Question

我正在寻找一个程序，我输入一个单词或一个短语，它计算输入中的元音数量，然后返回每个元音的数量。我已经这样做了，但我的代码很长，我想知道如何更有效地编写代码。

如何减少我使用的for循环量？

text = input("enter string:")

vowels = ["a","e","i","o","u"]
count = 0
for letters in text:
    if letters in vowels:
        count = count + 1


print ("vowel count is:",count)
numA = 0
numE = 0
numI = 0
numO = 0
numU = 0

for a in text:
    if a in "a":
        numA = numA + 1

for e in text:
    if e in "e":
        numE = numE + 1

for i in text:
    if i in "i":
        numI = numI + 1

for o in text:
    if o in "o":
        numO = numO + 1

for u in text:
    if u in "u":
        numU = numU + 1




print ("a occurs",numA)
print ("e occurs",numE)
print ("i occurs",numI)
print ("o occurs",numO)
print ("u occurs",numU)

Answer 1

使用dict：

vowels = ["a", "e", "i", "o", "u"]
# We use a dict to count the vowels.
# It is initialized with 0 for every vowel.
count = {v: 0 for v in vowels}

# Loop over every character in the input.
for c in "foo bar baz":
    # If the character is a vowel, increment its count.
    if c in vowels:
        count[c] = count[c] + 1

print(count)

输出：

{'i': 0, 'u': 0, 'a': 2, 'o': 2, 'e': 0}

Answer 2

作为@Lutz Horn答案的补充，您可以缩短它，并根据需要计算不同元音的数量：

text = input("enter string:")
# counts the number of different vowels
no_vowels = len(set(text).intersection("aeiou"))
# this is a dict comprehension, it constructs a dict with a vowel as key and the occurrence count as key
counts = {vowel: text.count(vowel) for vowel in "aeiou"}

# then you can print the number of occurrences from each item
print("vowel count is:", no_vowels)
for vowel in "aeiou":
    print(vowel, "is used", count[vowel], "times")

或者你可以将它减少到4个可读行，如果你不需要将计数保存在变量中，只需要打印它们：

text = input("enter string:")
print("vowel count is", len(set(text).intersection("aeiou")))
for vowel in "aeiou":
     print(vowel, "occurs", text.count(vowel), "times")

或者你可以使用内置的collections.Counter Python，它提供了最佳的性能和功能：

from collections import Counter
text = input("enter string:")
counter = Counter(text)

# counter now contains the counts for all letters, not only vowels
for vowel in "aeiou":
    print(vowel, "occurs", counter[vowel], "times")

# this one is trickier, it is a sum of boolean, which returns the number of True
print("vowel count is:", sum(vowel in counter for vowel in "aeiou"))

Answer 3

您可以使用count()：

number_of_a = text.count('a')
number_of_e = text.count('e')
...

和元音总数：

number_of_vowels = sum(1 for x in text if x in 'aeiou')

Answer 4

我喜欢上面显示的字典示例，但如果您想稍微改变，可以执行以下操作：

仅迭代输入字符串一次。

使用数组跟踪每个元音的计数。

vowelArray = [0, 0, 0, 0, 0]
inputStr = "This is the string to loop over"

for char in inputStr:
    if char == "a":
        vowelArray[0] += 1
    if char == "e":
        vowelArray[1] += 1
    if char == "i":
        vowelArray[2] += 1
    if char == "o":
        vowelArray[3] += 1
    if char == "u":
        vowelArray[4] += 1

print(vowelArray)

Answer 5

尝试尽可能保留原始代码，这是您的算法，使用dict - 来计算元音：

#text = input("enter string:")
text = "phraseanditcountsthenumberofvowels"

vowels = {"a":0,"e":0,"i":0,"o":0,"u":0}
for letter in text:
    letter = letter.lower()
    if letter in vowels:
        vowels[letter] += 1

print(vowels)

更有效地使用循环

5 个答案: