回答編集履歴
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後は純粋にMNISTのほうが意地悪なサンプルが割合多く含まれている可能性もありますが、上記の可能性を排除できないにはこのような結論を下すのは時期尚早でしょうね。
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---
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追記:
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気になったので試してみました。
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MNISTのほうがずっと難しいですね。
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digitsは8x8に対して、MNISTは28x28ですので、自由度がずっと高いですね。
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例えば、MNISTから2000だけ取り出して8x8にリサイズしてやると、正答率は
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digits:~98%、MNIST:~92%になります。
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```python
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from sklearn.model_selection import StratifiedKFold
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from sklearn.neighbors import KNeighborsClassifier
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from sklearn.metrics import accuracy_score
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from sklearn import datasets
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from keras.datasets import mnist
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from scipy.misc import imresize
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import numpy as np
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try:
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from tqdm import tqdm
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except (ImportError) as e:
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tqdm = lambda x:x
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def main(key='digits', random_state=2017):
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if key == 'digits':
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dataset = datasets.load_digits()
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X = dataset.data
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Y = dataset.target
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elif key == 'mnist':
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(X_train, y_train), (X_test, y_test) = mnist.load_data()
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kfold = StratifiedKFold(5, shuffle=True, random_state=0)
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tr, ts = next(kfold.split(X_test, y_test))
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X = X_test[ts]
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X = np.array([imresize(x, (8, 8)) for x in X])
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X = X.reshape(-1, np.prod(X.shape[1:]))
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Y = y_test[ts]
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Y = Y.reshape(-1)
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else:
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return [], []
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ks = np.linspace(1, 10, 5).astype('i')
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accuracy_scores = []
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for k in tqdm(ks):
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pY = np.zeros(Y.shape)
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kfold = StratifiedKFold(5, shuffle=True, random_state=random_state)
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for tr, ts in kfold.split(X, Y):
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x_tr = X[tr]
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y_tr = Y[tr]
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model = KNeighborsClassifier(n_neighbors=k, metric='manhattan')
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model.fit(x_tr, y_tr)
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py = model.predict(X[ts])
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pY[ts] = py
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score = accuracy_score(Y, pY)
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accuracy_scores.append(score)
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return ks, accuracy_scores
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if __name__ == '__main__':
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colors = ['red', 'blue']
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for ic, key in enumerate(['digits', 'mnist']):
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for i in np.linspace(1, 1000, 10).astype('i'):
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ks, accuracy_scores = main(key=key, random_state=2017+i)
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plt.plot(ks, accuracy_scores, marker='.', color=colors[ic])
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plt.xlabel('k')
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plt.ylabel('Accuracy')
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plt.grid()
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plt.xlim((0, np.max(ks)))
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plt.ylim((0.8, 1.0))
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plt.show()
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```
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