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

• Dumb chimp: I think readability counts here more than performance, but I could be wrong. performance
• Jjj: if statement and statement and statement : did you try this?
• Raymond C.: I could use 'and' but would need specific error codes for each condition. :(
• Dumb chimp: @RaymondC. This is probably related
• hiro protagonist: why is all of that in a while loop? neither a nor b change...

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

• Raymond C.: Interesting, I did not think of it this way. Thanks :)

Answer 2

Or another way :

 outcome = True
 if a != b:
 print("Error - 01")
 outcome &= False
 if a["test_1"] != "test_value":
 print("Error - 02")
 outcome &= False
 if "test_2" not in a:
 print("Error - 03")
 outcome &= False
 if a["test_3"] == "is long data string":
 print("Error - 04")
 outcome &= False

• Raymond C.: Hi, sorry for the late reply. Just wanted to say thanks :) Question: Why do you use &= instead of = ?
• Hank: Oh, I made it complicated, you can use = instead. Thanks

Answer 3

Hackiest way of doing this as per user request, not recommended tho, There are many ways to use structures if you are only looking to understand them. Functions like zip or list comprehension can be used once you convert to list Used list to keep positions intact.

a = {'test_1' : "test_value", 'test_4' : "None", 'test_3' : "long data string"}
b = {'test_1' : "test_value", 'test_4' : "None", 'test_3' : "long data string"}
# storing bool values, expecting True
test1 = [a==b,"test_2" in a,a["test_1"] == "test_value", a["test_3"] != "is long data string" ]
# storing error codes for False conditions 
result = ["Error - 01", "Error - 02", "Error - 03", "Error - 04"]
if False in test1:
 print(result[int(test1.index(False))])
else:
 print(True)

Error - 02
[Program finished]