Python-在单词列表中搜索字母

Question

I am absolutely new to programming atm. I have searched the forums but none of the results were applicable to my case (unless I am mistaken, but trust me the first thing I did was to google). Please help me if you have time.

Write a python program that prompts the user to enter a list of first names and stores them in a list. The program should display how many times the letter 'a' appears within the list.

Now I thought I have got the PERFECT code. But everytime the program only counts the number of "a"s in the first word on the list!?

terminate = False
position = 0
name_list = [ ]

while not terminate:
    name = str(input('Please enter name: '))
    name_list.append(name)

    response = input('Add more names? y/n: ')
    if response == 'n':
        terminate = True
        print(name_list)
        for t in name_list:
            tally = name_list[position].count('a')
            position = position + 1
        print("The numer of 'a' is: ", tally)

If anyone has time to help I would appreciate it.

Answer 1

A couple of points:

1. you meant to use tally as an accumulator but are actually not adding to it.
2. You don't need the position indexer, for t in name_list already iterates over all names in the list, at each iteration of the for loop t is a name from name_list .

fix the inner loop

terminate = False
# position = 0 <-- this is not needed anywhere
name_list = [ ]

while not terminate:
    name = str(input('Please enter name: '))
    name_list.append(name)

    response = input('Add more names? y/n: ')
    if response == 'n':
        terminate = True
        print(name_list)
        tally = 0  # initialise tally
        for t in name_list:
            tally += t.count('a')  # increment tally
    print("The numer of 'a' is: ", tally)

Answer 2

Your code is not fully understandable to me. Probably, because some stuff is missing. So, I am assuming, you already have the list of names. In the code you presented, you are overwriting tally each time you count the number of a in a name in the list. Besides other possible bugs in your code (maybe when creating the list of names - I'd use a while-loop for this), you should write

tally += name_list[position].count('a') which equals

tally = tally + name_list[position].count('a')

instead of your

tally = name_list[position].count('a')

Answer 3

I would suggest you get rid of the for loop, and position counter. If you want the total of all 'a' , you have to sum tally for all the names :

terminate = False
name_list = []
total = 0

while "Adding more names":
    name = str(input('Please enter name: '))
    name_list.append(name)

    tally = name.count('a')
    total += tally
    print("The numer of 'a' in current name is: ", tally)

    if input('Add more names? y/n: ') == 'n':
        break
print(name_list)
print("The numer of 'a' is: ", total)

Answer 4

terminate = False
position = 0
name_list = [ ]
tally = 0
while not terminate:
    name = str(input('Please enter name: '))
    name_list.append(name)

    response = input('Add more names? y/n: ')
    if response == 'n':
        terminate = True
        print(name_list)
        for t in name_list:
            #tally = name_list[position].count('a')
            tally += name_list[position].count('a')
            position = position + 1
        print("The numer of 'a' is: ", tally)

initilize tally=0 and insted of "tally = name_list[position].count('a')" use "tally += name_list[position].count('a')" if you have to count "a" in entire list you have to update the tally value