CODEWITHSUNDEEP
python
Kaita John
Kaita John

Reputation: 1109

How to check for Boolean without using For Loop?

I have this running well, however am unable to avoid the loop, how can I get an only True if present or False if not present without having to loop through the list using a for loop?

my_list = [[1, 2], [4, 6], [8, 3]]
combined = [3, 8]
for value in my_list:
    print(value)
    if set(combined) == set(value):
        print("present")
    else:
        print("absent")

Upvotes: 4

Views: 410

Answers (6)

Amal K
Amal K

Reputation: 4874

You could use recursion:

def is_present(seq, elem, n=None):
    if n is None:
       n = len(seq) - 1
    if n == -1:
       return False
    if set(seq[n]) == set(elem):
       return True
    return is_present(seq, elem, n-1)

if is_present(my_list, combined):
    print("Present")
else:
    print("Not present")

There is also a problem with your existing code. You shouldn't have an else inside the loop body because if the very first element in my_list is not equal to combined, it will print Not present even though it may be present in the list later. And it will do this for every element that is not equal to it. This can be fixed with a for-else statement:

my_list = [[1, 2], [4, 6], [8, 3]]
combined = [3, 8]
for value in my_list:
    print(value)
    if set(combined) == set(value):
        print("present")
        break
else:
  print("absent")

Or you can just wrap it in a function:

def is_present():
    my_list = [[1, 2], [4, 6], [8, 3]]
    combined = [3, 8]
    for value in my_list:
        print(value)
        if set(combined) == set(value):
            return True
    return False

Upvotes: 2

Tomerikoo
Tomerikoo

Reputation: 19414

To avoid the multiple absent printing, you can use a for/else construct:

for value in my_list:
    if set(combined) == set(value):
        print("present")
        break
else:
    print("absent")

But if you're looking to optimize/reduce the loop, you can "hide" the loop inside an any call:

any(set(combined) == set(value) for value in my_list)

Or alternatively use a list comprehension to convert all sub-lists to sets:

set(combined) in [set(sublist) for sublist in my_list]

But as you can see, some sort of looping is necessary. I believe that by removing the loop you meant to reduce its code size.

Upvotes: 3

Kunal Pandey
Kunal Pandey

Reputation: 21

From your problem statement I am able to understand is that you just want to find out that the list in combined inside the 2D list i.e my_list

Solution- you could directly use the the following block of code to get your result;

print("present") if set(combined) == set(value) else print("absent")

Upvotes: 2

U13-Forward
U13-Forward

Reputation: 71570

I would prefer a generator with next:

next(("present" for value in my_list if set(value) == set(combined)), 'absent')

Or with any:

['absent', 'present'][any(set(combined) == set(value) for value in my_list)]

Or even better with map and in:

['absent', 'present'][set(combined) in map(set, my_list)]

All output:

present

Upvotes: 3

user2390182
user2390182

Reputation: 73450

You can do this rather concisely, getting the short-circuiting of any and the brevity of map all while avoiding ugly lambda or barely readable conditional expressions:

if set(combined) in map(set, my_list):
    print("present")
else:
    print("absent")

Upvotes: 7

Jason Yang
Jason Yang

Reputation: 13061

Using built-in function any and class map,

def is_inside(item, items):
    return any(map(lambda x:set(x)==set(combined), my_list))

my_list = [[1, 2], [4, 6], [8, 3]]
combined = [3, 8]

if is_inside(combined, my_list):
    print("present")
else:
    print("absent")

Upvotes: 2

Related Questions