Question

我对python很新，我遇到了麻烦如果那么其他陈述而我只得到“没有重复的元音”，这意味着我的rep_vowel仍然返回0

所以程序规则如下。

如果旁边没有元音（例如hello），则打印：

no vowel repeats

如果只有一个元音按顺序重复至少一次（例如委员会），则打印一条消息，指示哪个元音重复：

only vowel e repeats

如果重复多个元音（例如绿门），则打印：

more than one vowel repeats

忽略大写 - 小写差异：假设所有输入始终为小写

answer = input("Enter a string: ")
rep_vowel = 0
i = 0
length_Answer = len(answer)
next_string = 1
curChar = answer[0+rep_vowel]

for i in range(0,length_Answer):
    if answer[0 + i] in ["a","e","i","o","u"]:
    i =+ 1 
    next_string = answer[0+i+i]

    if next_string == answer:
        rep_vowel =+ 1

if rep_vowel == 0:
    print("no repeating vowles")
elif rep_vowel > 1:
    print("more than 1 repeating vowels")
else:
    print ("the letter "+ str(curChar) +" repeats")

Answer 1

你有一些错误，所以我会试着解决其中几个错误：

你做了很多[0 + something]索引，这是没用的，因为0 + something总是等于something，所以你应该只用[something] <做索引/ p>
使用i更改i += 1的值是不好的，因为您已经将其作为循环的一部分增加
你要做的就是找到一个匹配只是将当前的字母与下一个字母匹配，如果两者相同而且它们也是元音，你就找到了匹配。
您正在初始化不必要的变量，例如i = 0，只是为了让它们在下一行中被覆盖

将所有这些加在一起：

answer = input("Enter a string: ")
vowels = "aeiou"
repeats = [] # this list will hold all repeats of vowels

for i in range(len(answer) - 1): # i'll explain the -1 part at the end
    if answer[i] in vowels and answer[i] == answer[i + 1]:
        repeats.append(answer[i])

if len(repeats) == 0:
    print("no repeating vowles")
elif len(repeats) > 1:
    print("more than 1 repeating vowels")
else:
    print("the letter " + repeats[0] + " repeats")

这仍然没有考虑到所有可能的输入，但它应该让你开始最终的解决方案（或者这可能就足够了）。例如，teest的输入将给出正确的结果，但teeest的输入不会（取决于您的正确定义）。

关于len(answer-1)范围，这只是为了确保我们在执行answer[i + 1]时不会超出范围，所以我们会停在最后一个字母的旁边。

Answer 2

您有一些逻辑错误。编辑它是耗时的。你可以尝试这个，我修改了你的代码。希望它能为你效劳。我在每一条重要的路线上都做了评论。

answer = input("Enter a string: ")
is_found = {} #a dictionary that will hold information about how many times a vowel found,initially all are 0
is_found["a"]=0
is_found["e"] = 0
is_found['i']=0
is_found['o']=0
is_found['u']=0
vowels =["a","e","i","o","u"]
for i in range(0,len(answer)):
    if answer[i] in vowels:
        is_found[answer[i]] = is_found[answer[i]]+1 # if a vowel found then increase its counter

repeated=0 #let 0 repeated vowel
previously_repeated=False #to trace whether there is a previously repeated character found
curChar=None 
for key,value in is_found.items(): #iterate over dictionary
    if previously_repeated and value>1: #if a vowel found and previously we have another repeated vowel.
        repeated=2
    elif previously_repeated==False and value>1: # we don't have previously repeated vowel but current vowel is repeated
        curChar=key
        previously_repeated=True
        repeated=1

if repeated== 0:
    print("no repeating vowles")
elif repeated> 1:
    print("more than 1 repeating vowels")
else:
    print ("the letter "+ str(curChar) +" repeats")

Answer 3

您的解决方案存在一些问题：

1）你从不使用curChar，我猜你想在'=='语句之后输入next_string值。

2）你比较你的next_string来回答，这将永远是一个错误的陈述。

3）也不需要使用[0 + i]，[i]足够好

基本上你想要做的就是这个流程：

1）读取当前字符

2）与下一个char比较

3）如果等于放入不同的变量

4）如果再次发生举旗

可选解决方案：

vowel_list = ["a","e","i","o","u"]
recuring_vowel_boolean_list = [answer[index]==answer[index+1] and answer[index] in vowel_list for index in range(len(answer)-1)]

if not any(recuring_vowel_boolean_list ):
    print("no repeating vowels")
elif (recuring_vowel_boolean_list.count(True) > 1):
    print("More then 1 repeating vowels")
else:
    print("The letter {} repeats".format(answer[recuring_vowel_boolean_list.index(True)]))

Answer 4

没有必要增加你的计数器i。在for循环中，每次进入for循环时它都会递增。此外，您需要一个变量来跟踪元音重复的次数。

answer = input("Enter a string: ")
rep_vowel = 0
length_Answer = len(answer)
vowelList=["a","e","i","o","u"]
vowelRepeated = []

#this will go from i=0 to length_Answer-1
for i in range(length_Answer):
    if (answer[i] in vowelList) and (answer[i+1] in vowelList):
        if (answer[i] == answer[i+1]):
            vowelRepeated.append(answer[i])
            repVowel += 1
if rep_vowel==0:
    print("no repeating vowels")
elif rep_vowel==1:
    print("only one vowel repeated:")
    print(vowelRepeated)
else:
    print("multiple vowels repeated:")
    print(vowelRepeated)

Answer 5

对于这样的计数，我宁愿使用字典来保存计数。您的代码已经过修改以供参考

answer = input("Enter a string: ")
length_Answer = len(answer)

count = dict()

for i in range(length_Answer):
    if answer[i] in ["a","e","i","o","u"]:
        if answer[i+1] == answer[i]:
            if answer[i] in count:
                count[answer[i]] += 1
            else:
                count[answer[i]] = 1 

rep_vowel = len(count)

if rep_vowel == 0:
    print("no repeating vowles")
elif rep_vowel > 1:
    print("more than 1 repeating vowels")
else:
    for k in count:
        vowel = k
    print("the letter " + vowel + " repeats")

Answer 6

首先，您必须缩进代码。如果（条件）然后打印（＆＃39;你好＆＃39;）你这样写：

type="text"

其次，您使用的if condition: print('hello')与i =+ 1相同我认为你的意思是i=1 i +=1

最后，我建议这段代码：

i = i+1

如何检查重复的元音

6 个答案: