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")
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")
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.
map (got my upvote)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
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")
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")
value here? Is it the loop variable? And this code should be inside the loop? In that case this is exactly the same as the original code...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
absenttwice andpresentonce. Is that intentional?for/elseconstruct. You basically need to add abreakin theifand unindent theelse. Then, the answers below still simplify the loop in a nice way.