计算 DataFrame 列中的标点符号

Question

我正在尝试计算此 DataFrame 的 content 列中的标点符号。我已经尝试 this 但它不起作用。我的 DataFrame 如下所示：

我希望结果是这样的：而是用情绪来统计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

给定以下输入：

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!.

试试这个代码：

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

如果要累加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     [,, ,, !, .]

如果你想累积字典中每一行的标点符号并将每个元素转置到数据框列中：

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

如果你想将新数据框与起始数据框合并，你只需要做：

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

注意：如果你不在意有字典的那一列可以通过df_res.drop('acc_punct_dict', axis=1)

删除

计算 DataFrame 列中的标点符号

Count punctuation in a DataFrame column

python

pandas

punctuation