如果不匹配子字符串，则从列表中删除项目，无论格式如何

Question

我有以下数据框：

df = pd.DataFrame()
df['full_string'] = [['apples and bananas', 'applesandbananasamongstothers', 'something else'], 
          ['ApplesandBananas', 'apples and Bananas', 'bananas']]
df['substring'] = ['apples and bananas', 'apples and bananas']

期望的结果是保留 df['full_string'] 中的项目，其中包含在 df['substring'] 中找到的文本，同时考虑到：

大小写，大小写无关紧要
字间距
这些词可能包含与 df['substring']

期望的结果：

df['outcome'] = [['apples and bananas', 'applesandbananasamongstothers'], 
      ['ApplesandBananas', 'apples and Bananas', 'bananas']]

我尝试的是获取 df['substring'] 的第一个关键字，将其用作 df['full_string'] 的匹配器，但是，这不允许我保留'bananas' 数据框第二行中的元素。

(这在虚拟数据上效果不佳):

first_keyword = []
for i in df['substring']:
    first_keyword.append(i.split(' ', 1)[0])

df['first_keyword'] = first_keyword

df['C'] = [x[0].lower() in (x[1].lower()) for x in zip(df['first_keyword'], df['full_string'])]

Answer 1

为了简化示例，我选择使用包含您的虚拟数据的列表。你需要让它适应你的问题。此外，我将你的句子 "The desired outcome is to keep the items in df['full_string'] which contain text that is found in df['substring']" 解释为 text = word.

full_str = ['apples and bananas', 'applesandbananasamongstothers', 'something else', 
           'ApplesandBananas', 'apples and Bananas', 'bananas']
sub_str = ['apples and bananas', 'red and blue']

# Extract words from sub strings
words_in_sub = [elt.split() for elt in sub_str]
# Flatten and remove duplicates
words_in_sub = list(set([item for sublist in words_in_sub for item in sublist]))

# Init output
output = list()
# Loop on the strings in full string
for full_s in full_str:
    # Loop on the words to look for
    for word in words_in_sub:
        if word.lower() in full_s.lower():
            output.append(full_s)
            break

输出：

In: output
Out: 
['apples and bananas',
 'applesandbananasamongstothers',
 'ApplesandBananas',
 'apples and Bananas',
 'bananas']

lower/upper 情况在 if 条件中得到处理。间距由 in 语句处理。 full_s 中其他文本的存在由 in 语句处理。 in 语句 return 如果单词出现在字符串中的某处则为真。唯一会 return False 而单词可能被认为存在于字符串中的情况是如果单词被 space 分成两个，例如 'bana naan dapp les'。此示例不会保留在输出列表中。

编辑： 多行。您也可以将列表展平并使用第一个代码。

full_str = [['apples and bananas', 'applesandbananasamongstothers', 'something else'], 
            ['ApplesandBananas', 'apples and Bananas', 'bananas']]
sub_str = [['apples and bananas'], ['apples and bananas']]

# Assuming same number of rows between full_str and sub_str
# And you want to keep element of full_str[k] according to sub strings in sub_str[k]
number_of_rows = len(full_str)
for k in range(number_of_rows):
    # Extract words from sub strings
    words_in_sub = [elt.split() for elt in sub_str[k]]
    # Flatten and remove duplicates
    words_in_sub = list(set([item for sublist in words_in_sub for item in sublist]))

    # Init output
    output = list()
    # Loop on the strings in full string
    for full_s in full_str[k]:
        # Loop on the words to look for
        for word in words_in_sub:
            if word.lower() in full_s.lower():
                output.append(full_s)
                break

如果不匹配子字符串，则从列表中删除项目，无论格式如何

remove item from list if it does not match substring, regardless of the formatting

python

string

substring

matching