Python：每日一题 164,Python交流,技术交流,鱼C论坛

冬雪雪冬 发表于 2018-3-15 21:51:40

Python：每日一题 164

本帖最后由冬雪雪冬于 2018-3-16 20:55 编辑

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题目：
元音字母有：a，e，i，o，u五个，写一个函数，交换字符串的元音字符位置。
假设，一个字符串中只有二个不同的元音字符。
二个测试用例：
输入 apple输出 eppla
输入machin输出 michan
不符合要求的字符串，输出None，比如：
输入 abca （两个元音相同）输出 None
输入 abicod （有三个元音）输入None

感谢 @shigure_takimi 提供题目，也希望大家有适宜的题目提供给我们，共同把每日一题这个项目做好。请 @shigure_takimi 多回一帖，给予出题奖励。
要求：
题目难易适度，以python语言的运用为主，算法为辅。
对于提供优质题目的鱼油给予鱼币奖励。
可以发私信给我或直接发帖。

sukiwhip 发表于 2018-3-15 22:17:48

本帖最后由 sukiwhip 于 2018-3-16 10:24 编辑

list_a = ['a','e','i','o','u']
list_b = list(input('输入一个字符串：'))

count_1 = []

for i in list_a:
if list_b.count(i) == 1:
   count_1.append(i)
   t = True

else: #字符串有多个同原音或完全没有的返回 None
   print('None')
   t = False
   break

if t:
if len(count_1) == 2: #只执行包含两个原音的字符串
   a = count_1
   b = count_1
   position_a = list_b.index(count_1)
   position_b = list_b.index(count_1)
   del list_b
   list_b.insert(position_a,b)
   del list_b
   list_b.insert(position_b,a)
   word = ''.join(list_b)
   print('结果是：',word)

else:
   print('None')

125zl 发表于 2018-3-15 22:25:44

list1 = ['a','e','i','o','u']
str1 = input('输入一个只有2个不同的元音字符：')
y = []
wz = []
num1 = 0
num2 = 0
for i in str1:
if i in list1:
   num1 += 1
   if i in y:
         pass
   else:
         y.append(i)
         num2 += 1
if num1 == num2 and num1 == 2:
for i in str1:
   if i in y:
         wz.append(str1.find(i))
str2 = str1[:wz] + str1] + str1 + 1:]
str2 = str2[:wz] + str1] + str2 + 1:]
print(str2)
else:
print('None')

shigure_takimi 发表于 2018-3-15 22:32:40

＃不考虑大小写，默认都是小写。
def func(string):
yuanyin =
if len(yuanyin)==2 and len(set(yuanyin)) == 2:
   i = string.index(yuanyin)
   j = string.index(yuanyin)
   string = list(string)
   string, string = string, string
   return ''.join(string)
else:
   return None

a = 'apple'
b = 'machin'
c = 'appla'
d = 'abicod'
e = 'abbbbbbb'
f = 'bdfssdfs'
g = 'abocda'

print(func(a))#-- > eppla
print(func(b))#-- > michan
print(func(c))#-- > None
print(func(d))#-- > None
print(func(e))#-- > None
print(func(f))#-- > None
print(func(g))#-- > None

shigure_takimi 发表于 2018-3-15 22:34:14

本帖最后由 shigure_takimi 于 2018-3-15 22:46 编辑

多回一贴，领奖励咯！
之前还写了一个比较墨迹的，粘贴于此。

def func(string):
yuanyin = 'aiueo'
container = []
firstYuanyin =
secondYuanyin =
for i in string:
   if i in yuanyin:
         if firstYuanyin == False:
            firstYuanyin, firstYuanyin = string.index(i), True
            container.append(i)
         elif secondYuanyin == False:
            if i in container:
               return None, '重复元音'
            else:
               secondYuanyin, secondYuanyin = string.index(i), True
         else:
            return None, '多于两个元音'
if firstYuanyin == False:
   return None, '没有元音'
elif secondYuanyin == False:
   return None, '只有一个元音'

newSring = list(string)
newSring], newSring] = newSring], newSring]
return ''.join(newSring)


a = 'apple'
b = 'machin'
c = 'appla'
d = 'abicod'
e = 'abbbbbbb'
f = 'bdfssdfs'
g = 'abocda'

print(func(a))#-- > eppla
print(func(b))#-- > michan
print(func(c))#-- > (None, '重复元音')
print(func(d))#-- > (None, '多于两个元音')
print(func(e))#-- > (None, '只有一个元音')
print(func(f))#-- > (None, '没有元音')
print(func(g))#-- > (None, '多于两个元音')

溯影发表于 2018-3-15 22:44:22

def check_vowel(str1):
standard = 'aeiou'
flag = {}

index_number = []
for i in range(len(str1)):
if str1 in standard:
if str1 in flag:
flag] += 1
else:
flag] = 0
index_number.append(str1)

if len(flag) != 2:
return 0
for each in flag.values():
if each == 1:
return 0

result = ''
cnt = 0
for each in str1:
if each not in standard:
result = result + each
else:
cnt = cnt + 1
result = result + index_number
return result

if __name__ == '__main__':
str1 = input('请输入字符串:')
if type(check_vowel(str1)) == str:
print(check_vowel(str1))
else:
print('None')

#################
请输入字符串:aeigh
None

***Repl Closed***
#################
请输入字符串:aadfg
None

***Repl Closed***
#################
请输入字符串:avbnm
None

***Repl Closed***
#################
请输入字符串:apple
eppla

***Repl Closed***
#################
请输入字符串:machin
michan

***Repl Closed***

o41003121 发表于 2018-3-15 23:39:44

vowel = 'aeiou'

def changev(l):
vlist = []#字符串中元音字母列表，用于对比出现元音字母是否相等
klist = []#字符串中元音字母索引列表，用于后面字母对换
vsum = 0#记录字符串中元音字母数量
for each in range(len(l)):
   if l in vowel:
         vsum += 1
         klist.append(each)
         vlist.append(l)
if vsum != 2:
   print('None')
elif vlist == vlist:
   print('None')
else:
   llist = list(l)
   temp = llist]
   llist] = llist]
   llist] = temp
   result = ''.join(llist)
   print(result)

str1 = input('输入字符串:')
changev(str1)

所向故往 发表于 2018-3-15 23:58:10

def new_del(s):
k = []
for i in s:
   if i not in k:
         k.append(i)
return k
while True:
word = input("请输入你想要输入的字符：")
L_1 = list(word)
proving = ['a','e','i','o','u']

num = 0 #元音字母出现的次数
L_2 = []
for i in new_del(L_1):
   if i in proving:
         num += 1
if num != 2:#判断当输入字符不满足条件时
   print('None')
   #下面运行的代码是当输入字符满足条件时
else:
   for i in L_1:
         if i in proving:
            L_2.append(L_1.index(i))
         #变换
   a = L_1]
   L_1] = L_1]
   L_1] = a
   print("".join(L_1))
result = input("是否还想验证:请输入'Y'or'N'")
if result == 'N':
   break

坑得飞起 发表于 2018-3-16 00:06:24

ss=input()
s1=['a','e','i','o','u']
s2=list(ss)
zs=0
jl=[]
for i in range(len(s2)) :
if s2 in s1:
   jl+=

if len(jl)==2 :
if jl==jl:
   print('None')
else :
   i=s2]
   s2]=s2]
   s2]=i
   ss=''
   for i in s2 :
         ss+=str(i)
   print(ss)
else :
print('None')

ouyunfu 发表于 2018-3-16 01:49:56

list1 = list(input('请输入一个字符串：'))
list2 = ['a','e','i','o','u']
n=0
list3 = []
for i in list1:
if i in list2:
   list3.append(list1.index(i))
   n = n + 1
def fun(n):
if n==2:
   list1],list1]=list1],list1]
   return (''.join(list1))
else:
   return None
print(fun(n))

scchapter 发表于 2018-3-16 09:02:27

while 1:
str1=input()
str1.replace('a','b')

x= 'auieo'
list1=[]
count=0
for i in range(len(str1)):
   for r in range(5):
         if str1==x:
            count+=1
            list1.append(i)


str2=list(str1)

if count==2 :
   a=str2]
   b=str2]
   if a!=b:
         str2]=b
         str2]=a
         str3=''.join(str2)#将列表转化为字符串
         print(str3)
   else:
         print('none')
else:
   print('none')
显示：
ersde
none
sdee
none
sdefa
sdafe
sdfeae
none

scchapter 发表于 2018-3-16 09:13:35

刚刚穿错了，加了个函数
def position():

   str1=input()
   x= 'auieo'
   list1=[]
   count=0
   for i in range(len(str1)):
         for r in range(5):
            if str1==x:
               count+=1
               list1.append(i)


   str2=list(str1)

   if count==2 :
         a=str2]
         b=str2]
         if a!=b:
            str2]=b
            str2]=a
            str3=''.join(str2)#将列表转化为字符串
            print(str3)
         else:
            print('none')
   else:
         print('none')

JessiFly 发表于 2018-3-16 09:25:57

def change(old):
vowel = 'aeiou'
#list1存储每个元音的个数
list1 = * 5
for i in range(5):
   list1 = old.count(vowel)
if list1.count(1)!= 2:
   return None
else:
   #list2存储两个元音在原字符串里的索引值
   list2 = []
   for i in range(5):
         if old.find(vowel)!= -1:
            list2.append(old.find(vowel))
   list2.sort()
   x = list2
   y = list2
   new = old[:x] + old + old + old + old
   return new
>>> change('apple')
'eppla'
>>> change('machin')
'michan'
>>> change('abca')
>>> change('abicod')

对面的小孩 发表于 2018-3-16 09:38:26

def vowels(n):
list1 = []
string = ''
for i in n:
   if i in ['a', 'e', 'i', 'o', 'u']:
         list1.append(i)

if len(set(list1)) == 2 and len(list1) == 2:
   list1.reverse()
   for i in n:
         if i not in ['a', 'e', 'i', 'o', 'u']:
            string += i
         else:
            for each in list1:
               if each not in string:
                     string += each
                     break

   return string

else:
   return None

temp = input('请输入有两个不同元音的单词：')
X = vowels(temp)
print(X)

inverseli 发表于 2018-3-16 09:53:08

list1 = ['a','e','i','o','u']
list2 = []
def swap(str1):
"""交换字符串中的元音字母位置"""
str1 = list(str1)
for i in str1:
   if i in list1:
         #保存元音字母在给定字符串中的下标
         sub = str1.index(i)
         #储存在list2中
         list2.append(sub)
#根据下标个数判断元音个数
if len(list2) == 2:
   #判断两个元音字母是否相等
   if str1] != str1]:
         #交换元音字母位置
         str1],str1] = str1],str1]
         #转换成字符串
         print(''.join(str1))
   else:
         print("None")
else:
   print("None")

str1 = input("输入进行测试的字符串：")
swap(str1)

yunying12321 发表于 2018-3-16 10:02:57

while True:
temp = input('输入字符串：')
mm = str(temp)
n = 0
x = ['a', 'e', 'i', 'o', 'u']
new_mm = ''

str_a = []
p = []
ss = ''

for i in range(len(mm)):
   str_a.append(mm)
   if mm in x:
         n += 1
         p.append(i)
   else:
         pass

if n == 2:
   ss = str_a]
   str_a] = str_a]
   str_a] = ss
   for each in str_a:
         new_mm += each

   print(new_mm)

else:
   print('None')

结果：
输入字符串：edit
idet
输入字符串：history
hostiry
输入字符串：hisiiii
None
输入字符串：hisdf
None
输入字符串：

大头目 发表于 2018-3-16 10:05:49

def check_a(alpha):
a = ''
b = ''
output = ''
count = 0
for i in alpha:
if i in check_str:
count +=1
if count < 3:
if not len(a):
a = i
else:
b = i
#print(a,b)
if a == b:
return None #a和b一样return None
else:
return None #元音超过3个return None

if len(a) == len(b) == 1:
beta = list(alpha)
x = beta.index(a)
y = beta.index(b)
beta, beta = b, a
for j in beta:
output += j
return output
else:
return None #其他不符合情况return None

check_str = 'aeiou'
alpha = input('请输入字符串：')
print(check_a(alpha))

graceasyi 发表于 2018-3-16 10:07:33

本帖最后由 graceasyi 于 2018-3-16 11:36 编辑

时间有限，先写了一个，有点复杂，有空再优化。

str1 = input("请输入字符串：")
vowel=['a', 'e', 'i', 'o', 'u']
num = []
for s in vowel:
num.append(str1.count(s))

if num.count(1)==2:
v = []
for i, c in enumerate(str1):
   if c in vowel:
         v.append(i)
         v.append(c)
tmp = list(str1)
tmp], tmp] = v, v
print(''.join(tmp))
else:
print("不符合条件")

luokaoge 发表于 2018-3-16 10:13:11

def aeiou():
#定义空列表
list1 = []
#让用户输入单词
a = input('Please input a word:')
#定义元音字符串
b = 'aeiou'
#判断输入单词中元音字母个数
for i in a:
   if i in b:
         list1.append(i)
   else:
         pass
#当输入单词仅有两个元音字母时互换两个元音字母的位置
if len(list1) == 2:
   s1 = list(a)
   #返回输入单词元音字母的索引值
   c = a.index(list1)
   d = a.index(list1)
   #交换元音字母的位置
   e = s1
   f = s1
   s1 = f
   s1 = e
   #赋值新的字符串并打印
   new =''.join(s1)
   print('New word is'+ ' ' + new)
#不为两个元音字母时打印None
else:
   print('None')

victor.xu 发表于 2018-3-16 10:39:00

def exchange(st):
count = 0
vowel = ['a', 'e', 'i', 'o', 'u']
collector = []
lis =
for j in vowel:
   if j in lis:
         count += 1
         collector.append(j)
if count >= 3 or count < 2 or len(set(collector)) == 1:
   return 'None'
else:
   i = lis.index(collector)
   j = lis.index(collector)
   lis, lis = lis, lis
   st = ''.join()
   return st
str1 = 'apple'
str2 = 'machin'
str3 = 'abca'
str4 = 'abicod'
print(exchange(str1))
print(exchange(str2))
print(exchange(str3))
print(exchange(str4))

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Python：每日一题 164