>>> import re
>>> re.sub("[^\d\.]", "", "$1,000,000.01")
'1000000.01'

Regular expression pattern says "anything that isn't a number or a decimal point". Anything matching that regex is replaced with "".

You may need to bear in mind that some European countries use . as the thousand seperator and , as the decimal mark.

So "one million euros and 5 cents" could be €1.000.000,05 or €1,000,000.05

Using @uingtea regex, but for pandas dataframe, you can do in this way:

import pandas as pd
import re

df = pd.DataFrame({'col':['Hello, world!', 'Warhammer 40,000', 'Codename 1,337']})
df['col'] = df['col'].apply(lambda x: re.sub(r'(\d+),(\d+)', r'\1\2', x))

First, make a function that can convert a single string element to a float:

valid = '1234567890.' #valid characters for a float
def sanitize(data):
    return float(''.join(filter(lambda char: char in valid, data)))

Then use the apply method to apply that function to every entry in the column. Reassign to the same column if you want to overwrite your old data.

df['column'] = df['column'].apply(sanitize)

Checkout https://docs.python.org/2/library/string.html or https://docs.python.org/3/library/string.html depending on the Python version you are using and use the "replace()" function:

int_price = int(price_result.replace(',',''))

This replaces all commas within the string and then casts it to an INT:

>>> price = "1,000,000"
>>> type(price)
<type 'str'>
>>> int_price = int(price.replace(',',''))
>>> type(int_price)
<type 'int'>
>>> 

Does your data look something like:

"123", "123,456", "123,456,789"

If so then try this

input = '"123", "123,456", "123,456,789"'

import re

reg = re.compile('"(\d{1,3}(,\d{3})*)"')

stringValues = [wholematch.replace(',', '') for wholematch, _endmatch 
                                                    in reg.findall(input)]

This regex should also work on thousands with decimal places as well.

re.compile('"(\d{1,3}(,\d{3})*(\.\d*)?)"')