How can we Check if something other than a value in a list in Python?-CodePudding

I want to know if there is a value other than a specific value in a list in python, for example: given a list

lst = [0,0,0,0,0,0,0]

How can I check if there is a value other than 0 in it like 1 or 2 or 3 or any number other than 0 and its position? Example :

lst = [0,0,0,0,1,0,0,0]

The check operation should bring a true value saying yes list have value other than 0, that value and its position as 4.

CodePudding user response：

to get the positions and values of nonzero elements you can do this:

lst = [0,0,0,0,1,0,0,0,2]
list(filter(lambda x: x[1]!=0, enumerate(lst)))  # [(4, 1), (8, 2)]

CodePudding user response：

You can use enumerate to get values plus their index in the list.

lst = [0,0,0,0,1,0,0,0]
baddies = [i for i,val in enumerate(lst)]
has_baddies = bool(baddies)

To short-circuit on the first bad apple a for loop would be a good option

for baddie, val in enumerate(lst):
    if has_baddies := val != 0:
        break
else:
    baddie = None

CodePudding user response：

Using the same output as @SergFSM, this is another tricky option using collections:

lst = [0, 0, 0, 0, 1, 0, 0, 0, 2]
print([f"({lst.index(f)}, {f})" for f in dict(collections.Counter(lst)) if f != 0])

['(4, 1)', '(8, 2)']

CodePudding user response：

Use itertools.compress from the standard library. Take advantage that 0 is cast to False:

from itertools import compress

lst = [0,0,0,0,1,0,99, 0,0]

res = list(compress(enumerate(lst), lst))

if res:
    print(*res)
else:
    print('Only 0s.')