使用lambda函数从列表中过滤元素值

Question

how can I rewrite the following code by using lambda function?the problem i have is to make x as variable so I can return different list based on input

s_list = [1,2,3,4,5,6]
def f(x):
    return [a for a in s_list if a not in x]

print(f([1,2]))#[3,4,5,6]
print(f([4,6]))#[1,2,3,5]

Answer 1

If you don't need s_list like a argument, you can use this:

F = lambda x: [a for a in [1,2,3] if a not in x]

and if you need s_list like an argument, you could make this:

F = lambda x, s_list: [a for a in s_list if a not in x]

Answer 2

Here's a way to write your function using filter :

def f2(x):
    return filter(lambda a: a not in x, s_list)

print(f2(x=[1,2]))
#[3, 4, 5, 6]
print(f2(x=[4,6]))
#[1, 2, 3, 5]
print(f2(x=[]))
#[1, 2, 3, 4, 5, 6]

Or if you wanted it to be a function of both s_list and x :

def f3(s_list, x):
    return filter(lambda a: a not in x, s_list)

print(f3(s_list, x=[1,2]))
#[3, 4, 5, 6]
print(f3(s_list, x=[4,6]))
#[1, 2, 3, 5]
print(f3(s_list, x=[]))
#[1, 2, 3, 4, 5, 6]

Answer 3

You can achieve that by using Filter method :

s_list = [1,2,3,4,5,6]

x_=[1,2]

print(list(filter(lambda x:x not in x_,s_list)))

output:

[3, 4, 5, 6]