值错误:sklearn.RFECV中不支持未知

2024-09-20 23:03:18 发布

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我试图使用rfecv缩小与分类器真正相关的特性的数量。这是我写的代码

import sklearn
import pandas as p
import numpy as np
import scipy as sp
import pylab as pl
from sklearn import linear_model, cross_validation, metrics
from sklearn.svm import SVC
from sklearn.feature_selection import RFECV
from sklearn.metrics import zero_one_loss
from sklearn import preprocessing
#from sklearn.feature_extraction.text import CountVectorizer
#from sklearn.feature_selection import SelectKBest, chi2

modelType = "notext"

# ----------------------------------------------------------
# Prepare the Data
# ----------------------------------------------------------
training_data = np.array(p.read_table('F:/NYC/NYU/SM/3/SNLP/Project/Data/train.tsv'))
print ("Read Data\n")

# get the target variable and set it as Y so we can predict it
Y = training_data[:,-1]

print(Y)

# not all data is numerical, so we'll have to convert those fields
# fix "is_news":
training_data[:,17] = [0 if x == "?" else 1 for x in training_data[:,17]]

# fix -1 entries in hasDomainLink
training_data[:,14] = [0 if x =="-1" else x for x in training_data[:,10]]

# fix "news_front_page":
training_data[:,20] = [999 if x == "?" else x for x in training_data[:,20]]
training_data[:,20] = [1 if x == "1" else x for x in training_data[:,20]]
training_data[:,20] = [0 if x == "0" else x for x in training_data[:,20]]

# fix "alchemy category":
training_data[:,3] = [0 if x=="arts_entertainment" else x for x in training_data[:,3]]
training_data[:,3] = [1 if x=="business" else x for x in training_data[:,3]]
training_data[:,3] = [2 if x=="computer_internet" else x for x in training_data[:,3]]
training_data[:,3] = [3 if x=="culture_politics" else x for x in training_data[:,3]]
training_data[:,3] = [4 if x=="gaming" else x for x in training_data[:,3]]
training_data[:,3] = [5 if x=="health" else x for x in training_data[:,3]]
training_data[:,3] = [6 if x=="law_crime" else x for x in training_data[:,3]]
training_data[:,3] = [7 if x=="recreation" else x for x in training_data[:,3]]
training_data[:,3] = [8 if x=="religion" else x for x in training_data[:,3]]
training_data[:,3] = [9 if x=="science_technology" else x for x in training_data[:,3]]
training_data[:,3] = [10 if x=="sports" else x for x in training_data[:,3]]
training_data[:,3] = [11 if x=="unknown" else x for x in training_data[:,3]]
training_data[:,3] = [12 if x=="weather" else x for x in training_data[:,3]]
training_data[:,3] = [999 if x=="?" else x for x in training_data[:,3]]

print ("Corrected outliers data\n")

# ----------------------------------------------------------
# Models
# ----------------------------------------------------------
if modelType == "notext":
    print ("no text model\n")
    #ignore features which are useless
    X = training_data[:,list([3, 5, 6, 7, 8, 9, 10, 14, 15, 16, 17, 19, 20, 22, 25])]
    scaler = preprocessing.StandardScaler()
    print("initialized scaler \n")
    scaler.fit(X,Y)
    print("fitted train data and labels\n")
    X = scaler.transform(X)
    print("Transformed train data\n")
    svc = SVC(kernel = "linear")
    print("Initialized SVM\n")
    rfecv = RFECV(estimator = svc, cv = 5, loss_func = zero_one_loss, verbose = 1)
    print("Initialized RFECV\n")
    rfecv.fit(X,Y)
    print("Fitted train data and label\n")
    rfecv.support_
    print ("Optimal Number of features : %d" % rfecv.n_features_)
    savetxt('rfecv.csv', rfecv.ranking_, delimiter=',', fmt='%f')

在调用“rfecv.fit(X,Y)”时,我的代码从metrices.py文件“ValueError:unknown is not supported”抛出一个错误

错误在sklearn.metrics.metrics中萌生:

# No metrics support "multiclass-multioutput" format
    if (y_type not in ["binary", "multiclass", "multilabel-indicator", "multilabel-sequences"]):
        raise ValueError("{0} is not supported".format(y_type))

这是一个分类问题,目标值只有0或1。 数据集可以在Kaggle Competition Data找到

如果有人能指出我错在哪里,我将不胜感激。


Tags: infromimportfordataifastraining
1条回答
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1楼 · 发布于 2024-09-20 23:03:18

RFECV检查目标/序列数据的类型为binarymulticlassmultilabel-indicatormultilabel-sequences

  • “binary”:y包含<;=2个离散值,是1d或列 矢量。
  • 'multiclass':y包含两个以上的离散值,不是 序列的序列,并且是一维或列向量。
  • 'mutliclass multioutput':y是一个包含更多 两个以上的离散值,不是一个序列序列,而且 尺寸为>;1。
  • “多标签指示器”:y是一个标签指示器矩阵,一个数组 具有至少两列且最多两个唯一的二维 价值观。

当你的Yunknown时,即

  • “unknown”:y与数组类似,但与上面的任何一个都不同,例如3d数组或非序列对象数组。

原因是您的目标数据是字符串(格式为"0""1"),并以read_table作为对象加载:

>>> training_data[:, -1].dtype
dtype('O')
>>> type_of_target(training_data[:, -1])
'unknown'

要解决此问题,可以转换为int

>>> Y = training_data[:, -1].astype(int)
>>> type_of_target(Y)
'binary'

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