Question

所以我有一个存储的单词。并且邀请用户检查他们选择的字母是否在这个单词中。我的代码是以下

storedword = "abcdeef"
word = list(germ)
print (word)
merge = input("letter please")

print ("your letter is", merge)

counter = int(0)
letterchecker = int(0)
listlength = len(word)
while counter < listlength and merge != word[counter]:
    counter +=1

if counter <listlength:
    print ("found")

else:
    print ("not found")

如何更改此代码以检查用户信函在此单词中的次数？我只能使用if和while循环而不使用.count

Answer 1

len([w for w in word if w == merge])

是

的缩写

x = []
for w in word:
    if w == merge:
        x.append(w)
len(x)

与while循环类似的方法：

i = x = 0
while i < len(word):
    if word[i] == merge:
        x += 1
    i += 1

Answer 2

您可以使用Counter

吗？

from collections import Counter

storedword = "abcdeef"

wordcounter = Counter(list(storedword))

merge = input("letter please ")

print("your letter is %s" % merge)
print('It occurs %d times' % wordcounter[merge])

Answer 3

counter = 0
letter_count = 0
while counter < len(word);
    if word[counter] == merge:
         letter_count +=1
    counter +=1

Answer 4

试试这个：

counter = 0
for c in word:
    if c == merge:
        counter += 1

如果您无法使用，请使用：

counter = 0
ind = 0
while ind < len(word):
    if word[ind] == merge:
        counter += 1
    ind +=1

在python上，如何检查字母在单词中的次数（列表）

4 个答案: