学习Python的第三天

1.读取小说内容

with open('./novel/threekingdom.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)

    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
).generate(text).to_file('TOP10.png')

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)

列表推导式，列表解析个字典解析

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})

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()