using a list of characters to search characters in a string

Question

I have not long been learning python and after going through the basics have been going through some tasks I have found online.

For this one I'm to use a list of characters in this case the alphabet to count the amount of those characters in a sentence.

I have got to the point where it is counting something but cuts out on character c giving me a value of 30 when there is only 1 c in there and the return:

  File "/home/pi/main.py", line 17, in <module>
    if statementLower[l] == alphabet[m]:
IndexError: string index out of range

Code attached

alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
statement = "This is my task I want to complete. But I am stuck!!!!!!"
statementLower = statement.lower()
m = 0
l=0
count = 0
print(statementLower)

for i in alphabet:
    frequency = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    #print("i for alphabet")
    #print("-" + i + "-")
    for k in statementLower:
        #print("k statement Lower")
        #print(k)
        if statementLower[l] == alphabet[m]:
            #print(count)
            count = count+1
            frequency[m] = count
            l+l+1
            continue
        else:
            #print("else continue loop")
            l=l+1
            continue
        
    if frequency[m] != 0:
        #print("frequency loop")
        print(str(alphabet[m]+ ":" + str(frequency[m])))
        m=m+1
        count = 0
        l=0
        continue

Can anyone give me any assistance in what I'm doing wrong I can see that I am constantly getting stuck in that count if statement but its not incrementing to the next index of the string even when I am increasing that value.

my only current returns are a:44 b:20 c:30 then the string index out of range

Answer 1

I've noticed two errors in your code:

• You are only resetting the counter variables if the current frequency value is not 0, meaning that it will error out eventually after encountering a character that isn't present in the target string. Resetting these values outside of the if statement makes your code go beyond "c", but will still result in invalid values.
• You have a l+l+1 statement in your code; probably a typo, and you meant l=l+1.

Correcting these errors result in your code working as intended. Below is your original source file (without the initializers at the top); I have left comments documenting what these errors are (and added some tips on how to make your code more "pythonic").

for i in alphabet:
    frequency = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    for k in statementLower:
        if statementLower[l] == alphabet[m]:
            count = count + 1 # Can be simplified to count += 1
            frequency[m] = count # You don't have to have the "count" variable at all; instead, you can simply use frequency[m]. That'll save you the hassle of keeping track of an extra variable.
            # If you want to stop using the "count" variable, simply replace the above two lines with "frequency[m] += 1". You can then delete all lines referencing "count".
            l+l+1 # This statement will do nothing; you probably meant the first "+" to be a "=".
            continue
        else:
            # These assignments can be moved outside the else-if; and, as this is the only statement in the else-if,
            # the else can be removed completely.
            l=l+1
            continue
        
    if frequency[m] != 0:
        print(str(alphabet[m]+ ":" + str(frequency[m])))
        # You are only resetting your counter variables if the current frequency value is not 0.
        # The reason why your code didn't execute correctly after "c" is because there are no
        # "d" characters in the string; so you're not updating m, count, and l after you encounter such a character.
        m=m+1
        count = 0
        l=0
        continue

Also, your script can be simplified to this:

alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
statement = "This is my task I want to complete. But I am stuck!!!!!!"
statementLower = statement.lower()
m = 0
l=0
count = 0
print(statementLower)

frequency = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

# For each letter index of the array "alphabet"
for i in range(len(alphabet)):
    # For each character of the string "statementLower"
    for k in statementLower:
        # If the character is equal to the character under the alphabet index i...
        if k == alphabet[i]:
            # ...add 1 to the frequency under that index.
            frequency[i] += 1
    
# Print out the frequencies
for i in range(len(alphabet)):
    if frequency[i] != 0:
        print(str(alphabet[i]+ ": " + str(frequency[i])))
        # bonus: f-strings! they're cool, fancy, and are more pythonic! :3
        # print(f"{alphabet[i]}: {frequency[i]}")

Best of luck in your Python endeavors!

Answer 2

I've come up with this, similar to first answer:

alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
statement = "This is my task I want to complete. But I am stuck!!!!!!"


statementLower = statement.lower()

print(statementLower)

setz = [i for i in statementLower if i in alphabet ] #list of char in statementLower if they are in alphabet
print(setz)

#freqs = [(item, setz.count(item)) for item in set(setz)] #freq as list, uses .count() method on set() of setz
                                                         
freqs = dict([(item, setz.count(item)) for item in set(setz)]) # freqs as dictionary, see above

print(freqs)

output:

this is my task i want to complete. but i am stuck!!!!!!
['t', 'h', 'i', 's', 'i', 's', 'm', 'y', 't', 'a', 's', 'k', 'i', 'w', 'a', 'n', 't', 't', 'o', 'c', 'o', 'm', 'p', 'l', 'e', 't', 'e', 'b', 'u', 't', 'i', 'a', 'm', 's', 't', 'u', 'c', 'k']
{'w': 1, 'm': 3, 'b': 1, 'n': 1, 'c': 2, 'a': 3, 'k': 2, 'h': 1, 'l': 1, 'u': 2, 'o': 2, 'e': 2, 'y': 1, 's': 4, 'i': 4, 'p': 1, 't': 7}

my bad that from Python 3.6 dictionary preserve inserting order while mine doesnt follow alphabet, see: Are dictionaries ordered in Python 3.6+?

Answer 3

As has already been said, collections.Counter is ideal for this. However, if you want to do this without importing anything, you could consider using a dictionary (which is essentially how Counter is implemented). The dictionary will be keyed on any letters in the range a-z found in the statement string.

statement = 'This is my task I want to complete. But I am stuck!!!!!!'

result = {}

for letter in statement.lower():
    if letter.isalpha():
        result[letter] = result.get(letter, 0) + 1

for key, value in sorted(result.items()):
    print(f'{key}={value}')

Output:

a=3
b=1
c=2
e=2
h=1
i=4
k=2
l=1
m=3
n=1
o=2
p=1
s=4
t=7
u=2
w=1
y=1

Answer 4

First option: use collections.Counter. As suggested by Steven Rumbalski:

frequency = collections.Counter(filter(str.isalpha, statement.lower()))

If you want a bit more work, then there is the str.count() method. Suggestion: let's make frequency a dict instead of a list

frequency = {}
for a in alphabet:
    frequency[a] = statementLower.count(a) 

If you want still more work, ok:

frequency = {a:0 for a in alphabet} #or use a defaultdict instead
for a in alphabet:
    for ch in statementLower:
        if a == ch:
            frequency[a] += 1