Question

我有一个单词列表，我想做if语句，下面是我的列表：

list = ['camera','display','price','memory'(will have 200+ words in the list)]

这是我的代码：

def check_it(sentences):
    if 'camera' in sentences and 'display' in sentences and 'price' in sentences:
        return "Camera/Display/Price"
    if 'camera' in sentences and 'display' in sentences:
        return "Camera/Display"
    ...
    return "Others"

h.loc[:, 'Category'] = h.Mention.apply(check_it)

这些组合将会太多，我也想让单词单独返回行。有谁知道如何制作此示例并单独返回单词，而不是执行“相机/显示器/价格”操作？

Answer 1

通过正则表达式使用str.findall-将列表的所有值与|联接，最后str.join个值由/联接：

df = pd.DataFrame({'Mention':['camera in sentences and display in sentences',
                              'camera in sentences price']})


L = ['camera','display','price','memory']
pat = '|'.join(r"\b{}\b".format(x) for x in L)
df['Category'] = df['Mention'].str.findall(pat).str.join('/')
print (df)
                                        Mention        Category
0  camera in sentences and display in sentences  camera/display
1                     camera in sentences price    camera/price

具有列表理解的另一种解决方案，也适用于join的列表生成器：

df['Category1'] = [[y for y in x.split() if y in L] for x in df['Mention']]
df['Category2'] = ['/'.join(y for y in x.split() if y in L) for x in df['Mention']]
print (df)
                                        Mention          Category1  \
0  camera in sentences and display in sentences  [camera, display]   
1                     camera in sentences price    [camera, price]   

        Category2  
0  camera/display  
1    camera/price

Answer 2

some_words = ['camera','display','price','memory']
def check_it(sentences, words):
   find_words = []
   for word in words:
      if word in sentences:
         find_words.append(word)
   return find_words
t = check_it('display has camera and price is', some_words)
print t

Answer 3

为什么不只检查每个句子中的单词？

wordsList = ['camera','display','price','memory'(will have 200+ words in the list)]

def check_it(sentence, wordsList):
    wordString = ''
    flag = False
    counter = 0
    for word in sentence.split():
        if word in wordsList:
            if counter != 0:
                wordString = wordString + '/' + word
            else:
                wordString = word
            flag = True
            counter += 1
    if flag:
        return wordString
    elif not flag:
        return 'Others'

python如果多个字符串返回句子中包含的单词

3 个答案: