如何在函数中创建并返回数据框？

Question

我有一个函数 pmi_count_phrase_create() ，我在其中输入 2 个列表，想要返回一个由我在函数中创建的列表组成的数据帧。现在的问题是我得到一个包含三个列表的元组。

这是我的代码：

def pmi_count_phrase_create(pmi_tups,freq_list):
    import pandas as pd

    """pmi_tups is result of running pmi_tups = [i for i in finder.score_ngrams(bigram_measures.pmi)] 
       freq_list is a result of running freq_list= finder.ngram_fd.items()
       
       -> 1 df made up of columns for  pmi list, count list, phrase list"""
    pmi3_list =[]
    count3_list =[]
    phrase3_list =[]
    for phrase, pmi in pmi_tups:
        for item in freq_list:
            quadgram,count = item
            if quadgram == phrase:
                pmi3_list.append(pmi)
                count3_list.append(count)
                phrase3_list.append(phrase)

                # create dataframe
    df = pd.DataFrame({'Phrase':phrase3_list,'PMI':pmi3_list,'Count':count3_list})
    return df

我当前的输出是以下形式的元组：(pmi3_list,count3_list,phrase3_list)

我想要的是一个数据框：

    Phrase                       PMI    Count
0   (activated, charcoal)     15.213655 12
1   (vocal, cords)            14.950620 10
2   (fossil, fuels)           14.872618 15
3   (simplified, explanation) 14.669334 12
4   (midnight, weekend)   14.181233 11

Answer 1

也许可以尝试：

def pmi_count_phrase_create(pmi_tups,freq_list):
    import pandas as pd

    """pmi_tups is result of running pmi_tups = [i for i in 
finder.score_ngrams(bigram_measures.pmi)] 
   freq_list is a result of running freq_list= finder.ngram_fd.items()
   
   -> 1 df made up of columns for  pmi list, count list, phrase list"""
    pmi3_list =[]
    count3_list =[]
    phrase3_list =[]
    for phrase, pmi in pmi_tups:
        for item in freq_list:
            quadgram,count = item
            if quadgram == phrase:
                pmi3_list.append(pmi)
                count3_list.append(count)
                phrase3_list.append(phrase)

            # create dataframe
    df = pd.DataFrame.from_dict({'Phrase':phrase3_list,'PMI':pmi3_list,'Count':count3_list})
    return df

在哪里使用

pd.Dataframe.from_dict

而不是

pd.Dataframe

。