X = np.array([3.4, 3.4, 3. , 2.8, 2.7, 2.9, 3.3, 3. , 3.8, 2.5])
y = np.array([0, 0, 0, 1, 1, 1, 2, 2, 2, 2])

     X  y
0  3.4  0
1  3.4  0
2  3.0  0
3  2.8  1
4  2.7  1
5  2.9  1
6  3.3  2
7  3.0  2
8  3.8  2
9  2.5  2

First we binarize the target

y = LabelBinarizer().fit_transform(y)

     X  y1  y2  y3
0  3.4   1   0   0
1  3.4   1   0   0
2  3.0   1   0   0
3  2.8   0   1   0
4  2.7   0   1   0
5  2.9   0   1   0
6  3.3   0   0   1
7  3.0   0   0   1
8  3.8   0   0   1
9  2.5   0   0   1
 

Then perform a dot product between feature and target, i.e. sum all feature values by class value

observed = y.T.dot(X) 
>>> observed 
array([ 9.8,  8.4, 12.6])

Next take a sum of feature values and calculate class frequency

feature_count = X.sum(axis=0).reshape(1, -1)
class_prob = y.mean(axis=0).reshape(1, -1)

>>> class_prob, feature_count
(array([[0.3, 0.3, 0.4]]), array([[30.8]]))
 

Now as in the first step we take the dot product, and get expected and observed matrices

expected = np.dot(class_prob.T, feature_count)
>>> expected 
array([[ 9.24],[ 9.24],[12.32]])

Finally we calculate a chi^2 value:

chi2 = ((observed.reshape(-1,1) - expected) ** 2 / expected).sum(axis=0)
>>> chi2 
array([0.11666667])
 

We have a chi^2 value, now we need to judge how extreme it is. For that we use a chi^2 distribution with number of classes - 1 degrees of freedom and calculate the area from chi^2 to infinity to get the probability of chi^2 be the same or more extreme than what we've got. This is a p-value. (using chi square survival function from scipy)

p = scipy.special.chdtrc(3 - 1, chi2)
>>> p
array([0.94333545])

Compare with SelectKBest:

s = SelectKBest(chi2, k=1)
s.fit(X.reshape(-1,1),y)
>>> s.scores_, s.pvalues_
(array([0.11666667]), [0.943335449873492])

總結起來就是chi_2=SUM((Obs-Exp)**2/Exp),其中各個類別下的該變數取值分別求和作為Obs，概率*(變數取值之和)作為Exp。計算方法類比了離散型變數。

來源：https://stackoverflow.com/questions/57273694/how-selectkbest-chi2-calculates-score