从第一个元素中获取某个字符串 - 数组答案

【问题标题】：Get a certain string from the first elements - array从第一个元素中获取某个字符串 - 数组
【发布时间】：2021-04-07 18:51:38
【问题描述】：

给定一个数组： arr = ['Get your face right in there.', 'You&#8217;re like, wait, though, isn&#8217;t that too close? PUH-LEEZ. You and I both know that a person can never be too close to those browned lasagna edges that are the perfect amount of chewy, and that&#8217;s exactly why we like to hang out together and get wild about all the foods. That&#8217;s this lasagna florentine.']

我想从数组中获取一个字符串，其中包含总共 15 个或更多单词的元素。因此，由于第一个元素没有 15 个单词，输出将转到下一个元素，并将输出整个数组（在这种情况下）。前任。如果第一个元素有 12 个单词，第二个元素有 20 个单词，则输出字符串将有 12+20 = 32 个单词。

这是我的代码：

def CleanText(text):
    text = str(text)
    forbidden = [r'\n', r'\t', r'.', r'?', r'!', r'(', r')', r'/', r'<', r'>', '\\', '\\r', '</div>', '</a>']
    for i in forbidden:
        text.replace(i, '')
    return text

def ExtractText(arr):
    for i in arr:
        #return i
        if len(i) >= 15:
            return CleanText(i)
        elif len(arr[i]) < 15:
            attachedString = CleanText(arr[i+1])
            return CleanText(arr[i]) + attachedString

arr = ['Get your face right in there.', 'You&#8217;re like, wait, though, isn&#8217;t that too close? PUH-LEEZ. You and I both know that a person can never be too close to those browned lasagna edges that are the perfect amount of chewy, and that&#8217;s exactly why we like to hang out together and get wild about all the foods. That&#8217;s this lasagna florentine.']
print(ExtractFirstPara(arr))

输出：Get your face right in there. 但是，当我不概括它时，它就可以了。也就是说，当我明确地说：

def ExtractText(arr):
    for i in arr[0]:
        if len(i) >= 15:
            return CleanText(arr[0])
        else:
            attachedString = CleanText(arr[1])
            return CleanText(arr[0]) + attachedString

输出：

Get your face right in there.You&#8217;re like, wait, though, isn&#8217;t that too close? PUH-LEEZ. You and I both know that a person can never be too close to those browned lasagna edges that are the perfect amount of chewy, and that&#8217;s exactly why we like to hang out together and get wild about all the foods. That&#8217;s this lasagna florentine.

但是，在某些情况下，如果字符串与前两个元素的长度加起来不等于 15，那么这种显式性（如果它甚至是一个词）是不可能的。为此，我需要概括一下。

【问题讨论】：

标签： python arrays string loops

【解决方案1】：

如果我理解正确，没有部分元素输出：如果数组的第一个 k 元素的总字数至少为 nmin=15，则返回。

如果这是正确的，并且您对找到与 r'(\w+)' 匹配的单词感到满意，那么以下操作会：

def get_elements(arr, nmin = 15):
    p = re.compile(r'(\w+)')
    words_per_element = [len(p.findall(s)) for s in arr]
    ntot = 0
    for i, n in enumerate(words_per_element):
        ntot += n
        if ntot >= nmin:
            return arr[:i+1]
    return []

测试：

>>> get_elements(['a'], 1)
['a']

>>> get_elements(['a'], 2)
[]

>>> get_elements(['a', 'b c'], 1)
['a']

>>> get_elements(['a', 'b c'], 2)
['a', 'b c']

>>> get_elements(['a', 'b c'], 4)
[]

>>> get_elements(['a', 'b c', 'd e f'], 2)
['a', 'b c']

【讨论】：

【解决方案2】：

在这里试试这个：

def CleanText(text):
    text = str(text)
    forbidden = [r'\n', r'\t', r'.', r'?', r'!', r'(', r')', r'/', r'<', r'>', '\\', '\\r', '</div>', '</a>']
    for i in forbidden:
        text.replace(i, '')
    return text
    
def ExtractText(arr):
    string = ''
    for i in arr:
        string += i
        if len(string.split()) >= 15:
            return string

【讨论】：