I} [i-\mu_1(t)]^2 \frac{P(i)}{q_2(t)}\f]
\f[q_1(t) = \sum_{i=1}^{t} P(i) \quad \& \quad q_2(t) = \sum_{i=t+1}^{I} P(i)\f]\f[\mu_1(t) = \sum_{i=1}^{t} \frac{iP(i)}{q_1(t)} \quad \& \quad \mu_2(t) = \sum_{i=t+1}^{I} \frac{iP(i)}{q_2(t)}\f]\f[\sigma_1^2(t) = \sum_{i=1}^{t} [i-\mu_1(t)]^2 \frac{P(i)}{q_1(t)} \quad \& \quad \sigma_2^2(t) = \sum_{i=t+1}^{I} [i-\mu_2(t)]^2 \frac{P(i)}{q_2(t)}\f]
It actually finds a value of t which lies in between two peaks such that variances to both classes
are minimum. It can be simply implemented in Python as follows: