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).

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.

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

That is because whatever regex pattern that you are using has an opening [ with no associated closing ]. Unless you really are trying to use a regex character set (see https://blog.finxter.com/python-character-set-regex-tutorial/), you probably just want to escape the first [ in your pattern - like \[. Seeing your code would help.

Something like this?

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

Edit:

Use regex, thanks to @mozway

use replace which applies on whole dataframe :

df
Out[14]: 
       Time      A1      A2
0  2.000255    1499    1592
1  2.176470    2096    1942
2  2.765405  *7639*  *8196*
3  2.765856  *7088*  *7542*
4  2.904896  *8736*  *8459*
5  2.905312  *7778*  *7704*
6  2.980718  *6612*  *6593*
7  3.058576  *8520*  *9122*

df=df.replace('\*','',regex=True).astype(float)

df
Out[16]: 
       Time    A1    A2
0  2.000255  1499  1592
1  2.176470  2096  1942
2  2.765405  7639  8196
3  2.765856  7088  7542
4  2.904896  8736  8459
5  2.905312  7778  7704
6  2.980718  6612  6593
7  3.058576  8520  9122

This should work (just change 0.5 to '.5'):

data[1].replace(regex=True,inplace=True,to_replace='½',value='.5')

First of all, you want the replacement value to be a string, not an number. Secondly, you mean to replace '½' with '.5', not '0.5'.

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}

Remove the / and g, this syntax is not valid in python:

df['column_name'] = df['column_name'].str.replace(r'[^a-zA-Z ]', '', regex=True)

output:

    id column_name
0  aaa         sam
1  bbb        joes