Python：创建过滤器函数

Question

I'm trying to create a function:

filter(delete,lst) 

When someone inputs:

filter(1,[1,2,1]) 

returns [2]

What I have come up with was to use the list.remove function but it only deletes the first instance of delete.

def filter(delete, lst):

"""

Removes the value or string of delete from the list lst

"""

   list(lst)

   lst.remove(delete)

   print lst

My result:

filter(1,[1,2,1])

returns [2,1]

Answer 1

Try with list comprehensions:

def filt(delete, lst):
    return [x for x in lst if x != delete]

Or alternatively, with the built-in filter function:

def filt(delete, lst):
    return filter(lambda x: x != delete, lst)

And it's better not to use the name filter for your function, since that's the same name as the built-in function used above

Answer 2

Custom Filter Function

def my_filter(func,sequence):
    res=[]
    for variable in sequence :
        if func(variable):
            res.append(variable)
    return res

def is_even(item):
    if item%2==0 :
        return True
    else :
        return False
 
    

seq=[1,2,3,4,5,6,7,8,9,10]
print(my_filter(is_even,seq))

Answer 3

Custom filter function in python:

def myfilter(fun, data):
    return [i for i in data if fun(i)]

Answer 4

I like the answer from Óscar López but you should also learn to use the existing Python filter function:

>>> def myfilter(tgt, seq):
        return filter(lambda x: x!=tgt, seq)

>>> myfilter(1, [1,2,1])
[2]