《Python语言程序设计基础》嵩天著-第6章程序部分练习题答案

友情提示：jieba库的安装在其他博客
记不住Python各种操作符，我要原地爆炸了，翻书翻得好烦！

文章目录

program practice 6.1 随机密码生成
program practice 6.2 重复元素判定
program practice 6.4 文本字符分析
program practice 6.5 生日驳论
program practice 6.6 红楼梦人物统计

program practice 6.1 随机密码生成

异常笨的方法，但是符合题意


import random as r
import random as r
a = ["1","2","3","4","5","6","7","8","9","0","q","w","e","r","t","y","u","i","o","p","a","s",\"d","f","g","h","j","k","l","z","x","c","v","b","n","m","Q","W","E","R","T","Y","U","I",\"O","P","A","S","D","F","G","H","J","K","L","Z","X","C","V","B","N","M"]
def code():mi=""global a #全局变量afor i in range(8):n = r.randint(0,61) # 从a列表中任取一个字符串mi = mi + a[n] # 字符串相加得字符串return mi
def main():for i in range(1,11):print("the {} code is {}".format(i,code()))
main()

program practice 6.2 重复元素判定

# program practice 6.2 重复元素判定
def repeat():n = []a = input("enter what you want:")while a != "":n.append(a)a = input("enter what you want:")else:if len(n) == len(set(n)): # set可以去除重复值return Falseelse:return True
repeat()

program practice 6.4 文本字符分析

思路比较简单，根据红楼梦改遍一下即可


def txt():counts = {}a = input("please enter anything you want: ")while a != "":a = input("please enter anything you want: ")counts[a] = counts.get(a,0) + 1 else:b = list(counts.items()) #字典转换成列表，元组为元素b.sort(key=lambda x:x[1], reverse=True)c = len(b)for i in range(c):word, count = b[i]print ("{0:<10}{1:>5}".format(word, count))
txt()

program practice 6.5 生日驳论

可以不直接给每个人随机设置生日的，可以直接用概论算就好了，我这是多此一举，画蛇添足了！


import random as r
p = 1
def birth(n):ls = []for i in range(n):a = r.randint(1,366) # 这里直接给每个人随机设置了生日ls.append(a) # 把所有的生日加到一个列表去if len(ls) != len(set(ls)): # 不相等即代表有生日相同的~global pfor i in range(n):p = p*(366-i) # 生日不相同的取法else:print("There is no same birth")return 1-p/366**n #新学的概论统计派上了用场~
birth(90)

program practice 6.6 红楼梦人物统计

忏愧，四大名著之一的红楼梦竟然没读过，谈起这些名字都不知道是谁……得一个个问度娘
比较简单，直接书上实例，再进行修改即可

# program practice 6.6 红楼梦人物统计
import jieba as j
excludes = {"什么","一个","我们","那里","如今","你们","起来","这里","说道","众人","他们","出来","姑娘","知道","自己","一面","只见","两个","怎么","没有","不是","不知","这个","不知","听见","这样","进来","告诉","东西","就是","咱们","回来","夫人","大家","只是","所以","出去","不敢","这些","只得","丫头","不过","的话","一时","不好","鸳鸯","过来","不能","心里","如此","今日","银子","二人","几个","答应","还有","罢了","一回","说话","只管","这么","那边","这话","外头","打发","自然","那些","今儿","听说","小丫头","屋里","奶奶"} #排除无用词
txt = open("红楼梦.txt", "r", encoding='utf-8').read() # 必须得把文件放在ipynb同源文件夹下才能这样打开
words  = j.lcut(txt) # 分词后某些字符串仍带有标点符号？
counts = {}
for word in words:if len(word) == 1:  #排除单个字符的分词结果，尽量确保分词有效为人名continue # 结束，进行下一个elif word == "王熙凤" or word == "凤姐" or word == "二奶奶" or word== "凤姐儿":rword = "王熙凤"elif word == "老太太" or word == "史太君" or word == "老祖宗":rword = "贾母"elif word == "宝钗" or word == "宝姐姐" or word == "宝姑娘":rword = "薛宝钗"elif word == "太太" or word == "王夫人" :rword = "王夫人"elif word == "花珍珠" or word =="袭人":rword = "花袭人"elif word == "湘云" or word == "云儿": rword = "史湘云"elif word == "黛玉" or word == "林妹妹" or word == "潇湘妃子": rword = "林黛玉"      elif word == "宝玉" or word == "宝二爷": rword = "林宝玉" else:rword = wordcounts[rword] = counts.get(rword,0) + 1 # i不在c内，返回1，在的话返回对应值+1，相当于给每一个i赋值，值即为次数，每次出现可更新
for word in excludes:del(counts[word])
items = list(counts.items()) #字典转换成列表，元组为元素
items.sort(key=lambda x:x[1], reverse=True) # 定义x函数为取第二个元素即次数
# reverse为逆向排序，x[1]为元组第二个数据即次数，key即为排序参数
for i in range(20):word, count = items[i] #键值对输出print ("{0:<10}{1:>5}".format(word, count))

码字不易，就这样吧~

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《Python语言程序设计基础》嵩天著-第6章程序部分练习题答案

文章目录

program practice 6.1 随机密码生成

program practice 6.2 重复元素判定

program practice 6.4 文本字符分析

program practice 6.5 生日驳论

program practice 6.6 红楼梦人物统计

《Python语言程序设计基础》嵩天著-第6章程序部分练习题答案相关推荐

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