LDA- 파이썬의 인식 패턴 (Sklearn)

Question

파이썬 에서이 코드를 실행하려고합니다. 이 코드는 Sklearn의 LDA를 나타냅니다.

import numpy as np
from sklearn.lda import LDA

X = np.array ([0.000000, 0.000000, 0.000000, 0.000000, 0.001550, 
               0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 
               0.000000, 0.000000, 0.201550, 0.011111, 0.077778,
               0.011111, 0.000000, 0.000000, 0.000000, 0.000000,
               0.000000, 0.092732, 0.000000, 0.000000, 0.000000,
               0.000000, 0.035659, 0.000000, 0.000000, 0.000000,
               0.000000, 0.066667, 0.000000, 0.000000, 0.010853,
               0.000000, 0.033333, 0.055556, 0.055556, 0.077778, 
               0.000000, 0.000000, 0.000000, 0.268170, 0.000000, 
               0.000000, 0.000000, 0.000000, 0.130233, 0.000000, 
               0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
               0.000000, 0.034109, 0.077778, 0.055556, 0.011111, 
               0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
               0.155388, 0.000000, 0.000000, 0.000000, 0.000000,
               0.181395, 0.000000, 0.000000, 0.000000, 0.000000,
               0.001550, 0.007752, 0.000000, 0.000000, 0.000000, 
               0.000000, 0.000000, 0.011111, 0.088889, 0.033333,
               0.000000, 0.000000, 0.142857, 0.000000, 0.000000,
               0.000000, 0.000000, 0.093023, 0.000000, 0.000000,
               0.000000, 0.000000, 0.000000, 0.009302, 0.010853, 
               0.000000, 0.100000, 0.000000, 0.000000, 0.000000,
               0.000000, 0.022222, 0.088889, 0.033333, 0.238095,
               0.000000, 0.000000, 0.000000, 0.000000, 0.032558,
               0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
               0.182946, 0.000000, 0.000000, 0.000000, 0.000000,
               0.000000, 0.000000, 0.022222, 0.077778, 0.055556,
               0.000000, 0.102757])

y = np.array ([0.000000, 0.000000, 0.008821, 0.000000, 0.000000, 
               0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
               0.000000, 0.000000, 0.179631, 0.010471, 0.036649,
               0.026178, 0.000000, 0.000000, 0.020942, 0.010471,
               0.000000, 0.109215, 0.000000, 0.000000, 0.060144, 
               0.000000, 0.042502, 0.000000, 0.005613, 0.000000,
               0.000000, 0.018444, 0.000000, 0.000000, 0.013633,
               0.020942, 0.031414, 0.083770, 0.015707, 0.041885,
               0.041885, 0.057592, 0.010471, 0.233788, 0.000000,
               0.000000, 0.018444, 0.000000, 0.000000, 0.000000,
               0.000000, 0.000000, 0.090617, 0.000000, 0.000000,
               0.000000, 0.104250, 0.005236, 0.020942, 0.031414,
               0.000000, 0.000000, 0.010471, 0.015707, 0.005236,
               0.056314, 0.000000, 0.000000, 0.026464, 0.000000,
               0.004010, 0.000000, 0.031275, 0.007217, 0.036889,
               0.007217, 0.013633, 0.000000, 0.000000, 0.005236,
               0.047120, 0.057592, 0.015707, 0.010471, 0.047120,
               0.062827, 0.005236, 0.262799, 0.000000, 0.000000,
               0.000000, 0.000000, 0.000802, 0.000000, 0.000000,
               0.000000, 0.001604, 0.000000, 0.052927, 0.000000,
               0.039294, 0.026178, 0.041885, 0.031414, 0.000000,
               0.000000, 0.041885, 0.073298, 0.000000, 0.308874,
               0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
               0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
               0.236568, 0.000000, 0.000000, 0.000000, 0.000000,
               0.000000, 0.000000, 0.000000, 0.020942, 0.015707,
               0.000000, 0.029010])

clf = lda () clf.fit (x, y) print (clf.predict ([0, 2]))

이 오류 메시지를 보여주세요.

clf.fit(X, y)
n_samples, n_features = X.shape
ValueError: need more than 1 value to unpack

내가 그것을 고치기 위해 무엇을 하는가? 문서 에서이 솔루션을 찾을 수 없었습니다.

Answer 1

배열은 1 차원입니다. 할 때 :

n_samples, n_features = X.shape

X. 셰이프는 샘플과 feautures의 매트릭스가 아니라 모양의 배열 (106,)입니다. 둘 이상의 샘플이 필요합니다. 그대로, 당신은 많은 기능과 하나의 샘플이 있습니다. 4 개의 기능을 갖는 4 개의 샘플의 매트릭스는 다음과 같이 정의됩니다.

featureMat = np.array([[ 10, 30, 40, 50],
                       [ 5,  6,  7,  8],
                       [ 54, 75, 6,  56],
                       [ 65, 34, 23, 22]])

Featuremat.hapape는 (4,4)입니다.