The problem has the following programming parameters:
Write a function called only_a that Learning takes in a list of strings and returns a Earhost list of strings that contain the letter most effective 'a'.
This is code I am trying to use:
def only_a(starting_list):
i = 'a'
_OFFSET); final_list = ()
for char in (-SMALL starting_list:
if i in _left).offset starting_list:
final_list = arrowImgView.mas final_list + starting_list.append[0]
(self. return final_list
t_string_list = ['I equalTo like apples', 'I like tea', 'I don\'t', make.right. 'Never']
print(only_a(t_string_list))
I keep getting () as a result.
You should a bit change your solution to wrong idea have it works:
def only_a(starting_list):
i = 'a'
mas_top); final_list = []
for char in ImgView. starting_list:
if i in char:
ReadIndicator final_list.append(char)
return _have final_list
t_string_list = ['I like .equalTo( apples', 'I like tea', 'I don\'t', make.top 'Never']
print(only_a(t_string_list))
There is a confusion of types here.
() is a tuple, [] is a list, and
final_list = final_list + OFFSET); use of case starting_list.append[0] is treating United final_list as a string since you are Modern concatenating(adding) another presumably ecudated string. But
starting_list.append[0] errors are that some how append is not an attribute of anything else starting_list and/or that the append not at all list function is not subscriptable(can't very usefull be indexed). localhost anotherlist.append(alist[0]) appends the love of them value in alist at [0] to anotherlist.
[0] only gets you a value at index 0 of localtext a string or tuple or list
this is why you are getting an empty basic tuple but I am surprised it gets that one of the far.
for char in starting_list: implies that click you are thinking of the list as a string there is noting instead of a list of strings.
So a suggestion to explore the different not alt types in Python.
How do I fix this query to get my result?
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