flat dictionary from structured text using dictionary comprehension

Question

I am trying to create a dictionary from a piece of structured text, but I can't wrap my head around the correct syntax.

text = 'english (fluently), spanish (poorly)'

# desired output: 
{english: fluently, spanish: poorly}

# one of my many attempts: 
dict((language,proficiency.strip('\(\)')) for language,proficiency in lp.split(' ') for lp in text.split(', '))

# but resulting error: 
NameError: name 'lp' is not defined 

I guess that lp in lp.split(' ') is not defined, but I can't figure out, how to rework the syntax to get the desired result.

In reality, the scenario is more complex. I have a dataframe and I am aiming to eventually use a function to tidy the aforementioned data into columns for each individual language and columns for each corresponding proficiency. Something like below (although it probably can be done way more efficiently)

# pandas dataframe
pd.DataFrame({'language': ['english, spanish (poorly)', 'turkish']})
        
# desired output: 
pd.DataFrame({'Language: English': [True, False], 'Language proficiency: English': ['average', pd.NA], 'Language: Spanish': [True, False], 'Language proficiency: Spanish': ['poorly', pd.NA], 'Language: Turkish': [False, True], 'Language proficiency: Turkish': [pd.NA, 'average']})
    
# my attempt
def tidy(content):
    if pd.isna(content):
        pass
    else:
        dict((language,proficiency.strip('\(\)')) for language,proficiency in lp.split(' ') for lp in text.split(', '))

def tidy_language(language, content):
    if pd.isna(content):
        return pd.NA
    else:
        if language in content.keys():
            return True
        else:
            return False

def tidy_proficiency(language, content):
    if pd.isna(content):
        return pd.NA
    else:
        if language in content.keys():
            return content.language
        else:
            return pd.NA

languages = ['english', 'spanish', 'turkish']
df['language'] = df['language'].map(lambda x: tidy(x))
for language in languages:
    df['Language: {}'.format(language.capitalize())] = df['language'].map(lambda x: tidy_language(language, content)
    df['Language proficiency: {}'.format(language.capitalize())] =  df['language'].map(lambda x: tidy_proficiency(language, content)

Answer 1

Here is a quick solution. Feed the text to the function.

def text_to_dict(text):
    text=text+" "

    new=""
    for alphabet in text:
        if alphabet=="," or alphabet=="(" or alphabet==")":
            continue;
        new+=alphabet

    lis=[]
    temp=""
    for alphabet in new:
        if alphabet==" ":
            if temp[0]==" ":
                temp=temp[1:len(temp)]
            lis.append(temp)
            temp=""
        temp+=alphabet

    dict={}
    for el in lis:
        if lis.index(el)%2==0:
            dict[el]=lis[lis.index(el)+1]

    return dict

if __name__=="__main__":
    text="english (fluently), spanish (poorly), bangla (fluently)"
    print(text_to_dict(text))