Python 将每一行与列表字典进行比较，并将新变量附加到数据帧

我想检查pandas dataframe string列的每一行，并附加一个新列，如果在列表字典中找到文本列的任何元素，该列将返回1

例如：

# Data
df = pd.DataFrame({'id': [1, 2, 3],
                   'text': ['This sentence may contain reference.',
                'Orange, blue cow','Does the cow operate any heavy machinery?']},
                 columns=['numbers', 'text'])

# Rule dictionary
rule_dict = {'rule1': ['Does', 'the'],
             'rule2':['Sentence','contain'],
             'rule3': ['any', 'reference', 'words']}

# List of variable names to be appended to df
rule_list = ['has_rule1','has_rule2','has_rule3']

# Current for loop
for Key in rule_dict:
    for i in rule_list:
        df[i] = df.text.apply(lambda x: (
            1 if any(ele in x for ele in rule_dict[Key]) == 1 and (len(str(x)) >= 3) 
            else 0))

# Current output, looks to be returning a 1 if text is found in ANY of the lists
df = pd.DataFrame({'id': [1, 2, 3],
                       'text': ['This sentence may contain reference.',
                    'Orange, blue cow','Does the cow operate any heavy machinery?'],
                    'has_rule1': [1,1,1],
                    'has_rule2': [0,0,0],
                    'has_rule3': [1,1,1]},
                     columns=['id', 'text','has_rule1','has_rule2','has_rule3'])

# Anticipated output
df = pd.DataFrame({'id': [1, 2, 3],
                       'text': ['This sentence may contain reference.',
                    'Orange, blue cow','Does the cow operate any heavy machinery?'],
                    'has_rule1': [0,0,1],
                    'has_rule2': [1,0,0],
                    'has_rule3': [1,0,1]},
                     columns=['id', 'text','has_rule1','has_rule2','has_rule3'])

---

假设您已经解决了评论中提到的有关dict理解的问题，您不应该使用嵌套的

for

循环。相反，使用单个

for

循环和

zip

：

for (k,v), n in zip(rule_dict.items(), rule_list):
    pat = rf'\b{"|".join(v)}\b'
    df[n] = df.text.str.contains(pat).astype(int)

输出：

      id  text                                         has_rule1    has_rule2    has_rule3
--  ----  -----------------------------------------  -----------  -----------  -----------
 0     1  This sentence may contain reference.                 0            1            1
 1     2  Orange, blue cow                                     0            0            0
 2     3  Does the cow operate any heavy machinery?            1            0            1

---

请注意，当您为输入规则dict执行

时，无法保证规则dict的排序。也就是说，您不知道
键
将按
“规则1”、“规则2”、“规则3”或
“规则2”、“规则3”、“规则1”等顺序出现。@Quanghaang我不知道，谢谢。我以为它是基于索引的。如何确保它保持我指定的顺序（即“rule1”、“rule2”、“rule3”）？一种方法是将
规则列表
也作为字典：
{'rule1'：'has_rule1'，}
并通过键访问这两个规则。另一种方法是使用。但这也取决于规则的名称。