如何计算熊猫数据框中最常见的 500 个单词

Question

我有一个数据框，在名为 Text 的列中有 500 个文本（每行 1 个文本），我想计算所有文本中最常见的单词。

到目前为止我尝试过（两种方法都来自 stackoverflow）：

pd.Series(' '.join(df['Text']).lower().split()).value_counts()[:100]

和

Counter(" ".join(df["Text"]).split()).most_common(100)

两者都给了我以下错误：

TypeError：序列项 0：预期的 str 实例，找到列表

我已经简单地尝试了计数器方法

df.Text.apply(Counter()) 

这给了我每个文本的字数 我还修改了 counter 方法，使其返回每个文本中最常见的单词

但我想要总体上最常用的词

这是数据框的示例（文本已经小写，标点符号已清除，标记化，停用词已删除）

    Datum   File    File_type                                         Text                         length    len_cleaned_text
Datum                                                   
2000-01-27  2000-01-27  _04.txt     _04     [business, date, jan, heineken, starts, integr...       396         220

编辑：“报告”它的代码

  for file in file_list:
    name = file[len(input_path):]
        date = name[11:17]
        type_1 = name[17:20] 


with open(file, "r", encoding="utf-8", errors="surrogateescape") as rfile:
                format
                text = rfile.read()
                text = text.encode('utf-8', 'ignore')
                text = text.decode('utf-8', 'ignore')
     a={"File": name, "Text": text,'the':count_the, 'Datum': date, 'File_type': type_1, 'length':length,}
        result_list.append(a)

新单元格

  df['Text']= df['Text'].str.lower()
    p = re.compile(r'[^\w\s]+')
    d = re.compile(r'\d+')
    for index, row in df.iterrows():
        df['Text']=df['Text'].str.replace('\n',' ')
        df['Text']=df['Text'].str.replace('################################ end of story 1 ##############################','')
        df['Text'] = [p.sub('', x) for x in df['Text'].tolist()]
        df['Text'] = [d.sub('', x) for x in df['Text'].tolist()]
    df['Text']=df['Text'].apply(word_tokenize)


    Datum   File    File_type   Text    length  the
Datum                       
2000-01-27  2000-01-27  0864820040_000127_04.txt    _04     [business, date, jan, heineken, starts, integr...   396     0
2000-02-01  2000-02-01  0910068040_000201_04.txt    _04     [group, english, cns, date, feb, bat, acquisit...   305     0
2000-05-03  2000-05-03  1070448040_000503_04.txt    _04     [date, may, cobham, plc, cob, acquisitionsdisp...   701     0
2000-05-11  2000-05-11  0865985020_000511_04.txt    _04     [business, date, may, swedish, match, complete...   439     0
2000-11-28  2000-11-28  1067252020_001128_04.txt    _04     [date, nov, intec, telecom, sys, itl, doc, pla...   158     0
2000-12-18  2000-12-18  1963867040_001218_04.txt    _04     [associated, press, apw, date, dec, volvo, div...   367     0
2000-12-19  2000-12-19  1065767020_001219_04.txt    _04     [date, dec, spirent, plc, spt, acquisition, co...   414     0
2000-12-21  2000-12-21  1076829040_001221_04.txt    _04     [bloomberg, news, bn, date, dec, eni, ceo, cfo...   271     0
2001-02-06  2001-02-06  1084749020_010206_04.txt    _04     [date, feb, chemring, group, plc, chg, acquisi...   130     0
2001-02-15  2001-02-15  1063497040_010215_04.txt    _04     [date, feb, electrolux, ab, elxb, acquisition,...   420     0

以及数据框的描述：

<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 557 entries, 2000-01-27 to 2017-10-06
Data columns (total 13 columns):
Datum               557 non-null datetime64[ns]
File                557 non-null object
File_type           557 non-null object
Text                557 non-null object
customers           557 non-null int64
grwoth              557 non-null int64
human               557 non-null int64
intagibles          557 non-null int64
length              557 non-null int64
synergies           557 non-null int64
technology          557 non-null int64
the                 557 non-null int64
len_cleaned_text    557 non-null int64
dtypes: datetime64[ns](1), int64(9), object(3)
memory usage: 60.9+ KB

提前致谢

Answer 1

您可以通过apply 和Counter.update 方法做到这一点：

from collections import Counter

counter = Counter()
df = pd.DataFrame({'Text': values})
_ = df['Text'].apply(lambda x: counter.update(x))

counter.most_common(10) 
Out:

[('Amy', 3), ('was', 3), ('hated', 2),
 ('Kamal', 2), ('her', 2), ('and', 2), 
 ('she', 2), ('She', 2), ('sent', 2), ('text', 2)]

df['Text'] 在哪里：

0    [Amy, normally, hated, Monday, mornings, but, ...
1    [Kamal, was, in, her, art, class, and, she, li...
2    [She, was, waiting, outside, the, classroom, w...
3              [Hi, Amy, Your, mum, sent, me, a, text]
4                         [You, forgot, your, inhaler]
5    [Why, don’t, you, turn, your, phone, on, Amy, ...
6    [She, never, sent, text, messages, and, she, h...
Name: Text, dtype: object

Answer 2

好的，我明白了。您的df['Text'] 包含文本列表。所以你可以这样做：

full_list = []  # list containing all words of all texts
for elmnt in df['Text']:  # loop over lists in df
    full_list += elmnt  # append elements of lists to full list

val_counts = pd.Series(full_list).value_counts()  # make temporary Series to count

此解决方案避免使用过多的列表推导，从而使代码易于阅读和理解。此外，不需要像 re 或 collections 这样的额外模块。

Answer 3

这是我的版本，我将列值转换为一个列表，然后我制作一个单词列表，清理它，你就有了你的计数器：

your_text_list = df['Text'].tolist()
your_text_list_nan_rm = [x for x in your_text_list if str(x) != 'nan']
flat_list = [inner for item in your_text_list_nan_rm for inner in ast.literal_eval(item)] 

counter = collections.Counter(flat_list)
top_words = counter.most_common(100)