You need to escape the plus sign:

In[10]:
df = pd.DataFrame({'a':['dsa^', '^++', '+++','asdasads']})
df

Out[10]: 
          a
0      dsa^
1       ^++
2       +++
3  asdasads

In[11]:
df['a'].str.count("\+")

Out[11]: 
0    0
1    2
2    3
3    0
Name: a, dtype: int64

Also when you do df['a'].str.count('^') this just returns 1 for all rows:

In[12]:
df['a'].str.count('^')

Out[12]: 
0    1
1    1
2    1
3    1
Name: a, dtype: int64

Again you need to escape the pattern:

In[16]:
df['a'].str.count('\^')

Out[16]: 
0    1
1    1
2    0
3    0
Name: a, dtype: int64

EDIT

Regarding the semantic difference between count on a normal string and on a Series, count on a python str just does a character count, but str.count takes a regex pattern. The ^ and + are special characters which need to be escaped with a backslash if you are searching for those characters

Just try pass regex

df = df.replace({'’':"'"}, regex=True)
           Name
0  abc' company
1  dfe' company

It does work, but you have to indicate the escape character for the embedded quotes:

In [1]: data = '''SEARCH_TERM,ACTUAL_URL
"bra tv bord","http://www.ikea.com/se/sv/catalog/categories/departments/living_room/10475/?se%7cps%7cnonbranded%7cvardagsrum%7cgoogle%7ctv_bord"
"tv p\xc3\xa5 hjul","http://www.ikea.com/se/sv/catalog/categories/departments/living_room/10475/?se%7cps%7cnonbranded%7cvardagsrum%7cgoogle%7ctv_bord"
"SLAGBORD, \\"Bergslagen\\", IKEA:s 1700-tals serie","http://www.ikea.com/se/sv/catalog/categories/departments/living_room/10475/?se%7cps%7cnonbranded%7cvardagsrum%7cgoogle%7ctv_bord"'''

In [2]: df = read_csv(StringIO(data), escapechar='\\', encoding='utf-8')

In [3]: df
Out[3]: 
                                      SEARCH_TERM                                         ACTUAL_URL
0                                     bra tv bord  http://www.ikea.com/se/sv/catalog/categories/d...
1                                      tv på hjul  http://www.ikea.com/se/sv/catalog/categories/d...
2  SLAGBORD, "Bergslagen", IKEA:s 1700-tals serie  http://www.ikea.com/se/sv/catalog/categories/d...

see this gist.

You can use the built-in function html.unescape:

import html
df['Country/Region'] = df['Country/Region'].astype(str).map(html.unescape)

Output:

>>> df
           Country/Region
0         Belgium (Dutch)
1             Saint Lucia
2     Trinidad and Tobago
3            Sierra Leone
4                    Mali
5  Svalbard and Jan Mayen

Thanks for posting the link to the Google Sheet. I downloaded it and loaded it via pandas:

df = pd.read_excel(r'~\relatorio_vendas_CRM.xlsx', encoding = 'utf-8')
df.columns = df.columns.str.replace('°', '')
df.columns = df.columns.str.replace('º', '')

Note that the two replace statements are replacing different characters, although they look very similar.

Help from: Why do I get a SyntaxError for a Unicode escape in my file path?

I solved this by using Pandas' regex replacer:

df = df.replace('\\\\', '\\\\\\\\', regex=True)

We need four slashes per final slash, because we are doing two layers of escaping. One for literal Python strings, and one to escape them in the regular expression. This will find-replace any \s in any column in the data frame, anywhere they appear in the string.

It is mind-boggling to me that this is still the default behavior.

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).

Something like this?

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

Edit:

Use regex, thanks to @mozway