I need to count how many excact letter in a word. E.g i need to count the letter "i" in these word. And i need to find which word has the most "i" letter in it.
list = ['box', 'computer', 'sheep', 'family', 'illegal', 'bubble', 'lamp', 'circuit', 'building', 'plant', 'these', 'cup']
def fgv(list):
counter= 0
maximal = 0
word= "x"
for e in list:
counter = 0
if e.__contains__("i"):
for i in range (len(e)):
if e[i] == "i":
couner += 1
if counter < maximal:
maximal = counter
word = e
return(word)
print(f"{fgv(list)}")
4 Answers 4
One more approach which is doing too much work for your problem, but shows off the stdlib's collections.Counter object which is very useful for a bunch of other applications!
from collections import Counter
words = ['box', 'computer', 'sheep', 'family', 'illegal', 'bubble', 'lamp', 'circuit', 'building', 'plant', 'these', 'cup']
word_counts = [Counter(word) for word in words]
# word_counts is now a list of Counter objects, which each look a little bit
# like {"b": 1, "o": 1, "x": 1} for box
# or {"i": 1, "l": 3, "e": 1, "g": 1, "a": 1} for illegal
# now we'll get the index for the word with the most "i's"
idx_max_by_i = max(enumerate(word_counts), key=lambda pair: pair[1]['i'])[0]
# and convert that back to the word itself by indexing into the words list
result = words[idx_max_by_i]
answered Jul 2, 2023 at 16:14
Adam Smith
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1 Comment
Adam Smith
There are definitely better ways to accomplish this, but I thought it was useful to show off what Counter can do. Note its more powerful methods like
most_common(n) which show the most common N number of elements, or elements which give back the "exploded" view of the elements in the counter (e.g. Counter('sheep').elements() produces something like "eehps")If you are only interested in one (the first) word that has a most amount of i's in case of a tie, then this functional one-liner may be desirable:
lst = ['box', 'computer', 'sheep', 'family', 'illegal', 'bubble', 'lamp', 'circuit', 'building', 'plant', 'these', 'cup']
print(max(lst, key=lambda x: x.count("i")))
answered Jul 2, 2023 at 14:30
juanpethes
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1 Comment
Adam Smith
@juanpethes consider storing the result of
word.count('i') for each word and iterating over the zip of lst and word_counts in your final list compdef count_letter(list):
counter = 0
maximal = 0
word = ""
for e in list:
counter = e.count("i")
if counter > maximal:
maximal = counter
word = e
return word
word_list = ['box', 'computer', 'sheep', 'family', 'illegal', 'bubble', 'lamp', 'circuit', 'building', 'plant', 'these', 'cup']
print(count_letter(word_list))
try this code
4 Comments
juanpethes
This only prints the first occurrence of a one with max amount of i's, if there is a tie. Not necessarily wrong, I just point that out.
NotTheDr01ds
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jackal
This is poor advice. 1) variable named list is a no-no. 2) If the input list is empty the return value will be an empty string. But there was no empty string in the list! 3) It will only ever return the first observed word matching the condition - there may be multiples
If you consider that dictionaries can be keyed on integers then how about:
_list = ['box', 'computer', 'sheep', 'family', 'illegal', 'bubble', 'lamp', 'circuit', 'building', 'plant', 'these', 'cup']
def get_count(list_of_words: list[str], letter: str) -> list[str]|None:
if list_of_words and len(letter) == 1:
d = {}
for word in list_of_words:
d.setdefault(word.count(letter), []).append(word)
return d[max(d)]
print(get_count(_list, 'i'))
Output:
['circuit', 'building']
answered Jul 2, 2023 at 15:49
jackal
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Comments
lang-py
[w.count('i') for w in list]→[0, 0, 0, 1, 1, 0, 0, 2, 2, 0, 0, 0]python, strings have acount()method which seems to be what you are looking for,