Python 解析时获取元素标记内的内容

鉴于我的代码：

browser.get(s_page_url)
soup = BeautifulSoup(browser.page_source, "html.parser")
s_image_element = soup.find('a', {'id': 'angle-3'})
s_image_href = s_image_element['href']
s_image_url = "http://www.zappos.com" + s_image_href
s_title_element = soup.find('h1', {'class': 'banner'})
print s_title_element

当前生成以下输出：

<h1 class="banner">
<a href="http://couture.zappos.com/a-testoni">a. testoni</a>
<meta content="a. testoni" itemtype="brand">
<a href="/p/a-testoni-sport-nappa-calf-sneaker/product/8835012"><span class="ProductName" itemprop="name">Sport Nappa Calf Sneaker</span></a>
</meta></h1>

但会收到以下错误消息：

  File "C:\Python27\lib\site-packages\bs4\element.py", line 958, in __getitem__
    return self.attrs[key]
KeyError: 'a'

get_text

将连接标记对象中的所有文本

strip=True

将删除开头和结尾的空白

def get_text(self, separator="", strip=False,
             types=(NavigableString, CData)):
    """
    Get all child strings, concatenated using the given separator.

---

谢谢，我如何在单词之间留出一个空格，因为我得到的是

a。testoniSport Nappa Calf运动鞋moment@methuselah这是源代码，重要的是参数。
print s_title_element.get_text(strip=True,separator=" " )

def get_text(self, separator="", strip=False,
             types=(NavigableString, CData)):
    """
    Get all child strings, concatenated using the given separator.