Try:

df['tumor-size'] = df['tumor-size'].replace("^'0-4'$", "'00-04'")

If you want to generalise, you can use str.replace with SOL/EOL anchors.

df['last_status'].str.replace(r'^(\(ID: \d+\))$', r'Planning: \1')

0     Planning: (ID: 10)
1     Planning: (ID: 20)
2     Planning: (ID: 40)
3     Planning: (ID: 50)
4     Planning: (ID: 60)
5     Planning: (ID: 80)
6      Assigned (ID: 30)
7      Assigned (ID: 40)
8     Assigning (ID: 30)
9     Cancelled (ID: 40)
10    Cancelled (ID: 80)
11      Planned (ID: 20)
12     Planning (ID: 10)
13    Processed (ID: 70)
14      Revoked (ID: 50)
15      Writing (ID: 50)
16      Written (ID: 60)
Name: last_status, dtype: object

If you want to replace a specific ID only, change your regex to -

r'^(\(ID: 10\))$'

Or,

r'^(\(ID: {}\))$'.format(number)

Where number is a variable that holds the ID value to perform replacement on.

You can use a regex to do this replacement, looking for one of

1. a / followed by another /;
2. a / at the start of the string; or
3. a / at the end of the string

df['a'] = df['a'].str.replace(r'/(?=/)|^/|/$', '')

Output:

       a
0    a/b
1       
2    a/b
3  a/b/c

In regex, some symbol have a meaning and trigger some functionality, when you want to explicitly match the symbol without triggering its function, you escape it.

Now re.escape is simply a method to avoid escaping a list of character manually.

instead of escaping (adding \) manually like this :

"\$\[\]\^"

You can simply do like the function you write.

pattern = "|".join(map(re.escape, "[|\[|\]|\^"

To see what do your code, simply print p.

list_of_strings = ['can we: remove', 'with @##%@/}\p special characters!!!!','EXACT']


p = '|'.join(map(re.escape, list_of_strings))
print(p)

As you will see all characters have been escaped \.

Discovered the solution:

df['Brand'] = df['Brand'].str.replace(r'(?i)stack\b', r'stackoverflow')

You can escape the $ (it's special character in regex) or use regex=False:

data = {
    "Column1": [
        "Income recorded on books this year not included on Schedule K, lines 1 through 11 (itemize):",
        "a Tax-exempt interest $ Statement #36",
        "Statement #36",
    ],
    "Column2": [254, 258, 356],
}

df = pd.DataFrame(data)

df["Column1"] = df["Column1"].str.replace(" $ Statement #36", "", regex=False)
print(df)

Prints:

                                             Column1  Column2
0  Income recorded on books this year not include...      254
1                              a Tax-exempt interest      258
2                                      Statement #36      356