Pass param thousands=',' to read_csv to read those values as thousands:

In [27]:
import pandas as pd
import io

t="""id;value
0;123,123
1;221,323,330
2;32,001"""
pd.read_csv(io.StringIO(t), thousands=r',', sep=';')

Out[27]:
   id      value
0   0     123123
1   1  221323330
2   2      32001

It seems there is problem with quoting, because if separator is , and thousands is , too, some quoting has to be in csv:

import pandas as pd
from pandas.compat import StringIO
import csv

temp=u"""'a','Base Amount'
'11','79,026,695.50'"""
#after testing replace 'StringIO(temp)' to 'filename.csv'
df = pd.read_csv(StringIO(temp), 
                 dtype = { 'Base Amount' : 'float64' },
                 thousands = ',' ,
                 quotechar = "'",
                 quoting = csv.QUOTE_ALL,
                 decimal = '.',
                 encoding = 'ISO-8859-1')

print (df)
    a  Base Amount
0  11   79026695.5

temp=u'''"a","Base Amount"
"11","79,026,695.50"'''
#after testing replace 'StringIO(temp)' to 'filename.csv'
df = pd.read_csv(StringIO(temp), 
                 dtype = { 'Base Amount' : 'float64' },
                 thousands = ',' ,
                 quotechar = '"',
                 quoting = csv.QUOTE_ALL,
                 decimal = '.',
                 encoding = 'ISO-8859-1')

print (df)
    a  Base Amount
0  11   79026695.5

Use thousands parameter.

df = pd.read_csv("file.csv",  parse_dates=['Date'], thousands=',')

I decided to change my code into the following:

df_uv = pd.read_csv(file, sep=',', parse_dates=[0, 1, 2], thousands=',').fillna(0)
df_uv = df_uv[columns_to_use]

Which is working completely fine.

As far as I can tell, if the right separators are entered, all numeric column types are inferred correctly; otherwise some columns will be strings. So to tell if you have the right separator arguments, you can use something like this:

all(df[col].dtype in ['int64', 'float64'] for col in numeric_columns)

At this point if you determine it's not right, your only option would be to try it again with switched separators. Here's kind of a contrived way to do this I came up with while testing, but it reads from a csv and doesn't capture the excel/html determination logic:

import pandas as pd

def read_data(data_file):
    numeric_columns = ['B', 'C']
    separators = {'thousands': ".", 'decimal': ","}
    df = pd.read_csv(data_file, **separators)
    if not parsed_correctly(df, numeric_columns):
        print('retry!')
        separators['thousands'], separators['decimal'] = separators['decimal'], separators['thousands']
        df = pd.read_csv(data_file, **separators)
        print(parsed_correctly(df, numeric_columns))

def parsed_correctly(df, numeric_columns):
    print('parsed_correctly', all(df[col].dtype in ['int64', 'float64'] for col in numeric_columns))
    return all(df[col].dtype in ['int64', 'float64'] for col in numeric_columns)

if __name__ == '__main__':
    files = ['csv_sep.csv', 'csv_sep2.csv']
    for f in files:
        read_data(f)
        print('\n')

If you have non-breaking spaces, I would suggest a more aggressive regular expression with str.replace:

df.col1 = df.col1.str.replace('[^\d.,e+-]', '')\
               .str.replace(',', '.').astype(float)

Regex

[       # character group
^       # negation - ignore everything in this character group
\d      # digit
.       # dot
e       # 'e' - exponent
+-      # signs 
]      

There is a thousands parameter for this. Try,

arq_pedido = pd.read_csv('Pedido.csv', delimiter=";", encoding = "ISO-8859-1", thousands=".")

You may also wish to set decimal="," to handle decimal numbers correcltly.