基于词典理解的结构化文本平面词典

Question

我正试图从一段结构化文本中创建一本词典，但我无法理解正确的语法

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 

我猜lp.split（“”）中的lp没有定义，但我不知道如何修改语法以获得所需的结果

事实上，情况更为复杂。我有一个dataframe，我的目标是最终使用一个函数将上述数据整理成每种语言的列和每种相应语言的列。如下所示（尽管可能可以更有效地完成）

# 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

这里有一个快速的解决方案。将文本馈送到函数

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))

Answer 2

1. 您需要反转列表理解中的两个for循环（for循环需要以与编写命令式代码相同的顺序出现）

2. 在.strip('\(\)')中不需要反斜杠

3. for language,proficiency in lp.split(' ')将尝试将lp.split(' ')的每个项解压到元组(language,proficiency)，因此，将lp.split(' ')包装到一个单元素列表中以实现您想要的：

dict((l,p.strip('()')) for lp in text.split(', ') for l,p in [lp.split(' ')])

{'english': 'fluently', 'spanish': 'poorly'}

以上内容可以写成dict理解：

{l: p.strip('()') for lp in text.split(', ') for l,p in [lp.split(' ')]}

读起来好一点

使用re的替代方法：

import re
dict(re.findall(r'(\w+) \((\w+)\),?', text))

{'english': 'fluently', 'spanish': 'poorly'}

Answer 3

虽然fferri为我的原始问题提供了一些完美的解决方案，但我在数据框架上下文中的最终解决方案更像SuperNoob的建议

我的最终解决方案：

# Create a parser function to form a dictionary of language: proficiency pairs from the values in the 'Speaks' column.
def parse_dictionary(content):
    if pd.isna(content):
        pass
    else:
        d = {}
        lps = content.split(', ')
        for lp in lps:
            if '(' not in lp:
                l = lp
                p = pd.NA
            else:
                l, p = lp.split('(')
                l = l.strip().capitalize()
                p = p.strip('()')
            d[l] = p
        return d
    
# Create a parser function to return the languages fom the dictionary in the 'Speaks' column.    
def parse_language(language, d):      
    if pd.isna(d):
        pass
    else:
        if language in d.keys():
            return True
        else:
            return False
        
# Create a parser function to return the language proficiencies fom the dictionary in the 'Speaks' column.
def parse_proficiency(language, d):   
    if pd.isna(d):
        pass
    else:
        if language in d.keys():
            return d[language]
        else:
            return pd.NA

# Parse the values in the 'Speaks' column to create a dictionary of language: proficiency pairs.
df['Speaks'] = df['Speaks'].map(lambda x: parse_dictionary(x))  

# Parse the values in the 'Speaks' column to create seperate 'language' columns with True-False values.
for language in languages:
    df['Language: {}'.format(language)] = df['Speaks'].apply(lambda d: parse_language(language, d))

# Parse the values in the 'Speaks' column to create seperate 'Language proficiency' columns with proficiency values.
for language in languages:
    df['Language proficiency: {}'.format(language)] = df['Speaks'].apply(lambda d: parse_proficiency(language, d))