I want to create a set of values that can divide a specific number. In my code below, I want factors of number 12.
x=12
a=set(map(lambda i:i if(x%i==0),range(1,x+1)))
print(a)
I expected output to be {1,2,3,4,6,12}.
but i have got some error:
File "<ipython-input-18-7f73bd4c5148>", line 2
a=set(map(lambda i:i if(x%i==0),range(1,x+1)))
^
SyntaxError: invalid syntax
I think your lambda is supposed to be lambda i : x%i==0. What you have now doesn't make sense. And I think you should be using filter instead of map.
>>> x = 12
>>> a = set(filter(lambda i:x%i==0, range(1,x+1)))
>>> print(a)
{1, 2, 3, 4, 6, 12}
Looks like you are confusing lambda notation with list/set comprehension notation. The correct way using lambda and filter (not map) has already been shown, so here's a version using a set comprehension:
>>> x = 12
>>> {i for i in range(1, x+1) if x % i == 0}
{1, 2, 3, 4, 6, 12}
As you see, the i if ... is present here (although distributed over different parts of the expression), but not in the lambda.
try
a=set(filter(lambda i: True if(x%i==0) else False,range(1,x+1)))
The condition in lambda function should be defined as explained here: Conditional statement in a one line lambda function in python?
I would recommend using a generator or function instead of a lambda as they are not PEP-8 compliant and typically make logic more complex. Using a generator will also allow you to save memory if you are going to be making very long sequences.
Example generator
gen = (x for x in range(1, 1000000000) if 12 % x == 0)
lambda i : x%i==0. What you have now doesn't make sense. And I think you should be usingfilterinstead ofmap.