DataFrame列中的标点符号

Question

我正在尝试在此DataFrame的content列中计算标点符号。我已经尝试过this，但是它不起作用。我的DataFrame看起来像这样：

我希望结果是这样的：而不是用情绪来计算每篇文章的标点符号。

In:

text_words = df.content.str.split()
punctuation_count = {}
punctuation_count[','] = 0
punctuation_count[';'] = 0
punctuation_count["'"] = 0
punctuation_count['-'] = 0

def search_for_single_quotes(word):
    single_quote = "'"
    search_char_index = word.find(single_quote)
    search_char_count = word.count(single_quote)
    if search_char_index == -1 and search_char_count != 1:
        return
    index_before = search_char_index - 1
    index_after = search_char_index + 1

    if index_before >= 0 and word[index_before].isalpha() and index_after == len(word) - 1 and word[index_after].isalpha():
        punctuation_count[single_quote] += 1

for word in text_words:
    for search_char in [',', ';']:
        search_char_count = word.count(search_char)
        punctuation_count[search_char] += search_char_count
    search_for_single_quotes(word)
    search_for_hyphens(word)

Out:
AttributeError: 'list' object has no attribute 'find'

Answer 1

提供以下输入：

df = pd.DataFrame(['I love, pizza, hamberget and chips!!.', 'I like drink beer,, cofee and water!.'], columns=['content'])

                                content
0   I love, pizza, hamberget and chips!!.
1   I like drink beer,, cofee and water!.

尝试此代码：

count = lambda l1,l2: sum([1 for x in l1 if x in l2])

df['count_punct'] = df.content.apply(lambda s: count(s, string.punctuation))

并给出：

                                 content  count_punct
0  I love, pizza, hamberget and chips!!.            5
1  I like drink beer,, cofee and water!.            4

如果要累积列表中每一行的标点符号：

accumulate = lambda l1,l2: [x for x in l1 if x in l2]

df['acc_punct_list'] = df.content.apply(lambda s: accumulate(s, string.punctuation))

并给出：

                                 content  count_punct   acc_punct_list
0  I love, pizza, hamberget and chips!!.            5  [,, ,, !, !, .]
1  I like drink beer,, cofee and water!.            4     [,, ,, !, .]

如果要在字典中累积每行的标点符号，并将每个元素转置为数据框列：

df['acc_punct_dict'] = df.content.apply(lambda s: {k:v for k, v in Counter(s).items() if k in string.punctuation})

                                 content            acc_punct_dict
0  I love, pizza, hamberget and chips!!.  {',': 2, '!': 2, '.': 1}
1  I like drink beer,, cofee and water!.  {',': 2, '!': 1, '.': 1}

现在在df的列中扩展字典：

df_punct = df.acc_punct_dict.apply(pd.Series)

   ,  !  .
0  2  2  1
1  2  1  1

如果您想将新数据框与起始数据框结合在一起，则只需要做：

df_res = pd.concat([df, df_punct], axis=1)

并给出：

                                 content            acc_punct_dict  ,  !  .
0  I love, pizza, hamberget and chips!!.  {',': 2, '!': 2, '.': 1}  2  2  1
1  I like drink beer,, cofee and water!.  {',': 2, '!': 1, '.': 1}  2  1  1

注：如果您不关心字典中的列，则可以通过df_res.drop('acc_punct_dict', axis=1)]将该列删除