RidgeCV gives a different result from running Ridge with manually implemented CV

sklearn_ridge_cv_issue.py

'''
Python 3.6.1 |Anaconda 4.4.0 (x86_64)| (default, May 11 2017, 13:04:09)

In [1]: import sklearn
In [2]: print(sklearn.__version__)
0.18.1
'''

import pandas as pd
from sklearn.model_selection import KFold
from sklearn.linear_model import Ridge, RidgeCV
import numpy as np
from operator import itemgetter

# Load data (Hitters dataset)
hitters_df = pd.read_csv('https://raw.githubusercontent.com/selva86/datasets/master/Hitters.csv')
hitters_clean_df = hitters_df.dropna()
hitters_clean_df = pd.get_dummies(hitters_clean_df, drop_first=True)

# Prepare the predictors and response for regression
X = hitters_clean_df.drop('Salary', axis = 1)
y = hitters_clean_df.Salary

# Method 1: Using RidgeCV
alphas = 10**np.linspace(-4, 2, 100)
regr = RidgeCV(alphas=alphas, cv=10, normalize=True, scoring='neg_mean_squared_error')
regr.fit(X.values, y.values)
print('Best alpha from RidgeCV:', regr.alpha_)  # ('Best alpha from RidgeCV:', 18.738174228603832)

# Method 2: Prepare data with KFold and apply Ridge for each fold
alphas = 10**np.linspace(-4, 2, 100)
K = 10  # Number of folds
kf = KFold(n_splits=K, shuffle=True, random_state=123) 
cv_scores = []  # to store CV scores

for alpha in alphas:
    for train_idx, test_idx in kf.split(X.values, y.values):  # kf is an iterator, which you can loop over
        X_train = X.values[train_idx]
        y_train = y.values[train_idx]
        X_test = X.values[test_idx]
        y_test = y.values[test_idx]
        
        model = Ridge(alpha=alpha, normalize=True, fit_intercept=True).fit(X_train, y_train)
        test_errors = model.predict(X_test) - y_test
        
    cv_scores.append( np.mean(test_errors**2) )  # mean squared error as the CV score

min_idx, min_cv_score = min( enumerate(cv_scores), key=itemgetter(1) )
best_alpha = alphas[min_idx]
print('Best alpha from KFold+Ridge: ', best_alpha)  # ('Best alpha from KFold+Ridge: ', 0.011497569953977356)