Python 朴素贝叶斯文本分类练习答案：12 道题与解释 | PyPractice 中文

Question 1

朴素贝叶斯的“朴素”来自哪个独立假设？

1. 标签总是平衡的
2. 给定类别后，各特征被假设为独立
3. 所有词长度相同
4. 每个类别先验相同

Accepted Answer

给定类别后，各特征被假设为独立

Question 2

为什么要给词频加入拉普拉斯平滑？

1. 让训练文档变长
2. 给词表排序
3. 删除先验
4. 避免没见过的词把概率变成零

Accepted Answer

避免没见过的词把概率变成零

Question 3

为什么要在对数空间给朴素贝叶斯计分？

1. 很多小概率连乘可能下溢
2. 这样就不需要平滑
3. 它把计数变成整数
4. 对数会让类别自动平衡

Accepted Answer

很多小概率连乘可能下溢

Question 4

运行结果是什么？

docs = [("good movie", "pos"), ("great film", "pos"), ("bad movie", "neg"), ("boring film", "neg")]
counts = {}
for text, label in docs:
    counts.setdefault(label, {})
    for word in text.split():
        counts[label][word] = counts[label].get(word, 0) + 1
print(counts["pos"].get("movie", 0))

1. 0
2. 4
3. 1
4. 2

Accepted Answer

1

Question 5

补全计数递增。

counts = {}
word = "movie"
counts[word] = counts.get(word, 0) + ____
print(counts[word])

Accepted Answer

1

Question 6

补全拉普拉斯平滑的分子。

counts = {"movie": 1}
total = 2
vocab_size = 3
probability = (counts.get("film", 0) + ____) / (total + vocab_size)
print(round(probability, 2))

Accepted Answer

1

Question 7

补全对数空间的分数计算。

import math
probability = 0.2
print(round(math.____(probability), 3))

Accepted Answer

log

Question 8

补全先验的分母。

import math
class_docs = {"pos": 2, "neg": 2}
docs = ["a", "b", "c", "d"]
print(math.log(class_docs["pos"] / ____(docs)))

Accepted Answer

len

Question 9

这段程序打印出什么？

import math

docs = [("good movie", "pos"), ("great film", "pos"), ("bad movie", "neg"), ("boring film", "neg")]
vocab = sorted({word for text, _ in docs for word in text.split()})
counts = {}
priors = {}
for text, label in docs:
    priors[label] = priors.get(label, 0) + 1
    counts.setdefault(label, {})
    for word in text.split():
        counts[label][word] = counts[label].get(word, 0) + 1

def predict(text):
    scores = {}
    for label in counts:
        total = sum(counts[label].values())
        score = math.log(priors[label] / len(docs))
        for word in text.split():
            score += math.log((counts[label].get(word, 0) + 1) / (total + len(vocab)))
        scores[label] = score
    return max(scores, key=scores.get)

print(predict("great film"))

Accepted Answer

pos

Question 10

代码的输出结果是什么？

docs = [("good movie", "pos"), ("great film", "pos"), ("bad movie", "neg"), ("boring film", "neg")]
counts = {}
for text, label in docs:
    counts.setdefault(label, {})
    for word in text.split():
        counts[label][word] = counts[label].get(word, 0) + 1
print(counts["neg"]["bad"])

Accepted Answer

1

Question 11

屏幕上会显示什么？

docs = ["good movie", "great film", "bad movie", "boring film"]
labels = ["pos", "pos", "neg", "neg"]
print(labels.count("pos") / len(docs))

Accepted Answer

0.5

Question 12

这段代码会输出什么？

counts = {"great": 0}
total = 4
vocab_size = 6
probability = (counts.get("great", 0) + 1) / (total + vocab_size)
print(round(probability, 1))

Accepted Answer

0.1