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Tariq Kfaery956 Points
what is wrong with my code here?
Same idea as the last one. My loopy function needs to skip an item this time, though. Loop through each item in items again. If the character at index 0 of the current item is the letter "a", continue to the next one. Otherwise, print out the current member. Example: ["abc", "xyz"] will just print "xyz".
def loopy(items): # Code goes here x =  for x in items: x.append(items) if x.index(0) == "a": continue else: print x
Stuart Wright41,115 Points
Here are some hints to get you started:
- The challenge doesn't ask you to create a new list and append items to it. You can delete both of these lines. You only need to check whether or not each item begins with 'a', and print it out if not.
- The correct syntax for getting the first character of a string is x (where x is the string).
- The print function needs (), so correct syntax is print(x).
Krishna Pratap Chouhan15,203 Points
lets go through the code and question, with couple of things to remember:
string can be treated as a List. so in string="Abrakadabra", string='A', string='b'...so on.
def loopy(items): # Code goes here x =  #creating an empty list for x in items: #for each string in items x.append(items) #Appending whole list of items into the list x. Now x is list of list. if x.index(0) == "a": #checking if index of '0' in the list is a continue else: #trying to print x, which is an array.[treehouse uses python3.0 so print function is print(). print x
Lets see what we need to do now:
def loopy(items): # Code goes here #Loop through each item in items for item in items: #If the character at index 0 of the current item is the letter "a", continue if item == 'a': continue; else: #else print the item print(item);
Hope this helps.