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chapter9.py
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138 lines (133 loc) · 3.38 KB
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#读入words.txt并且打印处那些长度超过20个字符的单词
def print20():
fin=open('E:\PythonTest\words.txt')
line=fin.readline()
for line in fin:
if len(line)>21:
word=line.strip()
print(word)
#写一个函数has_no_e,给定单词不包含字母e时返回True
def has_no_e(word):
if word.find('e')==-1:
return True
else:
return False
#打印出不含‘e’的单词，并计算这种单词在整个单词表中的百分比
def print_no_e():
fin=open('E:\PythonTest\words.txt')
line=fin.readline()
count=0
sum=0
for line in fin:
if has_no_e(line)==True:
word=line.strip()
print(word)
count=count+1
sum=sum+1
else:
sum=sum+1
bf=count/sum
print(bf)
#编写一个avoids待修改：
def avoids(word,x):
count=0
i=0
while i<len(x):
if word.find(x[i])==-1:
i=i+1
else:
count=count+1
i=i+1
if count==0:
return True
elif count!=0:
return False
def print_no():
fin=open('E:\PythonTest\words.txt')
line=fin.readline()
count=0
sum=0
str=input('请输入禁止字符串：')
for line in fin:
if avoids(line,str)==True:
word=line.strip()
print(word)
count=count+1
sum=sum+1
else:
sum=sum+1
bf=count/sum
print(bf)
print_no()
aueoi:0.0009401
ayeio:0.0089888
aeiouty:只输出4个单词
#接收一个单词以及由需要字母组成的字符串，当单词只由这些字母组成时返回true。空格问题
def uses_only(word,use):
count=0
i=0
while i<len(word):
if use.find(word[i])==-1:
count=count+1
i=i+1
else:
i=i+1
if count==0:
return True
elif count!=0:
return False
#参考答案
def has_no_e(word):
for letter in word:
if letter =='e':
return False
return True
def avoids(word,forbidden):
for letter in word:
if letter in forbidden:
return False
return True
def uses_only(word, available):
for letter in word:
if letter not in available:
return False
return True
#uses_all函数，接收一个单词以及由需要的字母组成的字符串，当单词中所有需要的字母都出现了至少一次时返回True
def uses_all(word,required):
for letter in required:
if letter not in word:
return False
return True
def uses_all(word,required):
return uses_only(required,word)
#编辑一个is_abecedarian的函数，如果单词中的字母是按照字母表顺序排列的（两个重复字母也ok），则返回True
def is_abecdarian(word):
previous=word[0]
for c in word:
if c<previous:
return False:
previous=c
return True
def is_abecdarian1(word):
if len(word)<=1:
return True
if word[0]>word[1]:
return False
return is_abecdarian1(word[1:])
def is_abecdarinan2(word):
i=0
while i<len(word)-1:
if word[i+1]<word[i]:
return False
i=i+1
return True
#判断是否是回文
def is_palindrome(word):
i=0
j=len(word)-1
while i<j:
if word[i]!=word[j]:
return False
i=i+1
j=j-1
return True