• In Python, you wouldn't use indexes for this at all, but just deal with the values—[value for value in a if value > 2]. Usually dealing with indexes means you're not doing something the best way.

• If you do need an API similar to Matlab's, you would use numpy, a package for multidimensional arrays and numerical math in Python which is heavily inspired by Matlab. You would be using a numpy array instead of a list.

   >>> import numpy
   >>> a = numpy.array([1, 2, 3, 1, 2, 3])
   >>> a
   array([1, 2, 3, 1, 2, 3])
   >>> numpy.where(a > 2)
   (array([2, 5]),)
   >>> a > 2
   array([False, False,  True, False, False,  True], dtype=bool)
   >>> a[numpy.where(a > 2)]
   array([3, 3])
   >>> a[a > 2]
   array([3, 3])
  

You could do it in a list comprehension. This is basically the same as your code but condensed into one line, and it builds a list of results that match the criteria.

The first way gets all the matches

Copymylist = [-1,-2,3,4,5,6]

results = [(i, el) for i, el in enumerate(mylist) if el > 0]

Another way would be to use a generator expression which is probably faster, and just unpack it. This gets the first one.

Copy*next((i, el) for i, el in enumerate(mylist) if el > 0))

This loops the list and checks the condition, then puts the index and element into a tuple. Doing this inside parentheses turns it into a generator, which is much faster because it hasn't actually got to hold everything in memory, it just generates the responses as you need them. Using next() you can iterate through them. As we only use next() once here it just generates the first match. Then we unpack it with *

As there are two other valid answers here I decided to use timeit module to time each of them and post the results. For clarity I also timed the OP's method. Here is what I found:

Copyimport timeit
# Method 1 Generator Expression
print(timeit.timeit('next((i, el) for i, el in enumerate([-1,-2,3,4,5,6]) if el > 0)', number=100000))
0.007089499999999999

# Method 2 Getting index of True
print(timeit.timeit('list(x > 0 for x in [-1,-2,3,4,5,6]).index(True)', number=100000))
0.008104599999999997

# Method 3 filter and lambda
print(timeit.timeit('myidx , myel = list(filter(lambda el: el[1] > 0, enumerate([-1,-2,3,4,5,6])))[0]', number=100000))
0.0155314

statement = """
for idx, el in enumerate([-1,-2,3,4,5,6]):
    if el > 0:
        myidx, myel = idx, el
        break
"""

print(timeit.timeit(statement, number=100000))
0.04074070000000002

I would just keep a dictionary of all of the owner data together:

    ownerData = { "Shaft" : {
              "carMake" : "Audi",
              "carModel" : "A8",
              "year" : "2015" },
         "JamesBond" : {
              "carMake" : "Aston",
              "carModel" : "DB8",
              "year" : "2012" },
         "JeffBezos" : {
              "carMake" : "Honda",
              "carModel" : "Accord"
              "year" : "1989"}
         }

Now you can loop through and query it something like this:

    for o in ownerData:
         if "Audi" in o["carMake"]:
              print("Owner %s drives a %s %s %s" % (o, o["year"], o["carMake"], o["carModel"]))

Should output:

    "Owner Shaft drives a 2015 Audi A8"

This way you can expand your data set for owners without creating multiple lists.