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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 iterate over each item, mark it for deletion (e.g., by setting it to None), and then filter the list with a comprehension. This requires iterating over the list twice and mutates the data. Is there a more idiomatic or efficient way to do this?
n = 3 def condition(x): return x < 5 data = [1, 10, 2, 9, 3, 8, 4, 7] out = do_remove(data, n, condition) print(out) # [10, 9, 8, 4, 7] (1, 2, and 3 are removed, 4 remains) 
849
The remove() method removes the first matching element (which is passed as an argument) from the list. The pop() method removes an element at a given index, and will also return the removed item. You can also use the del keyword in Python to remove an element or slice from a list.
The remove() function allows you to remove the first instance of a specified value from the list. This can be used to remove the list's top item. Pick the first member from the list and feed it to the remove() function.
You can use the pop() method to remove specific elements of a list. pop() method takes the index value as a parameter and removes the element at the specified index. Therefore, a[2] contains 3 and pop() removes and returns the same as output. You can also use negative index values.
One way using itertools.filterfalse and itertools.count:
from itertools import count, filterfalse data = [1, 10, 2, 9, 3, 8, 4, 7] output = filterfalse(lambda L, c=count(): L < 5 and next(c) < 3, data) Then list(output), gives you:
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