Scikit-learn 脚本给出的结果与教程大不相同，当我更改数据框时会出错答案

【问题标题】：Scikit-learn script giving vastly different results than the tutorial, and gives an error when I change the dataframesScikit-learn 脚本给出的结果与教程大不相同，当我更改数据框时会出错
【发布时间】：2017-01-04 09:36:10
【问题描述】：

我正在学习包含此部分的教程：

>>> import numpy as np
>>> import pandas as pd
>>> from sklearn.feature_extraction.text import TfidfVectorizer
>>> from sklearn.linear_model.logistic import LogisticRegression
>>> from sklearn.cross_validation import train_test_split, cross_val_score
>>> df = pd.read_csv('data/sms.csv')
>>> X_train_raw, X_test_raw, y_train, y_test = train_test_split(df['message'], df['label'])
>>> vectorizer = TfidfVectorizer()
>>> X_train = vectorizer.fit_transform(X_train_raw)
>>> X_test = vectorizer.transform(X_test_raw)
>>> classifier = LogisticRegression()
>>> classifier.fit(X_train, y_train)
>>> precisions = cross_val_score(classifier, X_train, y_train, cv=5, scoring='precision')
>>> print 'Precision', np.mean(precisions), precisions
>>> recalls = cross_val_score(classifier, X_train, y_train, cv=5, scoring='recall')
>>> print 'Recalls', np.mean(recalls), recalls

然后我复制了一些修改：

ddir = (sys.argv[1])
df = pd.read_csv(ddir + '/SMSSpamCollection', sep='\t', quoting=csv.QUOTE_NONE, names=["label", "message"])
X_train_raw, X_test_raw, y_train, y_test = train_test_split(df['label'], df['message'])


vectorizer = TfidfVectorizer()
X_train = vectorizer.fit_transform(X_train_raw)
X_test = vectorizer.transform(X_test_raw)
classifier = LogisticRegression()
classifier.fit(X_train, y_train)


precisions = cross_val_score(classifier, X_train, y_train, cv=5, scoring='precision')
recalls = cross_val_score(classifier, X_train, y_train, cv=5, scoring='recall')


print 'Precision', np.mean(precisions), precisions
print 'Recalls', np.mean(recalls), recalls

然而，尽管代码几乎没有差异，但书中的结果比我的要好得多：

书：Precision 0.992137651822 [ 0.98717949 0.98666667 1. 0.98684211 1. ] Recall 0.677114261885 [ 0.7 0.67272727 0.6 0.68807339 0.72477064]

我的：Precision 0.108435683974 [ 2.33542342e-06 1.22271611e-03 1.68918919e-02 1.97530864e-01 3.26530612e-01]Recalls 0.235220281632 [ 0.00152053 0.03370787 0.125 0.44444444 0.57142857]

回到脚本看看哪里出了问题，我以为第 18 行：

X_train_raw, X_test_raw, y_train, y_test = train_test_split(df['label'], df['message'])

是罪魁祸首，将(df['label'], df['message']) 更改为(df['message'], df['label'])。但这给了我一个错误：

Traceback (most recent call last):
  File "Chapter4[B-FLGTLG]C[Y-BCPM][G-PAR--[00].py", line 30, in <module>
    precisions = cross_val_score(classifier, X_train, y_train, cv=5, scoring='precision')
  File "/usr/local/lib/python2.7/dist-packages/sklearn/cross_validation.py", line 1433, in cross_val_score
    for train, test in cv)
  File "/usr/local/lib/python2.7/dist-packages/sklearn/externals/joblib/parallel.py", line 800, in __call__
    while self.dispatch_one_batch(iterator):
  File "/usr/local/lib/python2.7/dist-packages/sklearn/externals/joblib/parallel.py", line 658, in dispatch_one_batch
    self._dispatch(tasks)
  File "/usr/local/lib/python2.7/dist-packages/sklearn/externals/joblib/parallel.py", line 566, in _dispatch
    job = ImmediateComputeBatch(batch)
  File "/usr/local/lib/python2.7/dist-packages/sklearn/externals/joblib/parallel.py", line 180, in __init__
    self.results = batch()
  File "/usr/local/lib/python2.7/dist-packages/sklearn/externals/joblib/parallel.py", line 72, in __call__
    return [func(*args, **kwargs) for func, args, kwargs in self.items]
  File "/usr/local/lib/python2.7/dist-packages/sklearn/cross_validation.py", line 1550, in _fit_and_score
    test_score = _score(estimator, X_test, y_test, scorer)
  File "/usr/local/lib/python2.7/dist-packages/sklearn/cross_validation.py", line 1606, in _score
    score = scorer(estimator, X_test, y_test)
  File "/usr/local/lib/python2.7/dist-packages/sklearn/metrics/scorer.py", line 90, in __call__
    **self._kwargs)
  File "/usr/local/lib/python2.7/dist-packages/sklearn/metrics/classification.py", line 1203, in precision_score
    sample_weight=sample_weight)
  File "/usr/local/lib/python2.7/dist-packages/sklearn/metrics/classification.py", line 984, in precision_recall_fscore_support
    (pos_label, present_labels))
ValueError: pos_label=1 is not a valid label: array(['ham', 'spam'], 
      dtype='|S4')

这可能是什么问题？数据在这里：http://archive.ics.uci.edu/ml/datasets/SMS+Spam+Collection，以防有人想尝试。

【问题讨论】：

教程用来获取结果的数据是否相同？
是的。出处：We will use the SMS Spam Classification Data Set from the UCI Machine Learning Repository. The dataset can be downloaded from http://archive.ics.uci.edu/ml/datasets/SMS+Spam+Collection. First, let's explore the data set and calculate some basic summary statistics using pandas
为什么要包含sep="\t" 参数（以及read_csv 调用中的其他参数）？您是否检查过数据是否正确导入？如果教程使用相同的数据，但没有使用"\t"，那么数据可能是逗号分隔的，而不是制表符分隔的。
实际上，那条线有问题，所以我用这里的一条替换了书中的一条：radimrehurek.com/data_science_python。 SMSSpamCollection中的行是空格分隔的，但是内容和sms.csv一样，我没有。

标签： python csv pandas dataframe scikit-learn

【解决方案1】：

堆栈跟踪末尾的错误是了解这里发生了什么的关键。

ValueError: pos_label=1 is not a valid label: array(['ham', 'spam'], dtype='|S4')

您正在尝试使用精确度和召回率对模型进行评分。回想一下，这些评分方法是根据真阳性、假阳性和假阴性来制定的。但是sklearn 怎么知道什么是积极的，什么是消极的？是“火腿”还是“垃圾邮件”？我们需要一种方法来告诉sklearn，我们认为“垃圾邮件”是正面标签，而“垃圾邮件”是负面标签。根据sklearn 文档，精确度和召回率评分器默认期望1 的正标签，因此错误消息的pos_label=1 部分。

至少有 3 种方法可以解决此问题。

1.直接从数据源将“ham”和“spam”值编码为 0 和 1，以适应精确度/召回率评分器：

# Map dataframe to encode values and put values into a numpy array
encoded_labels = df['label'].map(lambda x: 1 if x == 'spam' else 0).values # ham will be 0 and spam will be 1

# Continue as normal
X_train_raw, X_test_raw, y_train, y_test = train_test_split(df['message'], encoded_labels)

2.使用 sklearn 的内置函数 (label_binarize) 将分类数据转换为编码整数，以适应精确度/召回率评分器：

这会将您的分类数据转换为整数。

# Encode labels
from sklearn.preprocessing import label_binarize
encoded_column_vector = label_binarize(df['label'], classes=['ham','spam']) # ham will be 0 and spam will be 1
encoded_labels = np.ravel(encoded_column_vector) # Reshape array

# Continue as normal
X_train_raw, X_test_raw, y_train, y_test = train_test_split(df['message'], encoded_labels)

3. 使用pos_label 的自定义参数创建记分器对象：

正如文档所述，默认情况下，精确度和召回率分数具有 pos_label 参数 1，但可以更改此参数以告知记分员哪个字符串代表正标签。您可以使用make_scorer 构造具有不同参数的记分器对象。

# Start out as you did originally with string labels
X_train_raw, X_test_raw, y_train, y_test = train_test_split(df['message'], df['label'])
# Fit classifier as normal ...


# Get precision and recall
from sklearn.metrics import precision_score, recall_score, make_scorer
# Precision
precision_scorer = make_scorer(precision_score, pos_label='spam')
precisions = cross_val_score(classifier, X_train, y_train, cv=5, scoring=precision_scorer)
print 'Precision', np.mean(precisions), precisions

# Recall
recall_scorer = make_scorer(recall_score, pos_label='spam')
recalls = cross_val_score(classifier, X_train, y_train, cv=5, scoring=recall_scorer)
print 'Recalls', np.mean(recalls), recalls

在对您的代码进行任何这些更改后，我的平均准确率和召回率得分约为 0.990 和 0.704，与书中的数字一致。

在所有 3 个选项中，我最推荐第 3 个选项，因为它更难出错。

【讨论】：

非常感谢，这真的困扰着我。不过，我可以问如何吗？数据没有任何标题：pastebin.com/rm3gyaGE
当您将 csv 作为 pandas 数据框导入并指定每列的名称时，数据会获取标题，如 names=["label", "message"]。第一列的标题为“标签”，第二列的标题为“消息”。
当您在 test_train_split 中切换标签和消息参数时，我不确定它为什么会起作用。这根本不应该工作。它基本上是要求模型预测，如果我有一封我知道是“垃圾邮件”的电子邮件，那么该邮件是“一些非常长的邮件”的可能性有多大。没有多大意义，对吧？那里正在发生一些非常糟糕的事情。更改后，您正确设置了参数。然而，目标标签的格式是记分员无法理解的。准确率和召回率评分者需要知道正面标签是什么。