Find and replace consecutive string values in a list

Question

Suppose I have a list that holds a list of column names:

columns = [
        'Village Trustee V Belleville', 'Unnamed: 74', 'Unnamed: 75',
        'Village President V Black Earth', 'Village Trustee V Black Earth',
        'Unnamed: 78', 'Unnamed: 79', 'Village President V Blue Mounds',
        'Village Trustee V Blue Mounds', 'Unnamed: 82',
        'Village Trustee V Cottage Grove', 'Unnamed: 94', 'Unnamed: 95',
        'Village President V Cross Plains', 'Village Trustee V Cross Plains',
        'Unnamed: 98', 'Unnamed: 99'
        ]

And I want to find all Unnamed: XX values and replace them consecutively with the first valid string. For example:

Input: 'Village Trustee V Belleville', 'Unnamed: 74', 'Unnamed: 75'

Output: 'Village Trustee V Belleville', 'Village Trustee V Belleville', 'Village Trustee V Belleville'

For this I have a function:

def rename_unnamed(columns):
    current_value = None
    pattern = re.compile(r'^Unnamed:\s\d+$')
    for column_index, column in enumerate(columns):
        if re.match(pattern, column):
            columns[column_index] = current_value
        else:
            current_value = column
    return columns

Which does the job pretty fine, however, I was wondering if this could be improved and/or simplified, as this looks like a mouthful for rather a simple task.

Answer 1

• Don't mutate input and return.

  If I need to pass columns unedited to two different functions, than your function will seem like it's not mutating columns and I wouldn't have to perform a copy of the data. But this isn't the case.

• You may want to potentially error if there are no values before the first "Unnamed" value.

• You may want to make pattern an argument, to allow reusability of the function. And instead use change it to a callback to allow even more reusability.

Overall your function's pretty good. I don't think it can be made much shorter.

def value_or_previous(iterator, undefined, default=None):
    prev_item = default
    for item in iterator:
        if not undefined(item):
            prev_item = item
        yield prev_item


def rename_unnamed(columns):
    return list(value_or_previous(columns, re.compile(r'^Unnamed:\s\d+$').match))

Answer 2

In case you need to modify the initial columns list (mutable data structure) you don't need to return it.

The crucial pattern can be moved out to the top making it reusable and accessible for other potential routines or inline conditions:

UNNAMED_PAT = re.compile(r'^Unnamed:\s\d+$')

As the function's responsibility is to back fill (backward filling) items matched by specific regex pattern I'd rename it appropriately like say backfill_by_pattern. The search pattern is passed as an argument.

I'll suggest even more concise solution based on replacement of search items by the item at the previous index:

def backfill_by_pattern(columns, pat):
    fill_value = None
    for i, column in enumerate(columns):
        if pat.match(column):
            columns[i] = columns[i - 1] if i else fill_value

Sample usage:

backfill_by_pattern(columns, pat=UNNAMED_PAT)
print(columns)

The output:

['Village Trustee V Belleville',
 'Village Trustee V Belleville',
 'Village Trustee V Belleville',
 'Village President V Black Earth',
 'Village Trustee V Black Earth',
 'Village Trustee V Black Earth',
 'Village Trustee V Black Earth',
 'Village President V Blue Mounds',
 'Village Trustee V Blue Mounds',
 'Village Trustee V Blue Mounds',
 'Village Trustee V Cottage Grove',
 'Village Trustee V Cottage Grove',
 'Village Trustee V Cottage Grove',
 'Village President V Cross Plains',
 'Village Trustee V Cross Plains',
 'Village Trustee V Cross Plains',
 'Village Trustee V Cross Plains']