You can perform this task by forming a |-separated string. This works because pd.Series.str.replace accepts regex:

Replace occurrences of pattern/regex in the Series/Index with some other string. Equivalent to str.replace() or re.sub().

This avoids the need to create a dictionary.

import pandas as pd

df = pd.DataFrame({'A': ['LOCAL TEST', 'TEST FOREIGN', 'ANOTHER HELLO', 'NOTHING']})

pattern = '|'.join(['LOCAL', 'FOREIGN', 'HELLO'])

df['A'] = df['A'].str.replace(pattern, 'CORP', regex=True)

#               A
# 0     CORP TEST
# 1     TEST CORP
# 2  ANOTHER CORP
# 3       NOTHING

Use replace with dict for replacing and regex=True:

df['url'] = df['url'].replace({'icashier.alipay.com': 'aliexpress.com'}, regex=True)
print (df)
                                          url
0  aliexpress.com/catalog/2758186/detail.aspx
1  aliexpress.com/catalog/2758186/detail.aspx
2  aliexpress.com/catalog/2758186/detail.aspx
3                                      vk.com

Use:

import pandas as pd

df = pd.DataFrame({'A': ['$5,756', '3434', '$45', '1,344']})
df['A'] = df['A'].str.replace('[$,]', '', regex=True)
print(df)

Output

      A
0  5756
1  3434
2    45
3  1344

The problem is that the character $ has a special meaning in regular expressions. From the documentation (emphasis mine):

$
Matches the end of the string or just before the newline at the end of the string, and in MULTILINE mode also matches before a newline. foo matches both ‘foo’ and ‘foobar’, while the regular expression foo$ matches only ‘foo’. More interestingly, searching for foo.$ in 'foo1\nfoo2\n' matches ‘foo2’ normally, but ‘foo1’ in MULTILINE mode; searching for a single $ in 'foo\n' will find two (empty) matches: one just before the newline, and one at the end of the string.mode; searching for a single $ in 'foo\n' will find two (empty) matches: one just before the newline, and one at the end of the string.

So you need to escape the character or put it inside a character class.

As an alternative use:

df['A'].str.replace('\$|,', '', regex=True)  # note the escaping \

You can use data frame replace method with regex=True, and use .*,.* to match strings that contain a comma (you can replace comma with other any other substring you want to detect):

str_cols = ['Answer']    # specify columns you want to replace
df[str_cols] = df[str_cols].replace('.*,.*', 'X', regex=True)
df
#Question   Answer
#0      1       A
#1      2       X
#2      3       C

or if you want to replace all string columns:

str_cols = df.select_dtypes(['object']).columns