[SOLVED] Count how many times a list of characters appear in a string – Python (no count or counter)

Issue

I’m a Python newbie, taking a class. I have to find out how many times each letter appears in a string. I’ve seen all the responses about .count and counter, but this has not been covered yet. I took my string and created a list of the characters so they only appeared once. Then I tried to count how many times the character from the list appears in the string, and I just can’t get it to work. I know how to check for a specific letter, but I can’t make it check for each letter in my list. Program is below (there are some extra codes, but I’m using them to make sure it’s counting properly).

# CONSTANTS and Blank Lists
letter_list = []
count = 0
letter_count = []
index = 0

# Begin programming
my_string = 'SUKI IS SWEET'

for ch in my_string:
    if ch not in letter_list:
        letter_list.append(ch)
print(letter_list)
print()

for ch in my_string:
    if letter_list[index] == ch:
        count = count + 1
letter_count.append(count)

# Double-checks to make sure running properly        
print(letter_list[0],'appears',letter_count[0], 'times')
print()
print(letter_count)
print()

When I run this, I get the proper answer:
[‘S’, ‘U’, ‘K’, ‘I’, ‘ ‘, ‘W’, ‘E’, ‘T’] (my letter_list)
S appears 3 times (S does appear 3 times)
[3] (my letter_count)

However, I KNOW that my letter_count.append is in the wrong place. But if I move it in, it gets jacked up. And the moment I try to move it along by increasing the index, it blows up (see below):

# CONSTANTS and Blank Lists
letter_list = []
count = 0
letter_count = []
index = 0

# Begin programming
my_string = 'SUKI IS SWEET'

for ch in my_string:
    if ch not in letter_list:
        letter_list.append(ch)
print(letter_list)
print()

for ch in my_string:
    if letter_list[index] == ch:
        count = count + 1
    letter_count.append(count)
    index = index + 1

# Double-checks to make sure running properly        
print(letter_list[0],'appears',letter_count[0], 'times')
print()
print(letter_count)
print()

RESULTS:
Traceback (most recent call last):
File “C:\Users\Rogue\Desktop\TEST_2.py”, line 30, in
main()
File “C:\Users\Rogue\Desktop\TEST_2.py”, line 18, in main
if letter_list[index] == ch:
IndexError: list index out of range

I know I’m messing this up bad. But no matter what I do, I can’t seem to get this to work. Please help!

Solution

The index out of range error comes from the fact that index is a large as your original string (13 chars), but you use it on letter_list, which is a large as the number of unique characters (8 unique chars).

Here’s a modified version of your code. It’s not efficient, but it works and doesn’t require dicts or Counters:

letter_list = []
letter_count = []

# Begin programming
my_string = 'SUKI IS SWEET'

# Create a list of unique characters
for ch in my_string:
    if ch not in letter_list:
        letter_list.append(ch)
        letter_count.append(0)

# Increment values in letter_count
for ch in my_string:
    for index in range(len(letter_list)):
        if letter_list[index] == ch:
            letter_count[index] += 1

# Double-checks to make sure running properly     
print(letter_list)
print(letter_count)

for index in range(len(letter_list)):
    print("'" + letter_list[index] + "'",'appears',letter_count[index], 'times')

It outputs:

['S', 'U', 'K', 'I', ' ', 'W', 'E', 'T']
[3, 1, 1, 2, 2, 1, 2, 1]

'S' appears 3 times
'U' appears 1 times
'K' appears 1 times
'I' appears 2 times
' ' appears 2 times
'W' appears 1 times
'E' appears 2 times
'T' appears 1 times

Answered By – Eric Duminil

Answer Checked By – David Goodson (BugsFixing Volunteer)