LDA - Python中的识别模式(sklearn)

时间:2013-11-07 23:16:42

标签: python scikit-learn lda

我正在尝试在Python上执行此代码。此代码指的是来自sklearn的LDA。

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]))

并告诉我此错误消息:

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

我该怎么做才能修好它?我在文档上找不到这个解决方案。

1 个答案:

答案 0 :(得分:1)

你的阵列是一维的。当你这样做时:

n_samples, n_features = X.shape

X.shape不是样本和特征的矩阵,而是一个形状的数组(106,)。您需要多个样本。原样,你有一堆功能和一个样本。具有4个特征的4个样本的矩阵将被定义为:

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

因此featureMat.shape将是(4,4)。