Multiple if conditions, without nesting

Question

I have a sample piece of code below that I need some help on. The code sets the 'outcome' variable to False in the beginning and only should become True if all the 'if' conditions are met. Is there a more efficient way of doing this? I am trying to avoid nested 'if' statements.

Thanks!

    outcome = False

    while True:
        if a != b:
            print("Error - 01")
            break

        if a["test_1"] != "test_value":
            print("Error - 02")
            break

        if "test_2" not in a:
            print("Error - 03")
            break

        if a["test_3"] == "is long data string":
            print("Error - 04")
            break

        outcome = True
        break

Answer 1

I would write it like this, so the function ends once it encounters an error, and the most likely error should be on top, and if it encounters that error, it will return False and end the function. Else, it will check for all the other errors and then eventually conclude that the outcome is indeed True.

# testOutcome returns a boolean
outcome = testOutcome(a,b)

# Expects a and b
# Breaks out of the function call once error happens
def testOutcome(a,b):
    # Most likely error goes here
    if a != b:
        print("Error - 01")
        return False

    # Followed by second most likely error
    elif a["test_1"] != "test_value":
         print("Error - 02")
         return False

    # Followed by third most likely error
    elif "test_2" not in a:
         print("Error - 03")
         return False

    # Least likely Error
    elif a["test_3"] == "is long data string":
         print("Error - 04")
         return False

    else:
        return True