返回熊猫单元中每个单词的列表，并在整列中返回该单词的总数

Question

我有一个熊猫数据框，DF，看起来像这样：

             column1
0   apple is a fruit
1        fruit sucks
2  apple tasty fruit
3   fruits what else
4      yup apple map
5   fire in the hole
6       that is true

我想产生一个列2，这是行中每个单词的列表，以及整列中每个单词的总数。因此，输出将是这样的。

    column1            column2
0   apple is a fruit   [('apple', 3),('is', 2),('a', 1),('fruit', 3)]
1        fruit sucks   [('fruit', 3),('sucks', 1)]

我尝试使用Sklearn，但无法实现上述。需要帮助。

from sklearn.feature_extraction.text import CountVectorizer
v = CountVectorizer()
x = v.fit_transform(df['text'])

Answer 1

这是一种给出您想要的结果的一种方法，尽管完全避免了sklearn：

def counts(data, column):
    full_list = []
    datr = data[column].tolist()
    total_words = " ".join(datr).split(' ')
    # per rows
    for i in range(len(datr)):
        #first per row get the words
        word_list = re.sub("[^w]", " ",  datr[i]).split()
        #cycle per word
        total_row = []
        for word in word_list:
            count = []
            count = total_words.count(word)
            val = (word, count)
            total_row.append(val)
        full_list.append(total_row)
    return full_list
df['column2'] = counts(df,'column1')
df
         column1                                    column2
0   apple is a fruit  [(apple, 3), (is, 2), (a, 1), (fruit, 3)]
1        fruit sucks                   [(fruit, 3), (sucks, 1)]
2  apple tasty fruit       [(apple, 3), (tasty, 1), (fruit, 3)]
3   fruits what else        [(fruits, 1), (what, 1), (else, 1)]
4      yup apple map           [(yup, 1), (apple, 3), (map, 1)]
5   fire in the hole  [(fire, 1), (in, 1), (the, 1), (hole, 1)]
6       that is true            [(that, 1), (is, 2), (true, 1)]

Answer 2

我不知道您是否可以使用scikit-learn进行此操作，但是您可以编写功能，然后使用apply()将其应用于DataFrame或Series。

这是您的示例的方法：

test = pd.DataFrame(['apple is a fruit', 'fruit sucks', 'apple tasty fruit'], columns = ['A'])
def a_function(row):
    splitted_row = str(row.values[0]).split()
    word_occurences = []
    for word in splitted_row:
        column_occurences = test.A.str.count(word).sum()
        word_occurences.append((word, column_occurences))
    return word_occurences
test.apply(a_function, axis = 1)
# Output
0    [(apple, 2), (is, 1), (a, 4), (fruit, 3)]
1                     [(fruit, 3), (sucks, 1)]
2         [(apple, 2), (tasty, 1), (fruit, 3)]
dtype: object

您可以看到，主要问题是test.A.str.count(word)将计算word的所有出现，而分配给word的模式在字符串中。这就是为什么"a"显示为4次的原因。这可能应该用一些言论轻松修复（我不太擅长）。

，如果您愿意丢失一些单词，可以在上面的功能中使用此解决方法：

if word not in ['a', 'is']:  # you can add here more useless words
    word_occurences.append((word, column_occurences))