我正在尝试在Python中使用垃圾邮件分类应用程序,但出现以下错误。不过我不明白,因为我正在使用.keys方法从字典中删除项目,所以这应该不是问题吗? 我尝试删除字典功能中的所有功能栏以尝试查找原因,但是我似乎无法将其包裹住
Python代码
import os
import numpy as np
from collections import Counter
from sklearn.naive_bayes import MultinomialNB
from sklearn.svm import LinearSVC
from sklearn.metrics import confusion_matrix
def make_Dictionary(train_dir):
emails = [os.path.join(train_dir,f) for f in os.listdir(train_dir)]
all_words = []
for mail in emails:
with open(mail) as m:
for i,line in enumerate(m):
if i == 2:
words = line.split()
all_words += words
dictionary = Counter(all_words)
list_to_remove = dictionary.keys()
for item in list_to_remove:
if item.isalpha() == False:
del dictionary[item]
elif len(item) == 1:
del dictionary[item]
dictionary = dictionary.most_common(3000)
return dictionary
def extract_features(mail_dir):
files = [os.path.join(mail_dir,fi) for fi in os.listdir(mail_dir)]
features_matrix = np.zeros((len(files),3000))
docID = 0;
for fil in files:
with open(fil) as fi:
for i,line in enumerate(fi):
if i == 2:
words = line.split()
for word in words:
wordID = 0
for i,d in enumerate(dictionary):
if d[0] == word:
wordID = i
features_matrix[docID,wordID] = words.count(word)
docID = docID + 1
return features_matrix
# Create a dictionary of words with its frequency
train_dir = r'.\train-mails'
dictionary = make_Dictionary(train_dir)
# Prepare feature vectors per training mail and its labels
train_labels = np.zeros(702)
train_labels[351:701] = 1
train_matrix = extract_features(train_dir)
# Training SVM and Naive bayes classifier and its variants
model1 = LinearSVC()
model1.fit(train_matrix,train_labels)
# Test the unseen mails for Spam
test_dir = r'.\test-mails'
test_matrix = extract_features(test_dir)
test_labels = np.zeros(260)
test_labels[130:260] = 1
result1 = model1.predict(test_matrix)
print (confusion_matrix(test_labels,result1))
print (confusion_matrix(test_labels,result2))
错误
RuntimeError: dictionary changed size during iteration
答案 0 :(得分:0)
这在Python 3.x中不起作用,因为keys
返回一个迭代器而不是列表。
另一种方法是使用列表来强制复制密钥。这个也可以在Python 3.x中使用:
for i in list(list_to_remove):
答案 1 :(得分:0)
dictionary.keys()
实际上是返回对原始字典键的引用。
您可以通过以下操作进行检查:
a_dict = {'a': 1}
keys = a_dict.keys() # keys is dict_keys(['a'])
a_dict['b'] = 2 # keys is dict_keys(['a', 'b'])
这就是为什么出现错误的原因:del dictionary[item]
实际上影响了list_to_remove
,在循环中是禁止的。
您可以通过在循环遍历原始密钥之前创建副本来避免这种情况。实现此目的的最简单方法是使用list
构造函数。所以改变你的线
list_to_remove = dictionary.keys()
具有:
list_to_remove = list(dictionary.keys())
解决了问题。
评论后的版本
请注意,此行为仅在python 3
中发生,在python 2
中,.keys()
方法返回一个普通列表,而没有引用字典:
a_dict = {'a': 1}
keys = a_dict.keys() # keys is ['a']
a_dict['b'] = 2 # keys is still ['a']
关于Python 3.0 changelog:
一些著名的API不再返回列表:
- dict方法dict.keys(),dict.items()和dict.values()返回“视图”而不是列表。