水浒传词频统计python

以下是使用Python进行《水浒传》词频统计的示例代码：

import jieba
from collections import Counter

# 读取文本文件
with open('input水浒传.txt', 'r', encoding='utf-8') as f:
    text = f.read()

# 分词
words = jieba.cut(text)

# 去除停用词
stopwords = ['的', '了', '和', '是', '在', '他', '她', '它']  # 停用词列表，可以根据实际情况进行修改
filtered_words = [word for word in words if word not in stopwords]

# 统计词频
word_count = Counter(filtered_words)

# 输出词频结果
for word, count in word_count.most_common(10):  # 只输出频率最高的10个词语
    print(word, count)

这段代码首先使用jieba库对文本进行分词，然后去除停用词，最后使用Counter类统计词频。最后，我们输出频率最高的10个词语及其出现次数。

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水浒传词频统计python123

import jieba
from collections import Counter
import matplotlib.pyplot as plt
from wordcloud import WordCloud

# 读取《水浒传》文本
with open('shuihuzhuan.txt', 'r', encoding='utf-8') as file:
    text = file.read()

# 使用jieba分词
words = jieba.lcut(text)

# 读取停用词库
with open('stopwords.txt', 'r', encoding='utf-8') as file:
    stopwords = file.read().splitlines()

# 进行词频统计
word_count = Counter([word for word in words if word not in stopwords])

# 生成词云
wordcloud = WordCloud(font_path="simhei.ttf").generate_from_frequencies(word_count)

# 可视化词云
plt.imshow(wordcloud, interpolation='bilinear')
plt.axis("off")
plt.show()

python水浒传词频统计

使用Python进行《水浒传》的词频统计是一种常见的文本分析方法。可以通过分词工具jieba对文本进行分词处理，然后统计每个词语出现的频率。

在给定的代码中，使用了jieba库对《水浒传》文本进行了分词处理，然后使用字典counts记录词语出现的频率。在统计过程中，排除了一些无效词语，比如“两个”、“一个”、“只见”等，同时对于相同对象的不同称谓进行了同一化处理。

最后，按照词语的频率进行排序，并输出前十个高频词汇。

因此，使用Python进行《水浒传》词频统计的代码如下所示：

import jieba

excludes = {"两个","一个","只见","如何","那里","哥哥","说道","军马","头领","众人","这里","兄弟","梁山泊","出来","小人","今日","这个","先锋","三个","因此","人马","问道","起来","便是","妇人","好汉","不是","不知","不曾","只是","如此","次日","我们","不得","如今","看时","不敢","来到","且说","一面","只得","山寨","原来","将军","却是"}

txt = open("水浒传.txt","r",encoding='gb18030', errors='ignore').read()
words = jieba.lcut(txt)

counts = {}
for word in words:
    if len(word) == 1:
        continue
    elif word == "宋江道" or word == "宋江":
        rword = "宋江"
    else:
        rword = word
    counts[rword = counts.get(rword, 0) + 1

for word in excludes:
    del(counts[word])

items = list(counts.items())
items.sort(key=lambda x:x<span class="em">1</span><span class="em">2</span>