Python-替换字符串列表中的字符

Question

I am trying to remove the chars '0' from chars 2 to 3 of each item in the following list:

list1 = ['0x001', '0x002', '0x0a3']

my desired output is:

list1 = ['0x1', '0x2', '0xa3']

this is the way I have tried so far but has not worked:

for i in list1:

   if i[2:3] == '0':
        i.replace('0', '')

But this makes no changes to the items in the list. Thank you in advance!

Answer 1

In Python, a string (be it byte or unicode) is immutable. That means that you cannot change it, you can only have the variable to reference a new value.

A list is a mutable object so you can change its individual elements.

Here you could do:

for index, i in enumerate(list1):
    if i[2:3] == '0':
        list1[index] = i[:2] + i[2:].replace('0', '')

(you cannot replace '0' in the whole string because you want to preserve the initial one)

---

To insist on mutability: above code modifies list1, while this:

list1 = [ i[:2] + i[2:].replace('0', '') for i in list1 ]

will create a new list.

Just look at those 2 example codes:

list1 = ['0x001', '0x002', '0x0a3']
list2 = list1

for index, i in enumerate(list1):
    if i[2:3] == '0':
        list1[index] = i[:2] + i[2:].replace('0', '')

print(list1, list2)

output is:

['0x1', '0x2', '0xa3'] ['0x1', '0x2', '0xa3']

because the list has been modified, so list2 points the the modified list. But with

list1 = ['0x001', '0x002', '0x0a3']
list2 = list1

list1 = [ i[:2] + i[2:].replace('0', '') for i in list1 ]
print(list1, list2)

output is:

['0x1', '0x2', '0xa3'] ['0x001', '0x002', '0x0a3']

because list1 is a new list while list2 still references the original one.

Answer 2

You can use re.sub :

import re
list1 = ['0x001', '0x002', '0x0a3']
new_result = [re.sub('(?<=x)[0]+', '', i) for i in list1]

Output:

['0x1', '0x2', '0xa3']