CODEWITHSUNDEEP
pythonlistsubstitution
Brando__
Brando__

Reputation: 365

Python substitute elements inside a list

I have the following code that is filtering and printing a list. The final output is json that is in the form of name.example.com. I want to substitute that with name.sub.example.com but I'm having a hard time actually doing that. filterIP is a working bit of code that removes elements entirely and I have been trying to re-use that bit to also modify elements, it doesn't have to be handled this way.

def filterIP(fullList):
    regexIP = re.compile(r'\d{1,3}.\d{1,3}.\d{1,3}.\d{1,3}$')
    return filter(lambda i: not regexIP.search(i), fullList)

def filterSub(fullList2):
    regexSub = re.compile(r'example\.com, sub.example.com')
    return filter(lambda i: regexSub.search(i), fullList2)

groups = {key : filterSub(filterIP(list(set(items)))) for (key, items) in groups.iteritems() }

print(self.json_format_dict(groups, pretty=True))

This is what I get without filterSub

"type_1": [
    "server1.example.com",
    "server2.example.com"
],

This is what I get with filterSub

"type_1": [],

This is what I'm trying to get

"type_1": [
    "server1.sub.example.com",
    "server2.sub.example.com"
],

Upvotes: 0

Views: 208

Answers (2)

Labrys Knossos
Labrys Knossos

Reputation: 393

If the examples are all truly as simple as those you listed, a regex is probably overkill. A simple solution would be to use string .split and .join. This would likely give better performance.

First split the url at the first period:

url = 'server1.example.com'
split_url = url.split('.', 1)  
# ['server1', 'example.com']

Then you can use the sub to rejoin the url:

subbed_url = '.sub.'.join(split_url)
# 'server1.sub.example.com'

Of course you can do the split and the join at the same time

'.sub.'.join(url.split('.', 1))

Or create a simple function: def sub_url(url): return '.sub.'.join(url.split('.', 1))

To apply this to the list you can take several approaches.

A list comprehension:

subbed_list = [sub_url(url)
               for url in url_list]

Map it:

subbed_list = map(sub_url, url_list)

Or my favorite, a generator:

gen_subbed = (sub_url(url)
               for url in url_list)

The last looks like a list comprehension but gives the added benefit that you don't rebuild the entire list. It processes the elements one item at a time as the generator is iterated through. If you decide you do need the list later you can simply convert it to a list as follows:

subbed_list = list(gen_subbed)

Upvotes: 0

K. A. Buhr
K. A. Buhr

Reputation: 50819

The statement:

regexSub = re.compile(r'example\.com, sub.example.com')

doesn't do what you think it does. It creates a compiled regular expression that matches the string "example.com" followed by a comma, a space, the string "sub", an arbitrary character, the string "example", an arbitrary character, and the string "com". It does not create any sort of substitution.

Instead, you want to write something like this, using the re.sub function to perform the substitution and using map to apply it:

def filterSub(fullList2):
    regexSub = re.compile(r'example\.com')
    return map(lambda i: re.sub(regexSub, "sub.example.com", i),
               filter(lambda i: re.search(regexSub, i), fullList2))

Upvotes: 1

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