我需要編寫一個程序，其中有一個縣名列表，我需要找到該列表中的 5 個元音中的每個元音的數(shù)量，并將這 5 個數(shù)字放入字典中。我想做一個 for 循環(huán)來遍歷每個元音，每次遍歷循環(huán)時，在字典中添加一個新條目，以元音為鍵，以計數(shù)為值。它應(yīng)該打?。簕'a':4, 'e':4, 'i':4, 'o':4, 'u':4}。我不知道有多少元音，所以我只為示例中的所有值寫了 4?？h的列表真的很長，所以我只是在這里粘貼了一個縮短的版本。counties = ['Autauga','Baldwin','Barbour','Bibb','Blount','Bullock','Butler','Calhoun','Chambers','Cherokee','Chilton','Choctaw','Clarke','Clay','Cleburne','Coffee','Colbert','Conecuh','Coosa','Covington','Crenshaw','Cullman','Dale','Dallas']letter = ('a', 'e', 'i', 'o', 'u') counter = 0 d={} for it in clist: def func(clist, letterlist, count): count += clist.count(letterlist) print("the number of vowels:" count) return count func(counties, letter, counter)如您所見，我對 Python 非常陌生，不知道自己在做什么。我不能讓它工作，絕對不能在字典中得到它。任何幫助，將不勝感激！
2 回答
長風(fēng)秋雁
TA貢獻(xiàn)1757條經(jīng)驗 獲得超7個贊
您可以使用嵌套for循環(huán)遍歷counties列表和每個縣的字符，如果字符在vowels列表中，則以字符為鍵繼續(xù)遞增輸出字典：
d = {}
for county in counties:
for character in county.lower():
if character in vowels:
d[character] = d.get(character, 0) + 1
d 變成：
{'a': 16, 'u': 10, 'i': 4, 'o': 14, 'e': 14}
或者，您可以使用collections.Counter從字符串列表中提取元音字符的生成器表達(dá)式：
from collections import Counter
Counter(c for county in counties for c in county.lower() if c in vowels)
慕的地8271018
TA貢獻(xiàn)1796條經(jīng)驗 獲得超4個贊
我相信您正在嘗試執(zhí)行我在下面列出的操作（我省略了您的國家/地區(qū)列表）。我嘗試添加注釋，您可以添加打印行以查看每段代碼在做什么。
vowels = ('a', 'e', 'i','o', 'u')
d={}
## select countries 1 at a time
for country in countries:
# convert each country to lowercase, to match list of vowels, otherwise, you need to deal with upper and lowercase
country = country.lower()
# select letter in country 1 at a time
for i in range(len(country)):
# check to see whether the letter selected in the country, the ith letter, is a vowel
if (country[i] == 'a') or (country[i] == 'e') or (country[i] == 'i') or (country[i] == 'o') or (country[i] == 'u'):
# if the ith letter is a vowel, but is not yet in the dictionary, add it
if (country[i]) not in d:
d[country[i]] = 1
# if the ith letter is a vowel, and is already in the dictionary, then increase the counter by 1
else:
d[country[i]] += 1
print(d) # {'a': 16, 'u': 10, 'i': 4, 'o': 14, 'e': 14}
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