CodeWars kata元音1
CodeWars kata元音1
提问于 2022-11-04 20:19:18
代码似乎没有替换字符串中的非元音字符。
def vowel_one(s):
not_vowel = '0'
vowels = {'a':'1',
'e':'1',
'i':'1',
'o':'1',
'u':'1'}
#dictionary with keys as vowels, and their values as 1
s = s.lower() #changing list to lowercase
vowelsList = list(s)
for index, item in enumerate(vowelsList): #counter, instead of doing i+=1, using enumerate
for key, value in vowels.items():
if item == key: #if character in string is equal to key in dictionary
vowelsList[index] = value #at the indexes of such characters, replace them with dictionary key values
if item != key and item != '1':
vowelsList[index] = not_vowel
return ("".join(vowelsList)) #removes brackets and ','例如，"vowelOne“的结果应该是: 01010101，而不是: v1w1l1n1，为什么另一个如果语句不工作呢？我设想，如果给定列表(vowelsList)中的项不等于字典(元音)中的任何键，那么它应该替换为'0‘。如果我不嵌套If语句，那么我只会得到0。
指向kata：https://www.codewars.com/kata/580751a40b5a777a200000a1/的链接
回答 1
Stack Overflow用户
回答已采纳
发布于 2022-11-04 20:27:33
如果所有的值都是1.
- You，则不需要字典，不需要两个
for loops，这只会增加complexity.
- You不需要将
string转换为list.
的时间
这是我的密码：
def vowel_one(s):
vowels = ['a', 'e','i', 'o', 'u'] # simple list
#dictionary with keys as vowels, and their values as 1
s = s.lower() #changing list to lowercase
for i in s: # Iterating through every value
if i in vowels: # Check if sub-string in vowels
s = s.replace(i, '1')
else: # If not vowel replace with 0
s = s.replace(i, '0')
return s
print(vowel_one("vowelOne"))编辑：，如果您想省略spaces，可以添加一个elif条件。尝试：
def vowel_one(s):
vowels = ['a', 'e','i', 'o', 'u'] # simple list
#dictionary with keys as vowels, and their values as 1
s = s.lower() #changing list to lowercase
for i in s: # Iterating through every value
if i in vowels: # Check if sub-string in vowels
s = s.replace(i, '1')
elif not i.isspace(): # If not vowel and space replace with 0
s = s.replace(i, '0')
return s
print(vowel_one("Mudith is my name"))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/74322549
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