The docs on pandas.DataFrame.replace says you have to provide a nested dictionary: the first level is the column name for which you have to provide a second dictionary with substitution pairs.

So, this should work:

>>> df=pd.DataFrame({'a': ['NÍCOLAS','asdč'], 'b': [3,4]})
>>> df
         a  b
0  NÍCOLAS  3
1     asdč  4

>>> df.replace({'a': {'č': 'c', 'Í': 'I'}}, regex=True)
         a  b
0  NICOLAS  3
1     asdc  4

Edit. Seems pandas also accepts non-nested translation dictionary. In that case, the problem is probably with character encoding, particularly if you use Python 2. Assuming your CSV load function decoded the file characters properly (as true Unicode code-points), then you should take care your translation/substitution dictionary is also defined with Unicode characters, like this:

dictionary = {u'í': 'i', u'á': 'a'}

If you have a definition like this (and using Python 2):

dictionary = {'í': 'i', 'á': 'a'}

then the actual keys in that dictionary are multibyte strings. Which bytes (characters) they are depends on the actual source file character encoding used, but presuming you use UTF-8, you'll get:

dictionary = {'\xc3\xa1': 'a', '\xc3\xad': 'i'}

And that would explain why pandas fails to replace those chars. So, be sure to use Unicode literals in Python 2: u'this is unicode string'.

On the other hand, in Python 3, all strings are Unicode strings, and you don't have to use the u prefix (in fact unicode type from Python 2 is renamed to str in Python 3, and the old str from Python 2 is now bytes in Python 3).

You can use apply with str.replace:

import re
chars = ''.join(map(re.escape, listOfSpecialChars))

df2 = df.apply(lambda c: c.str.replace(f'[{chars}]', '', regex=True))

Alternatively, stack/unstack:

df2 = df.stack().str.replace(f'[{chars}]', '', regex=True).unstack()

output:

  col1 col2
0    1    A
1    3    B
2    4    C

Something like this?

df.column_name = df.column_name.str.replace(r'["\']', '')

Edit:

Use regex, thanks to @mozway

You need to escape the $ character using \ in the old_values list:

old_values = ["\${z_mask0}", "\${z_mask1}", "\${z_mask2}"]

The above should be enough. Here is all the code:

old_values = ["\${z_mask0}", "\${z_mask1}", "\${z_mask2}"]
new_values = ["${z_00}", "${z_01}", "${z_02}"]
df = df.replace(old_values, new_values, regex=True)
print(df)

Output:

      col1
0   series ${z_00}
1   series ${z_01}
2   series ${z_02}

You can use a regex:

df.columns = df.columns.str.replace(r'[ \-&/:()]', '', regex=True)

Some characters need to be escaped in regexes, so to be safe you can get re to do it for you:

import re
regex = '['+re.escape(r' -&/:()')+']'
# '[\\ \\-\\&/:\\(\\)]'
df.columns = df.columns.str.replace(regex, '', regex=True)

I'm assuming you're using Python 2.x here and this is likely a Unicode problem. Don't worry, you're not alone--unicode is really tough in general and especially in Python 2, which is why it's been made standard in Python 3.

If all you're concerned about is the ñ, you should decode in UTF-8, and then just replace the one character.

That would look something like the following:

DF['name'] = DF['name'].str.decode('utf-8').replace(u'\xf1', 'n')

As an example:

>>> "sureño".decode("utf-8").replace(u"\xf1", "n")
u'sureno'

If your string is already Unicode, then you can (and actually have to) skip the decode step:

>>> u"sureño".replace(u"\xf1", "n")
u'sureno'

Note here that u'\xf1' uses the hex escape for the character in question.

Update

I was informed in the comments that <>.str.replace is a pandas series method, which I hadn't realized. The answer to this possibly might be something like the following:

DF['name'] = map(lambda x: x.decode('utf-8').replace(u'\xf1', 'n'), DF['name'].str)

or something along those lines, if that pandas object is iterable.

Another update

It actually just occurred to me that your issue may be as simple as the following:

DF['NAME']=DF['NAME'].str.replace(u"ñ","n")

Note how I've added the u in front of the string to make it unicode.

and i'm trying to replace the '-' sign with ''

These are (likely) not strings but numbers, the - means it is a negative number. Given that these are negative numbers, it perhaps is not a good idea to remove the minus, since -5 of course is something different than 5.

In case you want to map negative numbers on their positive counterpart, you can use pandas.DataFrame.abs(..) [pandas-doc] for that, to calculate the absolute value:

df_1['x'] = df_1['x'].abs()

Just replace everything excluding alphanumerics with nothing. Code below.

df['Col']= df['Col'].str.replace('[^\w]', '')

      Col
0     Beer
1  Whiskey
2     Wine
3    Tonic
4      Rum