You could use Series.str.replace:

import pandas as pd

df = pd.DataFrame(['$40,000*','$40000 conditions attached'], columns=['P'])
print(df)
#                             P
# 0                    $40,000*
# 1  $40000 conditions attached

df['P'] = df['P'].str.replace(r'\D+', '', regex=True).astype('int')
print(df)

yields

       P
0  40000
1  40000

since \D matches any character that is not a decimal digit.

Your regex is matching on all - characters:

In [48]:
df_raw.replace(['-','\*'], ['0.00','0.00'], regex=True)

Out[48]:
       A          B
0   1.00        1.0
1  0.001  0.0045.00
2    NaN       0.00

If you put additional boundaries so that it only matches that single character with a termination then it works as expected:

In [47]:
df_raw.replace(['^-$'], ['0.00'], regex=True)

Out[47]:
      A       B
0  1.00     1.0
1    -1  -45.00
2   NaN    0.00

Here ^ means start of string and $ means end of string so it will only match on that single character.

Or you can just use replace which will only match on exact matches:

In [29]:

df_raw.replace('-',0)
Out[29]:
      A       B
0  1.00     1.0
1    -1  -45.00
2   NaN       0

dfx.Brands.apply(lambda x: re.sub(r".*(asus|acer).*", lambda m: m.group(1).upper(), x, flags=re.IGNORECASE))

First, you have the wrong regex's in the wrong positions. The to_replace argument to .replace needs to match what to replace and what to delete. So you need a ^.* in front of and a .*$ behind your regex in this case since you want to trim the string outside the match:

^.*([A-Z]{2}[0-9]{3}_[0-9]{3}).*$

Demo

Second, the replace argument, if a regex, needs to be a capturing group or fixed string. In this case \1 will do.

Last, the Series form of .replace has a littler simpler syntax (at least for me) to understand.

So given:

>>> df
     Col1        Col2  Col3                                            Col4
0  SysLog  2016,09,17     1                        PD380_003 %LINK-3-UPDOWN
1  SysLog  2016,09,17     1                      NM380_005 %BGP-5-NBR_RESET
2  SysLog  2016,09,17     1                      NM380_005 %BGP-5-NBR_RESET
3  SysLog  2016,09,17     1  DO NOT TICKET LO380_004 %SYS-5-CONFIG_I Config

You can do:

>>> df['Col4'].replace(to_replace='^.*([A-Z]{2}[0-9]{3}_[0-9]{3}).*$', value=r'\1', regex=True) 
0    PD380_003
1    NM380_005
2    NM380_005
3    LO380_004
Name: Col4, dtype: object

You can also use a positional argument version if easier:

df['Col4'].replace('^.*([A-Z]{2}[0-9]{3}_[0-9]{3}).*$', r'\1', regex=True)

but you need to have regex=True since the replacement string is to be interpreted as a regex -- not just a static string.

Finally, assign directly into the original:

>>> df['Col4']=df['Col4'].replace('^.*([A-Z]{2}[0-9]{3}_[0-9]{3}).*$', r'\1', regex=True)
>>> df
     Col1        Col2  Col3       Col4
0  SysLog  2016,09,17     1  PD380_003
1  SysLog  2016,09,17     1  NM380_005
2  SysLog  2016,09,17     1  NM380_005
3  SysLog  2016,09,17     1  LO380_004