2019-05-12第三次课（缺课补习）

跨文件调用函数：

#caculate文件：
def caculateNum(num):
    '''
    计算1~num之间的累加和
    :param num: 累加和的末位
    :return: 累加和
    '''
    # res = 0
    # for i in range(1,num+1):
    #     res += i
    # return res

    return sum([i for i in range (1, num+1)])
    pass

#text1文件：
import caculate
res = caculate.caculateNum(100)
print(res)

必须参数和关键字参数

必须参数: 必须严格的遵循正确的顺序传入,调用的时候必须和声明的时候保持一致

 def f(name, age):
     print('I am %s , I am %d years old .'%(name, age))
#pass不加也没问题发生
f('eric', 18)

关键字参数: 使用关键字参数可以允许函数调用和声明时顺序不一致

#python 解释器能够用参数名字匹配参数值
# f(age= 18, name= 'eric')

默认参数

缺省的参数没有传入时,默认值会生效

#与.get()类似
def f(name, age, sex = 'male'):
    print('I am %s , I am %d years old .'%(name, age))
    print('Sex is %s'%sex)

f(name= '李四', age= 19)
f('张三',88,'female')
#至于是否显示指定参数,以方便之后阅读为准

匿名函数

语法:
lambda 参数: 表达式
lambda 参数1, 参数2,......(冒号前的参数可以有多个)
后面的是表达式,只能是一个表达式,不写return,返回值就是表达式的结果；
优点：
减少代码量, 代码看起来"优雅"

 def rect(x,y):
     return x*y

 area = rect(3, 5)
 print(area)

#使用lambda表达式
 res = lambda x, y: x*y
 print(res(4, 5))

 store = ['33',88]
 s = "当当自营" if len(store) == 0 else store[0]
 print(s)

 def cal(x,y):
     if x > y:
         return x*y
     else:
         return x/y

#使用lambda表达式
 calc = lambda x, y:x*y if x>y else x/y
 print('使用lambda: ',calc(5, 4))
 print('使用lambda: ',calc(2, 4))

• 列表的排序中使用lambda表达式

stus = [
    {'name':'zhangsan','age':33},
     {'name':'lisi','age':22},
      {'name':'wangwu','age':43},
       {'name':'zhaolui','age':18},
        {'name':'tangqi','age':9},
]
print('排序前',stus)
#key值是按照哪个元素为依据进行排序,reverse为True为由大到小
res = sorted(stus,key= lambda x: x['age'], reverse= True)
print('排序前',res)
res = sorted(stus,key= lambda x: x['name'])
print('name排序后',res)

案例 三国小说人物出场词频统计

• 代码需要jieba和WordCloud

import jieba# 用于分词
from collections import Counter # 用于统计
from wordcloud import WordCloud#生成词云

#jieba分词
txt = '我来到北京清华大学'
#将字符串分割成等量的中文
seg_list = jieba.lcut(txt)
print(seg_list)

def parse():
    """三国小说人物出场词频统计"""
    #定义无关词的集合
    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","次曰","引兵","大喜","天下",
                "东吴","于是","今日","不敢","魏兵","陛下","都督","人马","不知",
                "玄德曰","孔明曰","刘备","关公"}
    
    with open('threekingdom.txt','r',encoding='utf-8')as f:
        txt = f.read()

    #print(txt)
    words = jieba.lcut(txt)
    print(words)
    #字典内容:'曹操':555
    counts = {}
    for word in words:
        if len(word) == 1:
            continue
        else:
            # 往字典里添加元素
            # count['key'] = 次数+1
            counts[word] = counts.get(word, 0) + 1
            #如果字典中存在值,则加一,如果不存在,则新建并给予默认值0,再加一.
    print(counts)
    
    #将与有关词同义的无关词加到有关词中
    counts['孔明'] = counts.get('孔明') + counts.get('孔明曰')
    counts['玄德'] = counts.get('玄德') + counts.get('玄德曰') + counts.get('刘备')
    counts['关公'] = counts.get('关公') + counts.get('云长')
    #删除无关词
    for word in excludes:
        del counts[word]

    #统计出现频次最高的前10个词(方法一)
    items = list(counts.items())
    #print('排序前',items)
    items.sort(key = lambda x: x[1], reverse=True)
    #print('排序后',items)
    for i in range(10):
        character,count = items[i]
        print(character, count)
        
    #统计出现频次最高的前10个词(方法二)
    #需要collections库文件
    # roles =  Counter(counts)
    # role = roles.most_common(10)
    
    #构造词云字符串
    li = []
    for i in range(10):
        character,count = items[i]
        for _ in range(count):
            li.append(character)
    cloud_txt = ",".join(li)
        
    wc = WordCloud(
        background_color = 'white',#背景色
        font_path = 'msyh.ttc',#文字编码格式
        #是否包含两个词的搭配,默认是T
        collocations = False
    ).generate(cloud_txt)
    wc.to_file('三国中出现前十的人物.png')