[英]count letter in words using Python
I am working with Python to write a function Here is an example: lettercount(['hello','it','is','me'],'i')
should return ['it-1','is-1']
.我正在与 Python 一起写一个 function 这是一个例子: lettercount(['hello','it','is','me'],'i')
应该返回['it-1','is-1']
。
Here is my code:这是我的代码:
def lettercount(list, letter):
result = [] #create an empty list
for word in list: #traverse through all the words in the provided list
if word not in result: #from this line I don't know what I'm really doing though
if letter in word:
wrd = "{}-".format(word).append(word.count(letter))
#tried to achieve the same format as provided but failed
result.append(wrd) #this is wrong because str doesn't have the attribute "append"
return result
Can someone give me a hint on this problem?有人可以给我提示这个问题吗? Thanks so much!非常感谢!
The problem is the way you construct wrd
.问题是你构造wrd
的方式。 Change this line -更改此行 -
wrd = "{}-".format(word).append(word.count(letter))
To -到 -
wrd = f"{word}-{word.count(letter)}"
Ps please.请附言。 avoid using list
as a name of a variable and use, for example, lst
instead since list
is a protected word in Python.避免使用list
作为变量名,而是使用lst
等,因为list
在 Python 中是受保护的字。
def lettercount(lst, letter): out = [] for word in lst: cnt = sum(1 for l in word if l == letter) if cnt: out.append(f'{word}-{cnt}') return out
Try:尝试:
def lettercount(lst, letter):
result = []
for word in lst:
if letter in word:
result.append(word + '-' + str(word.count(letter)))
And I don't recommend naming variables "list" because it's an existing keyword in Python.而且我不建议将变量命名为“list”,因为它是 Python 中的现有关键字。
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