python语言程序设计基础第二版第六章答案-Python语言程序设计基础（第2版）课后题第六章...

心已赠人 2022-11-21 14:19 276阅读 0赞

课后题是配套的参考答案

\#e10.1CalHamlet.py

def getText():

txt = open("hamlet.txt", "r").read()

txt = txt.lower()

for ch in '!"\#$%&()\*+,-./:;<=>?@\[\\\]^\_"\{|\}~':

txt = txt.replace(ch, " ") \#将文本中特殊字符替换为空格

return txt

hamletTxt = getText()

words = hamletTxt.split()

counts = \{\}

for word in words:

counts\[word\] = counts.get(word,0) + 1

items = list(counts.items())

items.sort(key=lambda x:x\[1\], reverse=True)

for i in range(10):

word, count = items\[i\]

print ("\{0:<10\}\{1:>5\}".format(word, count))

the 1086

and 967

to 757

of 675

you 555

a 554

i 551

my 520

in 433

it 419

\#e10.2CalHamlet.py

excludes = \{"the","and","of","you","a","i","my","in"\}

def getText():

txt = open("hamlet.txt", "r").read()

txt = txt.lower()

for ch in '!"\#$%&()\*+,-./:;<=>?@\[\\\]^\_"\{|\}~':

txt = txt.replace(ch, " ") \#将文本中特殊字符替换为空格

return txt

hamletTxt = getText()

words = hamletTxt.split()

counts = \{\}

for word in words:

counts\[word\] = counts.get(word,0) + 1

for word in excludes:

del(counts\[word\])

items = list(counts.items())

items.sort(key=lambda x:x\[1\], reverse=True)

for i in range(10):

word, count = items\[i\]

print ("\{0:<10\}\{1:>5\}".format(word, count))

to 757

it 419

that 388

ham 358

is 346

not 314

his 304

this 298

with 278

but 273

\#e10.3CalThreeKingdoms.py

import jieba

excludes = \{\}\#\{"将军","却说","丞相"\}

txt = open("三国演义.txt", "r", encoding='utf-8').read()

words = jieba.lcut(txt)

counts = \{\}

for word in words:

if len(word) == 1: \#排除单个字符的分词结果

continue

else:

counts\[word\] = counts.get(word,0) + 1

for word in excludes:

del(counts\[word\])

items = list(counts.items())

items.sort(key=lambda x:x\[1\], reverse=True)

for i in range(15):

word, count = items\[i\]

print ("\{0:<10\}\{1:>5\}".format(word, count))

Building prefix dict from the default dictionary ...

Loading model cost 1.020 seconds.

Prefix dict has been built succesfully.

曹操 953

孔明 836

将军 772

却说 656

玄德 585

关公 510

丞相 491

二人 469

不可 440

荆州 425

玄德曰 390

孔明曰 390

不能 384

如此 378

张飞 358

\#e10.4CalThreeKingdoms.py

import jieba

excludes = \{"将军","却说","荆州","二人","不可","不能","如此"\}

txt = open("三国演义.txt", "r", encoding='utf-8').read()

words = jieba.lcut(txt)

counts = \{\}

for word in words:

if len(word) == 1:

continue

elif word == "诸葛亮" or word == "孔明曰":

rword = "孔明"

elif word == "关公" or word == "云长":

rword = "关羽"

elif word == "玄德" or word == "玄德曰":

rword = "刘备"

elif word == "孟德" or word == "丞相":

rword = "曹操"

else:

rword = word

counts\[rword\] = counts.get(rword,0) + 1

for word in excludes:

del(counts\[word\])

items = list(counts.items())

items.sort(key=lambda x:x\[1\], reverse=True)

for i in range(5):

word, count = items\[i\]

print ("\{0:<10\}\{1:>5\}".format(word, count))

曹操 1451

孔明 1383

刘备 1252

关羽 784

张飞 358

\#6.1

from random import randint

def randd():

pword=''

for i in range(8):

u = randint(0,62)

if u>=10:

if 90<(u+55)<97:

pword+=chr(u+62)

else:

pword+=chr(u+55)

print("\{\} ".format(u+55),end="")

else:

pword+='%d'%u

return pword

def main():

for i in range(1,11):

print("生成的第\{\}个密码是：\{\}".format(i,randdd()))

main()

115 99 65 75 生成的第1个密码是：7scA5K69

100 98 97 76 71 72 81 109 生成的第2个密码是：dbaLGHQm

115 67 67 65 112 104 66 生成的第3个密码是：sCCAphB3

116 113 72 90 75 108 109 生成的第4个密码是：tq9HZKlm

66 114 71 66 80 生成的第5个密码是：6BrG6B2P

89 103 95 114 80 87 生成的第6个密码是：Y3gfrP2W

95 108 111 71 87 115 生成的第7个密码是：floG3Ws0

101 75 81 75 71 生成的第8个密码是：eKQ52K4G

84 77 93 82 107 116 70 生成的第9个密码是：T4MdRktF

84 83 79 77 104 89 89 生成的第10个密码是：TSOM5hYY

\#6.2

def main():

num=\[\]

n=input("请输入一组数字(或者直接按回车结束程序):")

while n!="":

num.append(eval(n))

n=input("请输入一组数字(或者直接按回车结束程序):")

else:

print("正在处理，请稍等")

judge(num)

def judge(n):

if len(n) == len(set(n)):

print("鉴定完毕，没有重复的元素")

else:

print("有重复的元素，总共有\{\}个".format(len(n)-len(set(n))))

main()

请输入一组数字(或者直接按回车结束程序):56

请输入一组数字(或者直接按回车结束程序):25

请输入一组数字(或者直接按回车结束程序):56

请输入一组数字(或者直接按回车结束程序):22

请输入一组数字(或者直接按回车结束程序):11

请输入一组数字(或者直接按回车结束程序):

正在处理，请稍等

有重复的元素，总共有1个

\#6.4

txt=input("请输入您想输入的英文句子：")

counts=\{\}

ex=\[',','.','?','!',':','"',';'\]

for i in txt:

if i == " " or i in ex:

continue

else:

if ord(i)<97:

i=chr(ord(i)+32)

counts\[i\]=counts.get(i,0)+1

items=list(counts.items())

items.sort(key=lambda x:x\[1\],reverse=True)

for u in range(len(items)):

alpha,count=items\[u\]

print("\{\} -> \{\}".format(alpha,count))

请输入您想输入的英文句子：hsgstge

s -> 2

g -> 2

h -> 1

t -> 1

e -> 1

\#6.6

import jieba.posseg as ps

txt = open('红楼梦.txt','r',encoding = 'utf-8').read()

exclude = \['明白'\]

counts = \{\}

def countFigures():

words = ps.cut(txt)

for w in words:

if len(w.word) == 1:

continue

if w.flag == 'nr':

counts\[w.word\] = counts.get(w.word, 0) + 1

for key in exclude:

del(counts\[key\])

items = list(counts.items())

items.sort(key = lambda x:x\[1\], reverse = True)

for i in range(20):

word, count = items\[i\]

print('\{0:<10\}\{1:>5\}'.format(word,count))

countFigures()

宝玉 3748

贾母 1252

凤姐 1129

王夫人 1011

老太太 966

黛玉 870

宝钗 747

贾琏 679

凤姐儿 470

薛姨妈 453

贾政 433

探春 432

紫鹃 406

小丫头 287

贾珍 284

邢夫人 284

林黛玉 280

尤氏 267

薛蟠 237

贾蓉 176

标签：count,word,words,Python,items,课后,第六章,counts,txt

来源： https://blog.csdn.net/qq\_41318400/article/details/89424349

python语言程序设计基础第二版第六章答案-Python语言程序设计基础（第2版）课后题第六章...

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