Can I compare IF and FOR in python?-CodePudding

I have a problem with this code, Can i Use a IF beetwen a FOR in python?

In this code I want know if the first letter is a consonant or a vowel, but the code do a error

word = input("Enter a word: ")
word = word.lower()
vowel="aeiou"
if word[0] == for i in vowel:
        word = word "way"
        print(word)
else:
        word = word[1:len(word)]  "ay"
        print(word)

Input In [16] if word[0] == for i in vowel: ^ SyntaxError: invalid syntax

Can I compare FOR with an IF? What is the correct sintax for that?

CodePudding user response：

You don't need a for loop at all. Instead, you can use the in operator to check if a string contains a substring, like so:

word = input("Enter a word: ")
word = word.lower()
vowel = "aeiou"

if word[0] in vowel:
  word = word   "way"
  print(word)
else:
  word = word[1:len(word)]   "ay"
  print(word)

For example, if you ran this code and entered "hello", the result would be "elloay". If you entered "aardvark", the result would be "aardvarkway".

CodePudding user response：

What you need to do is move the for loop outside the if statement. It should look something like this:

word = input("Enter a word: ")
word = word.lower()
vowel="aeiou"
for i in vowel:
    if word[0] == i:
            word = word "way"
            print(word)
    else:
            word = word[1:len(word)]  "ay"
            print(word)

CodePudding user response：

other way to do this maybe more simple or efficient

word = input("Enter a word: ")
word = word 'way' if word.lower()[0] in ['a','e','i','o','u'] else word[1:len(word)] 'ay'
print(word)