1、三国人物top10分析
import jieba
from wordcloud import WordCloud
import imageio
mask = imageio.imread('china.jpg')
# 1.读取小说内容
with open('threeking.txt', 'r', encoding='utf-8') as f:
words = f.read()
counts = {} # {‘曹操’:234，‘回寨’:56}
excludes = {"将军", "却说", "丞相", "二人", "不可", "荆州", "不能", "如此", "商议",
"如何", "主公", "军士", "军马", "左右", "次日", "引兵", "大喜", "天下",
"东吴", "于是", "今日", "不敢", "魏兵", "陛下", "都督", "人马", "不知",
"孔明曰","玄德曰","刘备","云长"}
# 2. 分词
words_list = jieba.lcut(words)
# print(words_list)
for word in words_list:
if len(word) <= 1:
continue
else:
# 更新字典中的值
# counts[word] = 取出字典中原来键对应的值 + 1
# counts[word] = counts[word] + 1 # counts[word]如果没有就要报错
# 字典。get（k） 如果字典中没有这个键 返回 NONE
counts[word] = counts.get(word, 0) + 1
print(len(counts))
# 3. 词语过滤,删除无关词，重复词
counts['孔明'] = counts['孔明'] + counts['孔明曰']
counts['玄德'] = counts['玄德'] + counts['玄德曰'] +counts['刘备']
counts['关公'] = counts['关公'] +counts['云长']
for word in excludes:
del counts[word]
# 4.排序 [(), ()]
items = list(counts.items())
print(items)
# def sort_by_count(x):
# return x[1]
# items.sort(key=sort_by_count, reverse=True)
items.sort(key=lambda x: x[1], reverse=True)
li = [] # ['孔明'， 孔明， 孔明，孔明...., '曹///操'。。。。。]
for i in range(10):
# 序列解包
role, count = items[i]
print(role, count)
# _ 是告诉看代码的人，循环里面不需要使用临时变量
for _ in range(count):
li.append(role)
# 5得出结论
text = ' '.join(li)
WordCloud(
font_path='msyh.ttc',
background_color='white',
width=800,
height=600,
# 相邻两个重复词之间的匹配
collocations=False,
mask=mask
).generate(text).to_file('TOP10.png')
2、lambda表达式
- lambda表达式，通常是在需要一个函数，但是又不想费神去命名一个函数的场合下使用，也就是指匿名函数。
add = lambda x, y : x+y
add(1,2) # 结果为3
- 需求：将列表中的元素按照绝对值大小进行升序排列
list1 = [3,5,-4,-1,0,-2,-6]
sorted(list1, key=lambda x: abs(x))
#匿名函数
#结构
lambda x1, x2....xn: #表达式
sum_num = lambda x1, x2: x1+x2
print(sum_num(2, 3))
# 参数可以是无限多个，但是表达式只有一个
name_info_list = [
('张三',4500),
('李四',9900),
('王五',2000),
('赵六',5500),
]
name_info_list.sort(key=lambda x:x[1], reverse=True)
print(name_info_list)
stu_info = [
{"name":'zhangsan', "age":18},
{"name":'lisi', "age":30},
{"name":'wangwu', "age":99},
{"name":'tiaqi', "age":3},
]
stu_info.sort(key=lambda i:i['age'])
print(stu_info)
3、列表推导式
- 格式
[表达式 for 变量 in 列表] 或者 [表达式 for 变量 in 列表 if 条件]
过滤条件可有可无，取决于实际应用，只留下表达式；
# 之前我们使用普通for 创建列表
li = []
for i in range(10):
li.append(i)
print(li)
# # 使用列表推导式
# # [表达式 for 临时变量 in 可迭代对象 可以追加条件]
print([i for i in range(10)])
# 列表解析
# # 筛选出列表中所有的偶数
li = []
for i in range(10):
if i%2 == 0:
li.append(i)
print(li)
# # 使用列表解析
#
# print([i for i in range(10) if i%2 == 0])
# 筛选出列表中 大于0 的数
from random import randint
num_list = [randint(-10, 10) for _ in range(10)]
print(num_list)
print([i for i in num_list if i>0])
# 字典解析
# 生成100个学生的成绩
stu_grades = {'student{}'.format(i):randint(50, 100) for i in range(1, 101)}
print(stu_grades)
# 筛选大于 60分的所有学生
print({k: v for k, v in stu_grades.items() if v >60})
4、Matplotlib绘图库
# matplotlib
# 导入
from matplotlib import pyplot as plt
plt.rcParams["font.sans-serif"] = ['SimHei']
plt.rcParams['axes.unicode_minus'] = False
import numpy as np
# # 使用100个点 绘制 [0 , 2π]正弦曲线图
# #.linspace 左闭右闭区间的等差数列
# x = np.linspace(0, 2*np.pi, num=100)
# print(x)
# y = np.sin(x)
# # 正弦和余弦在同一坐标系下
# cosy = np.cos(x)
# plt.plot(x, y, color='g', linestyle='--',label='sin(x)')
# plt.plot(x, cosy, color='r',label='cos(x)')
# plt.xlabel('时间（s）')
# plt.ylabel('电压（V）')
# plt.title('欢迎来到python世界')
# # 图例
# plt.legend()
# plt.show()
# 柱状图
# import string
# from random import randint
# # print(string.ascii_uppercase[0:6])
# # ['A', 'B', 'C'...]
# x = ['口红{}'.format(x) for x in string.ascii_uppercase[:5] ]
# y = [randint(200, 500) for _ in range(5)]
# print(x)
# print(y)
# plt.xlabel('口红品牌')
# plt.ylabel('价格（元）')
# plt.bar(x, y)
# plt.show()
#饼图
# from random import randint
# import string
# counts = [randint(3500, 9000) for _ in range(6)]
# labels = ['员工{}'.format(x) for x in string.ascii_lowercase[:6] ]
# # 距离圆心点距离
# explode = [0.1,0,0, 0, 0,0]
# colors = ['red', 'purple','blue', 'yellow','gray','green']
# plt.pie(counts,explode = explode,shadow=True, labels=labels, autopct = '%1.1f%%',colors=colors)
# plt.legend(loc=2)
# plt.axis('equal')
# plt.show()
# 散点图
# 均值为 0 标准差为1 的正太分布数据
# x = np.random.normal(0, 1, 100)
# y = np.random.normal(0, 1, 100)
# plt.scatter(x, y)
# plt.show()
x = np.random.normal(0, 1, 1000000)
y = np.random.normal(0, 1, 1000000)
# alpha透明度
plt.scatter(x, y, alpha=0.1)
plt.show()
5、三国人物top10饼图
import jieba
from matplotlib import pyplot as plt
plt.rcParams["font.sans-serif"] = ['SimHei']
plt.rcParams['axes.unicode_minus'] = False
# 1.读取小说内容
with open('threeking.txt', 'r', encoding='utf-8') as f:
words = f.read()
counts = {} # {‘曹操’:234，‘回寨’:56}
excludes = {"将军", "却说", "丞相", "二人", "不可", "荆州", "不能", "如此", "商议",
"如何", "主公", "军士", "军马", "左右", "次日", "引兵", "大喜", "天下",
"东吴", "于是", "今日", "不敢", "魏兵", "陛下", "都督", "人马", "不知",
"孔明曰","玄德曰","刘备","云长"}
# 2. 分词
words_list = jieba.lcut(words)
# print(words_list)
for word in words_list:
if len(word) <= 1:
continue
else:
# 更新字典中的值
# counts[word] = 取出字典中原来键对应的值 + 1
# counts[word] = counts[word] + 1 # counts[word]如果没有就要报错
# 字典。get（k） 如果字典中没有这个键 返回 NONE
counts[word] = counts.get(word, 0) + 1
print(len(counts))
# 3. 词语过滤,删除无关词，重复词
counts['孔明'] = counts['孔明'] + counts['孔明曰']
counts['玄德'] = counts['玄德'] + counts['玄德曰'] +counts['刘备']
counts['关公'] = counts['关公'] +counts['云长']
for word in excludes:
del counts[word]
# 4.排序 [(), ()]
items = list(counts.items())
print(items)
# def sort_by_count(x):
# return x[1]
# items.sort(key=sort_by_count, reverse=True)
sum_num = lambda x1, x2: x1+x2
print(sum_num(2, 3))
items.sort(key=lambda x: x[1], reverse=True)
li = [] # ['孔明'， 孔明， 孔明，孔明...., '曹操'。。。。。]
count1 = []
role1 = []
for i in range(10):
# 序列解包
role, count = items[i]
count1.append(count)
role1.append(role)
print(role1)
counts = count1
labels = role1
plt.pie(counts, shadow=True, labels=labels, autopct='%1.1f%%')
plt.legend(loc=2)
plt.axis('equal')
plt.show()
6、红楼梦人物top10
from wordcloud import WordCloud
import jieba
import imageio
mask = imageio.imread('china.jpg')
with open('all.txt', 'r', encoding='utf-8') as f:
excludes = {"什么", "一个", "我们", "你们", "如今", "说道", "知道", "起来", "这里",
"出来", "众人", "那里", "自己", "一面", "只见", "太太", "两个", "没有",
"怎么", "不是", "不知", "这个", "听见", "这样", "进来", "咱们", "就是",
"老太太", "东西", "告诉", "回来", "只是", "大家", "姑娘", "奶奶", "凤姐儿"}
counts = {}
words = f.read()
words_list = jieba.lcut(words)
for word in words_list:
if len(word) <= 1:
continue
else:
# 更新字典中的值
# counts[word] = 取出字典中原来键对应的值 + 1
# counts[word] = counts[word] + 1 # counts[word]如果没有就要报错
# 字典。get（k） 如果字典中没有这个键 返回 NONE
counts[word] = counts.get(word, 0) + 1
counts['贾母'] = counts['贾母'] + counts['老太太']
counts['黛玉'] = counts['黛玉'] + counts['林黛玉']
counts['宝玉'] = counts['宝玉'] + counts['贾宝玉']
counts['宝钗'] = counts['宝钗'] + counts['薛宝钗']
counts['老爷'] = counts['老爷'] + counts['贾政']
counts['王夫人'] = counts['王夫人'] + counts['太太']
counts['凤姐'] = counts['凤姐儿'] + counts['凤姐'] + counts['王熙凤']
for word in excludes:
del counts[word]
items = list(counts.items())
print(items)
sum_num = lambda x1, x2: x1 + x2
print(sum_num(2, 3))
items.sort(key=lambda x: x[1], reverse=True)
li = [] # ['孔明'， 孔明， 孔明，孔明...., '曹操'。。。。。]
for i in range(10):
# 序列解包
role, count = items[i]
print(role, count)
# _ 是告诉看代码的人，循环里面不需要使用临时变量
for _ in range(count):
li.append(role)
# 5得出结论
text = ' '.join(li)
wc = WordCloud(
font_path='msyh.ttc',
background_color='white',
width=800,
height=600,
# 相邻两个重复词之间的匹配
collocations=False,
mask=mask
).generate(text).to_file('红楼梦词云.png')
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