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]
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]
lambda functions. Thus, you can do lambda x=[]: ... if you want a default argument. Note that default arguments need to be last in list, so either you have to have s_list as the first argument or just use s_listfrom the global (outer) scope.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]
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]
