Remove the first N items that match a condition in a Python list

Question

If I have a function matchCondition(x), how can I remove the first n items in a Python list that match that condition?

One solution is to itera

Answer 1

Write a generator that takes the iterable, a condition, and an amount to drop. Iterate over the data and yield items that don't meet the condition. If the condition is met, increment a counter and don't yield the value. Always yield items once the counter reaches the amount you want to drop.

def iter_drop_n(data, condition, drop):
    dropped = 0

    for item in data:
        if dropped >= drop:
            yield item
            continue

        if condition(item):
            dropped += 1
            continue

        yield item

data = [1, 10, 2, 9, 3, 8, 4, 7]
out = list(iter_drop_n(data, lambda x: x < 5, 3))

This does not require an extra copy of the list, only iterates over the list once, and only calls the condition once for each item. Unless you actually want to see the whole list, leave off the list call on the result and iterate over the returned generator directly.

Answer 2

Straightforward Python:

N = 3
data = [1, 10, 2, 9, 3, 8, 4, 7]

def matchCondition(x):
    return x < 5

c = 1
l = []
for x in data:
    if c > N or not matchCondition(x):
        l.append(x)
    else:
        c += 1

print(l)

This can easily be turned into a generator if desired:

def filter_first(n, func, iterable):
    c = 1
    for x in iterable:
        if c > n or not func(x):
            yield x
        else:
            c += 1

print(list(filter_first(N, matchCondition, data)))

Answer 3

If mutation is required:

def do_remove(ls, N, predicate):
    i, delete_count, l = 0, 0, len(ls)
    while i < l and delete_count < N:
        if predicate(ls[i]):
           ls.pop(i) # remove item at i
           delete_count, l = delete_count + 1, l - 1 
        else:
           i += 1
    return ls # for convenience

assert(do_remove(l, N, matchCondition) == [10, 9, 8, 4, 7])

Answer 4

Using list comprehensions:

n = 3
data = [1, 10, 2, 9, 3, 8, 4, 7]
count = 0
def counter(x):
    global count
    count += 1
    return x

def condition(x):
    return x < 5

filtered = [counter(x) for x in data if count < n and condition(x)]

This will also stop checking the condition after n elements are found thanks to boolean short-circuiting.

Answer 5

The accepted answer was a little too magical for my liking. Here's one where the flow is hopefully a bit clearer to follow:

def matchCondition(x):
    return x < 5


def my_gen(L, drop_condition, max_drops=3):
    count = 0
    iterator = iter(L)
    for element in iterator:
        if drop_condition(element):
            count += 1
            if count >= max_drops:
                break
        else:
            yield element
    yield from iterator


example = [1, 10, 2, 9, 3, 8, 4, 7]

print(list(my_gen(example, drop_condition=matchCondition)))

It's similar to logic in davidism answer, but instead of checking the drop count is exceeded on every step, we just short-circuit the rest of the loop.

Note: If you don't have yield from available, just replace it with another for loop over the remaining items in iterator.

Answer 6

Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), we can use and increment a variable within a list comprehension:

# items = [1, 10, 2, 9, 3, 8, 4, 7]
total = 0
[x for x in items if not (x < 5 and (total := total + 1) <= 3)]
# [10, 9, 8, 4, 7]

This:

• Initializes a variable total to 0 which will symbolize the number of previously matched occurrences within the list comprehension
• Checks for each item if it both:
  • matches the exclusion condition (x < 5)
  • and if we've not already discarded more than the number of items we wanted to filter out by:
    • incrementing total (total := total + 1) via an assignment expression
    • and at the same time comparing the new value of total to the max number of items to discard (3)