Python / (no regex) How to make a list from words of a string after removing numbers and punctiations or how can I make a new string after removing?

user_12095959 0
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I need to make a list from a string and I need to remove the numbers and/or punctiations before adding them to the list. I have written something but it adds only number words as " " spaces. So I have added an "elif" statement to pass all number words but somehow I am uneasy. I can't use regex so is there another aproach or how can I improve this code? Appriciate for help. Here is my codes:

    my_text = " This3 is a sentence6 that1 I have written 44 to see 5 if this code works so that it makes a list of words from using this code"

new_text = my_text.split()
cleared_text = []

for word in new_text:
    if word.isalpha():
        cleared_text.append(word)
    elif word.isnumeric():
        pass
    else:
        containeer = ""
        for letter in word:
            if letter.isalpha():
                containeer += letter
        cleared_text.append(containeer)

print(cleared_text)

My main target is to store words from the string in a dictionary then show how many times a word repeated so is it better to do do this task from list to dictionary instead of list to list then to dictionary.

Thank you all, my regards

python
2 Answers

Here is how I would do it:

my_text = " This3 is a sentence6 that1 I have written 44 to see 5 if this code works so that it makes a list of words from using this code"

new_text = my_text.split()
cleared_text = []

for word in new_text:
    new_word = "".join([char for char in word if char.isalpha()])
    cleared_text += [new_word] if new_word != "" else []
print(cleared_text)

We can also use filter() to remove empty strings:

for word in new_text:
    new_word = "".join([char for char in word if char.isalpha()])
    cleared_text.append(new_word)
cleared_text = list(filter(None, cleared_text))
print(cleared_text)

Output:

['This', 'is', 'a', 'sentence', 'that', 'I', 'have', 'written', 'to', 'see', 'if', 'this', 'code', 'works', 'so', 'that', 'it', 'makes', 'a', 'list', 'of', 'words', 'from', 'using', 'this', 'code']

We can then count the number of occurrences of each word using list comprehension and count():

occurences = dict((word, cleared_text.count(word)) for word in set(cleared_text))
print(occurences)

Output:

{'I': 1, 'of': 1, 'makes': 1, 'is': 1, 'so': 1, 'sentence': 1, 'a': 2, 'this': 2, 'works': 1, 'it': 1, 'words': 1, 'code': 2, 'list': 1, 'if': 1, 'to': 1, 'This': 1, 'see': 1, 'written': 1, 'have': 1, 'from': 1, 'that': 2, 'using': 1}
user_141597960

To stick closer to your own code, you could add a test to your else branch to only append non-empty words and drop the isnumeric branch. For example:

my_text = " This3 is a sentence6 that1 I have written 44 to see 5 if this code works so that it makes a list of words from using this code"

new_text = my_text.split()
cleared_text = []

for word in new_text:
    if word.isalpha():
        cleared_text.append(word)
    else:
        container = ''.join([l for l in word if l.isalpha()])
        if container: cleared_text.append(container)

print(cleared_text)

(Edit) This works, because if container resolves the value of container to a boolean; an empty string gets resolved to false, hence cleared_text.append(container) is only executed for non-empty container values.

user_114113420
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