我正在研究包含以下部分的教程：
>>> 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
然后我做了一些修改就复制了：
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
但是，尽管代码几乎没有差异，但本书中的结果比我的要好得多：
图书：
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，看看出了什么问题，我认为第18行：
X_train_raw, X_test_raw, y_train, y_test = train_test_split(df['label'], df['message'])
是罪魁祸首，并将
(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，以防有人尝试。
最佳答案
stacktrace末尾的错误是了解此处发生情况的关键。
ValueError：pos_label = 1不是有效的标签：array（['ham'，'spam']，
dtype ='| S4'）
您正在尝试对模型进行精确度和召回评分。回想一下，这些计分方法是根据真阳性，假阳性和假阴性制定的。但是sklearn如何知道什么是积极的，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“ ham”是负面标签。根据sklearn文档，默认情况下，精度和召回评分者期望的是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.直接从数据源中将“ ham”和“ spam”值分别编码为0和1，以适应精度/召回评分器：
# 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的整数，以适应精度/调用记分器：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的自定义参数创建评分器对象：如文档所述，默认情况下，精度和查全率得分的
pos_label参数为1，但是可以更改此值以通知计分员哪个字符串代表肯定标签。您可以使用make_scorer构造具有不同参数的得分器对象。# 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之后，我得到的平均准确度和召回得分分别在
0.990和0.704左右，与本书的编号一致。在这三个选项中，我最推荐＃3，因为它更容易出错。
关于python - Scikit-learn脚本提供的结果与本教程截然不同，并且在更改数据框时出现错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/39187875/