連接除 list_pandas 中的字符串之外的兩列

Question

對於列“單詞”中的字符串不存在於“句子”列中的情況，我將兩列連接起來。 我的代碼是：

def check(row):
     df['sentence'] = df['sentence'].astype(str)
     df['words'] = df['words'].astype(str)
     left = row['sentence'].split()
     right = row['words'].split()
     unmatched = []
     for y in left:
        word = "".join([x for x in y.lower() if x not in string.punctuation])
        if word not in [v.lower() for v in right]:
           unmatched.append(y)
     return " ".join(unmatched)
mask = df['type'] == 'Is there a match with the Words?'
df.loc[mask, 'new'] = df.loc[mask, :].apply(check, axis=1)
df['new'] = np.where(c, df['new'] + ' ' + df['words'], df['new'])
df['new'] = df['new'].str.replace('nan', '')
df['new'] = df['new'].fillna("")

此外，如果在“單詞”列中我有此列表中存在的字符串，我想限制每行的串聯：

restricted = ['not present', 'for sale', 'unknown']

這是一個結果應該如何的示例

      words             sentence                    output
0   unknown  This is a new paint       This is a new paint
1     brown   This is a new item  This is a new item brown
2  for sale   The product is new        The product is new

上面代碼給出的 Output 是：

output
 This is a new paint unknown
 This is a new item brown
 The product is new for sale

Answer 1

鑒於：

      words             sentence
0   unknown  This is a new paint
1     brown   This is a new item
2  for sale   The product is new

正在做：

restricted = ['not present', 'for sale', 'unknown']
mask = df.words.str.contains('|'.join(restricted))
df['output'] = df.sentence.where(mask, df.sentence + ' ' + df.words)
print(df)

Output：

      words             sentence                    output
0   unknown  This is a new paint       This is a new paint
1     brown   This is a new item  This is a new item brown
2  for sale   The product is new        The product is new