'Columns must be same length as key' error when trying .Split

The code below just runs fine with Python 3.8.10 but does not run in Python 3.10. Any idea what could be the problem?

import pandas as pd
import requests

url = "https://coinmarketcap.com/new/"
page = requests.get(url,headers={'User-Agent': 'Mozilla/5.0'}, timeout=1)
pagedata = page.text
usecols = ["Name", "Symbol", "1h", "24h", "MarketCap"]


df = pd.read_html(page.text)[0]
df[["Name", "Symbol"]] = df["Name"].str.split(r"\d+", expand=True)

df = (df.rename(columns={"Fully Diluted Market Cap": "MarketCap"})[usecols]
          .sort_values("24h", ascending=False, key=lambda ser: ser.str.replace("%", "").astype(float))
          .replace(r"^$", "", regex=True)
     )

numcols = df.columns[~df.columns.isin(['Name'])]
df = df.head(5).to_markdown(index=True)
print (df)

Current Output:

Traceback (most recent call last):
  df[["Name", "Symbol"]] = df["Name"].str.split(r"\d+", expand=True)
  ....
  ....
  ValueError: Columns must be same length as key

Correct Output: (Output in Python 3.8)

|    | Name        | Symbol    | 1h     | 24h      | MarketCap   |
|---:|:------------|:----------|:-------|:---------|:------------|
|  3 | Shrekt      |4HREK      | 23.82% | 2536.51% | 342,357     |
|  8 | BLAZE       |TOKEN9BLZE | 1.07%  | 106.71%  | 3,828,088   |
| 26 | Goner27     |GONER      | 6.32%  | 88.09%   | 1,094,010   |
| 14 | Party Hat15 |PHAT       | 13.34% | 81.64%   | 60,136      |
| 29 | PepeChat    |30PPC      | 48.01% | 78.25%   | 431,159     |

I think it has to do with one of the values (NOOT (BRC-20)4NOOT) found in the column Name.

To handle this, we can try to split on the last number found in each row of this column.

Replace this :

df[["Name", "Symbol"]] = df["Name"].str.split(r"\d+", expand=True)

By this :

df[["Name", "Symbol"]] = df["Name"].str.split(r"\d+(?!.*\d)", expand=True)

> Regex [demo]

Output :

print(df)

|    | Name        | Symbol   | 1h     | 24h      | MarketCap   |
|---:|:------------|:---------|:-------|:---------|:------------|
|  5 | Shrekt      | HREK     | 54.61% | 1124.57% | 159,013     |
| 10 | BLAZE TOKEN | BLZE     | 2.40%  | 109.53%  | 3,880,242   |
|  8 | CMC DOGE    | CMCDOGE  | 12.93% | 102.76%  | 169,492     |
| 28 | Goner       | GONER    | 1.37%  | 88.66%   | 1,050,089   |
|  4 | nomeme      | NOMEME   | 53.86% | 86.14%   | 4,603,393   |

Do you have to use keep using df as dataframe. I think you can create a new one and use that one after the function. Try just doing this

newDF = df["Name"].str.split(r"\d+", expand=True)
print(newDF)

Edit Fixed Code:

df["Name"] = df["Name"].str.replace("\(BRC-20\)","")

add this line to your code which will replace anything that has (BRC-20) in it.
So the problem wasn't being about the version of your python.

I think you need to split the name by the value of columns '#'.
You can do it like this:

Create a function for splitting:

def splitting(num, strng):
    splitted = strng.split(str(num))
    print (num)
    return [splitted[0], splitted[1]]

Then apply the function and explode the new column into new columns:

df["split"] = df.apply(lambda x: splitting(x['#'], x['Name']), axis=1)
df[['OnlyName','Symbol']] = pd.DataFrame(df.split.tolist(), index= df.index)

Let me know if this helps.