如何在 python 中编写“过滤器”生成器？

Question

I'm studying the topic of generators in python and I can't solve an exercise.

I have to write a "filter" generator that takes a predicate and a sequence and produces a sequence that has no elements for which the predicate is true. (similar to the built-in filter function)

I know how to solve an exercise through a function, but I do not know how to solve it through "yield".

This is my code:

def filter(x, lst):
    i = 0
    while i < len(lst):
        if lst[i] != x:
            yield lst[i]
            i += 1
        else:
            continue

I'd be glad to get help with the task.

Answer 1

you may use:

def my_filter(pred, seq):
    for item in seq:
        if not pred(item):
            yield item

my_filter generator is iterating over your sequence and when an element applied over the predicate it is False then will yield that element

example:

list(my_filter(lambda x: x >4, [0, 1, 5, 2]))

# [0, 1, 2]

Answer 2

You can use a generator literal to accomplish this in one line

def my_filter(x, lst):
    yield from (i for i in lst if i != x)

This uses the return keyword instead of yield, but is actually equivalent as it also returns a generator which yields the same items.

def my_filter(x, lst):
    return (i for i in lst if i != x)