Find character occurance percentage from a List of words

You can use collections' Counter for this, and then divide by the total number of characters:

from collections import Counter

word_list = ['THE', 'ZEN', 'OF', 'PYTHON', 'BY', 'TIM', 'PETERS', 'BEAUTIFUL', 'IS', 'BETTER', 'THAN', 'UGLY', 'EXPLICIT', 'IS', 'BETTER', 'THAN', 'IMPLICIT', 'SIMPLE', 'IS', 'BETTER', 'THAN', 'COMPLEX', 'COMPLEX', 'IS', 'BETTER', 'THAN', 'COMPLICATED', 'FLAT', 'IS', 'BETTER', 'THAN', 'NESTED', 'SPARSE', 'IS', 'BETTER', 'THAN', 'DENSE', 'READABILITY', 'COUNTS', 'SPECIAL', 'CASES', 'ARENT', 'SPECIAL', 'ENOUGH', 'TO', 'BREAK', 'THE', 'RULES', 'ALTHOUGH', 'PRACTICALITY', 'BEATS', 'PURITY', 'ERRORS', 'SHOULD', 'NEVER', 'PASS', 'SILENTLY', 'UNLESS', 'EXPLICITLY', 'SILENCED', 'IN', 'THE', 'FACE', 'OF', 'AMBIGUITY', 'REFUSE', 'THE', 'TEMPTATION', 'TO', 'GUESS', 'THERE', 'SHOULD', 'BE', 'ONE', 'AND', 'PREFERABLY', 'ONLY', 'ONE', 'OBVIOUS', 'WAY', 'TO', 'DO', 'IT', 'ALTHOUGH', 'THAT', 'WAY', 'MAY', 'NOT', 'BE', 'OBVIOUS', 'AT', 'FIRST', 'UNLESS', 'YOURE', 'DUTCH', 'NOW', 'IS', 'BETTER', 'THAN', 'NEVER', 'ALTHOUGH', 'NEVER', 'IS', 'OFTEN', 'BETTER', 'THAN', 'RIGHT', 'NOW', 'IF', 'THE', 'IMPLEMENTATION', 'IS', 'HARD', 'TO', 'EXPLAIN', 'ITS', 'A', 'BAD', 'IDEA', 'IF', 'THE', 'IMPLEMENTATION', 'IS', 'EASY', 'TO', 'EXPLAIN', 'IT', 'MAY', 'BE', 'A', 'GOOD', 'IDEA', 'NAMESPACES', 'ARE', 'ONE', 'HONKING', 'GREAT', 'IDEA', '', 'LETS', 'DO', 'MORE', 'OF', 'THOSE']

# Combine all words together
chars = "".join(word_list)
n_chars = len(chars)
# Count all characters
counter = Counter(chars)
# Get frequency and convert frequency into percentages
occ_pct = [(char, occ / n_chars * 100) for char, occ in counter.most_common()]
print(occ_pct)

[('E', 13.58936484490399), ('T', 11.669128508124077), ('I', 7.828655834564254), ('A', 7.828655834564254), ('S', 6.794682422451995), ('O', 6.3515509601181686), ('N', 6.20384047267356), ('R', 4.874446085672083), ('L', 4.874446085672083), ('H', 4.579025110782865), ('P', 3.2496307237813884), ('B', 3.10192023633678), ('U', 3.10192023633678), ('Y', 2.511078286558346), ('C', 2.511078286558346), ('D', 2.511078286558346), ('M', 2.363367799113737), ('F', 1.7725258493353029), ('G', 1.6248153618906942), ('X', 0.8862629246676514), ('V', 0.7385524372230428), ('W', 0.5908419497784343), ('K', 0.29542097488921715), ('Z', 0.14771048744460857)]

This can easily be printed out nicely:

for occ, pct in occ_pct:
    print(occ, f"{pct:.2f}%")

E 13.59%
T 11.67%
I 7.83%
A 7.83%
S 6.79%
O 6.35%
N 6.20%
R 4.87%
L 4.87%
H 4.58%
P 3.25%
B 3.10%
U 3.10%
Y 2.51%
C 2.51%
D 2.51%
M 2.36%
F 1.77%
G 1.62%
X 0.89%
V 0.74%
W 0.59%
K 0.30%
Z 0.15%

Or sorted alphabetically by character instead:

for occ, pct in sorted(occ_pct, key=lambda x: x[0]):
    print(occ, f"{pct:.2f}%")

A 7.83%
B 3.10%
C 2.51%
D 2.51%
E 13.59%
F 1.77%
G 1.62%
H 4.58%
I 7.83%
K 0.30%
L 4.87%
M 2.36%
N 6.20%
O 6.35%
P 3.25%
R 4.87%
S 6.79%
T 11.67%
U 3.10%
V 0.74%
W 0.59%
X 0.89%
Y 2.51%
Z 0.15%

Edit: As requested, without collections:

# Mapping from character to usage frequency
usage = {}
# The total number of characters
total = 0
for word in word_list:
    for char in word:
        # Add 1 to the value corresponding to the `char` key.
        # (and set to 1 if it doesn't exist yet)
        usage[char] = usage.get(char, 0) + 1
        total += 1
# usage.items() returns a list of characters/occurrences, so we can use
# that alongside `total` to compute percentages.
occ_pct = [(char, occ / total * 100) for char, occ in usage.items()]

word_list = ['THE', 'ZEN', 'OF', 'PYTHON', 'BY', 'TIM', 'PETERS', 'BEAUTIFUL', 'IS', 'BETTER', 'THAN', 'UGLY', 'EXPLICIT', 'IS', 'BETTER', 'THAN', 'IMPLICIT', 'SIMPLE', 'IS', 'BETTER', 'THAN', 'COMPLEX', 'COMPLEX', 'IS', 'BETTER', 'THAN', 'COMPLICATED', 'FLAT', 'IS', 'BETTER', 'THAN', 'NESTED', 'SPARSE', 'IS', 'BETTER', 'THAN', 'DENSE', 'READABILITY', 'COUNTS', 'SPECIAL', 'CASES', 'ARENT', 'SPECIAL', 'ENOUGH', 'TO', 'BREAK', 'THE', 'RULES', 'ALTHOUGH', 'PRACTICALITY', 'BEATS', 'PURITY', 'ERRORS', 'SHOULD', 'NEVER', 'PASS', 'SILENTLY', 'UNLESS', 'EXPLICITLY', 'SILENCED', 'IN', 'THE', 'FACE', 'OF', 'AMBIGUITY', 'REFUSE', 'THE', 'TEMPTATION', 'TO', 'GUESS', 'THERE', 'SHOULD', 'BE', 'ONE', 'AND', 'PREFERABLY', 'ONLY', 'ONE', 'OBVIOUS', 'WAY', 'TO', 'DO', 'IT', 'ALTHOUGH', 'THAT', 'WAY', 'MAY', 'NOT', 'BE', 'OBVIOUS', 'AT', 'FIRST', 'UNLESS', 'YOURE', 'DUTCH', 'NOW', 'IS', 'BETTER', 'THAN', 'NEVER', 'ALTHOUGH', 'NEVER', 'IS', 'OFTEN', 'BETTER', 'THAN', 'RIGHT', 'NOW', 'IF', 'THE', 'IMPLEMENTATION', 'IS', 'HARD', 'TO', 'EXPLAIN', 'ITS', 'A', 'BAD', 'IDEA', 'IF', 'THE', 'IMPLEMENTATION', 'IS', 'EASY', 'TO', 'EXPLAIN', 'IT', 'MAY', 'BE', 'A', 'GOOD', 'IDEA', 'NAMESPACES', 'ARE', 'ONE', 'HONKING', 'GREAT', 'IDEA', '', 'LETS', 'DO', 'MORE', 'OF', 'THOSE']

word_str = ''.join(word_list)
word_length = len(word_str)

character_count = dict()
for character in word_str:
    if character in character_count:
        # character has occured at least once already  
        character_count[character] += 1
    else:
        # character occurs for the first time
        character_count[character] = 1

for character, count in sorted(character_count.items()):
    print(f'{character}: {round(count / word_length * 100, 2)}%')

This is a fairly straight-forward solution. First, I make a single string out of your list of strings to simply things a bit and the individual strings do not appear to be necessary for this. ''.join(word_list) concatenates the strings with an empty string between them.

word_str -> 'THEZENOFPYTHON...'

The total number of characters len(word_str) is necessary to calculate the percentage occurance of a character.

There are probably more elegant ways to count the characters, but I find a dictionary (character_count) easier to read/explain.

In the for-loop, I check if a counter for a character already exists in the dictionary. If it does, increment the counter. If it doesn't, initiate the counter with the value 1.

Now, the dictionary contains a count of every occuring character:

character_count -> {
    'T': 10,
    'H': 5,
    ...
}

Next, to print the values the way you want, we can iterate over the keys and values of the dictionary with character_count.items(). However, use sorted() to make them appear in lexicographic order.

I calculate the percentage directly in the formatted string. count / word_length would be something like 0.018912347, so I multiply by 100 and then use round( ... , 2) to only display up to two decimal digits.

You can use dictionary to count characters in all words and count character. then go through whole dict and calculate percentage of character.

word_list = ['THE', 'ZEN', 'OF', 'PYTHON', 'BY', 'TIM', 'PETERS', 'BEAUTIFUL', 'IS', 'BETTER', 'THAN', 'UGLY', 'EXPLICIT', 'IS', 'BETTER', 'THAN', 'IMPLICIT', 'SIMPLE', 'IS', 'BETTER', 'THAN', 'COMPLEX', 'COMPLEX', 'IS', 'BETTER', 'THAN', 'COMPLICATED', 'FLAT', 'IS', 'BETTER', 'THAN', 'NESTED', 'SPARSE', 'IS', 'BETTER', 'THAN', 'DENSE', 'READABILITY', 'COUNTS', 'SPECIAL', 'CASES', 'ARENT', 'SPECIAL', 'ENOUGH', 'TO', 'BREAK', 'THE', 'RULES', 'ALTHOUGH', 'PRACTICALITY', 'BEATS', 'PURITY', 'ERRORS', 'SHOULD', 'NEVER', 'PASS', 'SILENTLY', 'UNLESS', 'EXPLICITLY', 'SILENCED', 'IN', 'THE', 'FACE', 'OF', 'AMBIGUITY', 'REFUSE', 'THE', 'TEMPTATION', 'TO', 'GUESS', 'THERE', 'SHOULD', 'BE', 'ONE', 'AND', 'PREFERABLY', 'ONLY', 'ONE', 'OBVIOUS', 'WAY', 'TO', 'DO', 'IT', 'ALTHOUGH', 'THAT', 'WAY', 'MAY', 'NOT', 'BE', 'OBVIOUS', 'AT', 'FIRST', 'UNLESS', 'YOURE', 'DUTCH', 'NOW', 'IS', 'BETTER', 'THAN', 'NEVER', 'ALTHOUGH', 'NEVER', 'IS', 'OFTEN', 'BETTER', 'THAN', 'RIGHT', 'NOW', 'IF', 'THE', 'IMPLEMENTATION', 'IS', 'HARD', 'TO', 'EXPLAIN', 'ITS', 'A', 'BAD', 'IDEA', 'IF', 'THE', 'IMPLEMENTATION', 'IS', 'EASY', 'TO', 'EXPLAIN', 'IT', 'MAY', 'BE', 'A', 'GOOD', 'IDEA', 'NAMESPACES', 'ARE', 'ONE', 'HONKING', 'GREAT', 'IDEA', '', 'LETS', 'DO', 'MORE', 'OF', 'THOSE'] 



character_list = []
printable =''
character_count = 0
char_dict = {}
per_dict = {}
for word in word_list:
   for character in word:
      character_count=character_count+1
      if(character in char_dict):
         char_dict[character]=char_dict[character]+1
      else:
         char_dict[character]=1

for k in char_dict.keys():
    per_dict[k]=round((char_dict[k]/character_count)*100,2)

print(per_dict)