在Python中计算字符串中的元音数量

Question

Okay so what I did was

def countvowels(st):
    result=st.count("a")+st.count("A")+st.count("e")+st.count("E")+st.count("i")+st.count("I")+st.count("o")+st.count("O")+st.count("u")+st.count("U")
    return result

This works(I'm aware indentation might be wrong in this post, but the way I have it indented in python, it works).

Is there a better way to do this? Using for loops?

Answer 1

I would do something like

def countvowels(st):
  return len ([c for c in st if c.lower() in 'aeiou'])

Answer 2

There's definitely better ways. Here's one.

   def countvowels(s):
      s = s.lower()
      return sum(s.count(v) for v in "aeiou")

Answer 3

You can do that using list comprehension

def countvowels(w):
    vowels= "aAiIeEoOuU"
    return len([i for i in list(w) if i in list(vowels)])

Answer 4

You could use regex pattern to do this easily. But it looks to me, that you want to do it without. So here is some code to do so:

string = "This is a test for vowel counting"
print [(i,string.count(i)) for i in list("AaEeIiOoUu")]

Answer 5

you can do it in various ways, first look in google before asking, i had copy pasted 2 of them

def countvowels(string):
    num_vowels=0
    for char in string:
        if char in "aeiouAEIOU":
           num_vowels = num_vowels+1
    return num_vowels

---

data = raw_input("Please type a sentence: ")
vowels = "aeiou"
for v in vowels:
    print v, data.lower().count(v)

Answer 6

You can also try Counter from collections (only available from Python 2.7+) as shown below . It'll show how many times each letter has been repeated.

from collections import Counter
st = raw_input("Enter the string")
print Counter(st)

But you want vowels specifically then try this.

import re

def count_vowels(string):
    vowels = re.findall('[aeiou]', string, re.IGNORECASE)
    return len(vowels)

st = input("Enter a string:")
print count_vowels(st)

Answer 7

Here is a version using map:

phrase=list("This is a test for vowel counting")
base="AaEeIiOoUu"
def c(b):
    print b+":",phrase.count(b)
map(c,base)