Python 樸素貝葉斯文字分類練習答案：12 道題與解釋 | PyPractice 繁體

Question 1

樸素貝葉斯的「樸素」來自哪個獨立假設？

1. 給定類別後，各特徵被假設為獨立
2. 每個類別先驗相同
3. 所有詞長度相同
4. 標籤總是平衡的

Accepted Answer

給定類別後，各特徵被假設為獨立

Question 2

為什麼要給詞頻加入拉普拉斯平滑？

1. 給詞表排序
2. 讓訓練文件變長
3. 刪除先驗
4. 避免沒見過的詞把機率變成零

Accepted Answer

避免沒見過的詞把機率變成零

Question 3

為什麼要在對數空間給樸素貝葉斯計分？

1. 這樣就不需要平滑
2. 它把計數變成整數
3. 很多小機率連乘可能下溢
4. 對數會讓類別自動平衡

Accepted Answer

很多小機率連乘可能下溢

Question 4

執行結果是什麼？

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. 2
2. 0
3. 1
4. 4

Accepted Answer

1

Question 5

補全計數遞增。

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

Accepted Answer

1

Question 6

補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

補全對數空間的分數計算。

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

Accepted Answer

log

Question 8

補全先驗的分母。

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

這段程式列印出什麼？

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

程式碼的輸出結果是什麼？

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

螢幕上會顯示什麼？

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

這段程式碼會輸出什麼？

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