Python check if list in a dict satisfies two conditions

bayman :

Given list1 below, how do I return a new list where the values of 'codes' contain the string 'Two' but does not contain the string 'One'?

# Example, given list1:
list1 = [{'id': 11, 'codes': ['OneSeven', 'TwoThree']}, {'id': 22, 'codes': ['FiveTwoSix', 'Four', 'FourFive']}, {'id': 33, 'codes': ['SixSeven', 'OneSix']}]

# Return list with element where 'id': 22 since the string 'Two' is in codes but 'One' isn't.
list2 = [{'id': 22, 'codes': ['FiveTwoSix', 'Four', 'FourFive']}]

Boseong Choi :

# Example, given list1:
list1 = [{'id': 11, 'codes': ['OneSeven', 'TwoThree']}, {'id': 22, 'codes': ['FiveTwoSix', 'Four', 'FourFive']}, {'id': 33, 'codes': ['SixSeven', 'OneSix']}]

list2 = [
    d for d in list1
    if any(
        'Two' in word
        for word in d['codes']
    ) and all(
        'One' not in word
        for word in d['codes']
    )
]

print(list2)

output:

[{'id': 22, 'codes': ['FiveTwoSix', 'Four', 'FourFive']}]

---

If you don't like redundant traversing lists, you can variable alternatives.

1. For python 3.8 or later, you can do this:

# Example, given list1:
list1 = [{'id': 11, 'codes': ['OneSeven', 'TwoThree']}, {'id': 22, 'codes': ['FiveTwoSix', 'Four', 'FourFive']}, {'id': 33, 'codes': ['SixSeven', 'OneSix']}]

list2 = [
    d for d in list1
    if 'Two' in (text := ' '.join(d['codes'])) and 'One' not in text
]

print(list2)

2. Of course, using for-statement instead of comprehension, you can do this with 3.7 or earlier version.

list2 = []
for d in list1:
    text = ' '.join(d['codes'])
    if 'Two' in text and 'One' not in text:
        list2.append(d)

3. Or use function + comprehension:

def check_condition(data: dict) -> bool:
    text = ' '.join(data['codes'])
    return'Two' in text and 'One' not in text

list2 = [
    d for d in list1 if check_condition(d)
]

Last one is quite readable, but someone might dislike declaring function which is used in only one place.
Choose a method fits your situation.