think python学习心得-（6）利用words.txt来进行填字游戏

练习9-1

fin = open('C://words.txt')
for line in fin:
    word = line.strip()
    if len(word) >20:
            print word

words.txt被保存在了C盘根目录下。

练习9-2

统计字母表中含有e的单词占有的百分比

fin = open('C://words.txt')
counter1 = 0
counter2 = 0
def has_no_e(wo):
    for i in range(len(wo)):
        if wo[i-1]=='e':
            return True

for line in fin:
    word = line.strip()
    flag = has_no_e(word)
    counter1=counter1+1
    if flag == True:
        counter2=counter2+1

counter1=float(counter1)
counter2=float(counter2)
print 100-counter2/counter1*100

练习9-3

利用下标循环实现输入文字中是否存在禁用字母的鉴别

def avoids(word,forbiden):
    for i in range(len(forbiden)):
        for j in range(len(word)):
            if word[j-1]==forbiden[i-1]:
                return False
    return True

print 'Plesase input forbiden alphabet:'
ban=raw_input('>')
print 'Please input desired word:'
test=raw_input('>>')
print avoids(test,ban)

练习9-4

检测是否是由可用字符组成的

def use_only(word,avaliable):
    for i in range(len(word)):
        flag=0
        for j in range(len(avaliable)):
            if word[i-1]==avaliable[j-1]:
                flag =1
        if flag==0:
            return False
    return True

print 'Plesase input avaliable alphabet:'
ban=raw_input('>')
print 'Please input desired word:'
test=raw_input('>>')
print use_only(test,ban)

练习9-5

检测alphebat是否被全部使用

def uses_all(word,avaliable):
    counter=0
    for i in range(len(avaliable)):
        flag=0
        for j in range(len(word)):
            if word[j-1]==avaliable[i-1]:
                flag =1
        if flag==0:
            return False
        else:
            counter=counter+1
    
    if counter == len(avaliable):
        return True
    else:
        return False

print 'Plesase input avaliable alphabet:'
ban=raw_input('>')
print 'Please input desired word:'
test=raw_input('>>')
print uses_all(test,ban)

练习9-6

def is_abecedarian():
    fin = open ('C://words.txt')
    counter=0
    for line in fin:
        flag = 0
        for i in range(len(line)-1):
            if line[i]<=line[i+1]:
                flag=flag+1
                if flag==len(line)-2:
                    counter=counter+1
    return counter
print 'The number of abecedarian words in directionary is'
print is_abecedarian()

练习9-7

QNMD车迷天下。。。。不做了