从段落中获取单词的最大长度

Question

我正在处理一个文本问题，我的 pandas 数据框包含许多列，其中一个由段落组成。我在输出中需要的是定义的 3 列 -

• 最大单词的长度
• 最大单词的数量（如果有任何相似的长度）
• 此类长度相似的单词的总数。

如果一个单词用空格隔开，我就说明它。使用pythonapply-map寻找答案。

这是一个示例输入数据 -

df = pd.DataFrame({'text':[
    "that's not where the biggest opportunity is - it's with heart failure drug - very very huge market....",
    "Of course! I just got diagnosed with congestive heart failure and type 2 diabetes. I smoked for 12 years and ate like crap for about the same time. I quit smoking and have been on a diet for a few weeks now. Let me assure you that I'd rather have a coke, gummi bears, and a bag of cheez doodles than a pack of cigs right now. Addiction is addiction.",
    "STILLWATER, Okla. (AP) ? Medical examiner spokeswoman SpokesWoman: Oklahoma State player Tyrek Coger died of enlarged heart, manner of death ruled natural."
]})

df

    text                                                
0   that's not where the biggest opportunity is - ...   
1   Of course! I just got diagnosed with congestiv...   
2   STILLWATER, Okla. (AP) ? Medical examiner spok...   

这是预期的输出 -

    text                                               word_count   word_length     words
0   that's not where the biggest opportunity is - ...   1           11             opportunity
1   Of course! I just got diagnosed with congestiv...   1           10              congestive
2   STILLWATER, Okla. (AP) ? Medical examiner spok...   2           11              spokeswoman SpokesWoman

Answer 1

使用apply-map 的一种可能解决方案-

import nltk
import pandas as pd

# Reading df and proceeding with code

expanded_text = df.text.apply(lambda x: ' '.join(nltk.word_tokenize(x))).str.split(" ", expand=True)

df.word_length = expanded_text.applymap(lambda x: len(str(x)) if x != None else 0).max(axis=1)

i = 1
for idx, val in enumerate(expanded_text.itertuples()):
    temp = expanded_text.iloc[idx:idx + i, :].applymap(lambda x: True if len(str(x)) == df.loc[idx, 'word_length'] else False if x != None else False).T
    idx_ = temp.index[temp[idx] == True].values 
    words = " ".join(expanded_text.iloc[idx:idx + i, idx_].values.tolist()[0])
    df.loc[idx, 'words'] = words
    df.loc[idx, 'word_count'] = len(words.split())
    i += 1    

Answer 2

下面的代码应该可以解决问题：

def get_values(text):
    tokens = text.split() # Splitting by whitespace
    max_word_length = -1
    list_words = [] # Initializing list of max length words

    for token in tokens:
        if len(token) > max_word_length:
           max_word_length = len(token)
           list_words = [] # Clearning the list, since there's a new max
           list_words.append(token)
        elif len(token) == max_word_length:
           list_words.append(token)

     words_string = ' '.join(list_words) if len(list_words) > 1 else list_words[0] # Concatenating list into string

     return [len(list_words), max_word_length, list_words]

df['word_count'], df['word_length'], df['words'] = zip(*df['text'].map(get_values))

编辑：忘记连接列表