《Python语言程序设计基础》嵩天著

2023-11-09 22:47| 来源: 网络整理| 查看: 265

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a for 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 False else: 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 ("{0:5}".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 p for 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 = word counts[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 ("{0:5}".format(word, count))

码字不易，就这样吧~

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