Extract year contents from html code and save them as dataframe - python-3.x

Question

Given a section of html source code named li as follows:
[<li>Project construction cycle</li>,
<li>
Start date: 2019...
Completion date: 2021... <a class="login-btn" href="javascript:">Click to view details</a>
</li>,
<li>Preliminary preparation progress</li>,
<li>
The project has been completed by... <a class="login-btn" href="javascript:">Click to view details</a>
</li>,
<li>Progress in design work</li>,
<li>
The project design has... <a class="login-btn" href="javascript:">Click to view details</a>
</li>,
<li>Procurement of equipment</li>,
<li>
The project equipment... <a class="login-btn" href="javascript:">Click to view details</a>
</li>,
<li>Project construction progress</li>,
<li>
The project is in... <a class="login-btn" href="javascript:">Click to view details</a>
</li>]
How could we extract Start date and Completion date and convert them to a dataframe?
PS: I convert it to dataframe because I need to concatenate it with other columns.
The expected result:
Start date Completion date
0 2019 2021
Thanks.
Updates:
li = str(li)
s = re.compile('Start date：[0-9]{4}').findall(li)
df1 = pd.DataFrame([x.split('：')for x in s ]).set_index(0).T
e = re.compile('Completion date：[0-9]{4}').findall(li)
df2 = pd.DataFrame([x.split('：')for x in e ]).set_index(0).T
# df = pd.concat([df1, df2], axis = 1)
New update:
rmktxt2 = soup.find("table", attrs={"id":"mse_new"}).find("ul", attrs={"class":"rmktxt2"})
li = rmktxt2.find_all("li")
li = str(li)
li = " ".join(li.split())
regex = r"(Start date：\d{4}|Completion date：\d{4})"
data = re.findall(regex, li)
df = pd.DataFrame([x.split('：')for x in data]).set_index(0).T
print(df)
Out:
0 Start date Completion date
1 2019 2021
Now how can I set index 0 starting from row of 2019 2021?
Updates:
regex = r"Start date：(\d{4}).*Completion date：(\d{4})"
data = re.findall(regex, li)[0]
out['Start date'] = data[0]
out['Completion date'] = data[1]
df = pd.DataFrame([out])
Out:
Start date Completion date
0 2019 2021

Answer 1

You may try:
(Start date: \d{4}|Completion date: \d{4})
Explanation of the above regex:
(Start date: \d{4}) - Represents first capturing group matching Start date: literally along with digits appearing exactly 4 times.
| - Represents alternation.
Completion date: \d{4}) - Matches Completion date: literally along with digits appearing exactly 4 times.
You can find the demo of the above regex in here.
Code Demo