Python: Values being assigned for wrong IF conditional

Question

I was messing around with Python and was making a little planet simulator. I have one section that is supposed to receive a value and based on that value return a second value. I spent a few hours troubleshooting this, but I'm absolutely stumped. I added some print messages to help me track the values and it seems like it should be working, but isn't. I'm almost sure it is something obvious. What am I doing wrong?

Here is the code in question:

def assignTerrain(pType):

print("Assigning terrain values to planet type: ", planetType[pType], pType)

if pType == 0:
    print("Value 0 assigned", pType)
    return 0

if pType == 1 or 2:
    temp = random.randint(1, 11)
    print("Value %d assigned" % temp, pType)
    return temp

if pType == 3 or 4:
    print("Value 12 assigned", pType)
    return 12

print("There was an error with pType: ", pType)

As a sample, here is an example of the output I'm getting:

Assigning terrain values to planet type:  Asteroid Belt 4
Value 3 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 5 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 4 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 7 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 8 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 2 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 4 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 1 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 9 assigned 4
Assigning terrain values to planet type:  Asteroid Belt 4
Value 8 assigned 4

It seems to me that pType 4 should be skipping the first two IF statements and receive a value from the third, but it looks like it is getting caught by the 1 or 2. Any thoughts?

Instead of if pType == 1 or 2 try if pType in (1, 2) or if pType == 1 or pType == 2. Comparators in Python don't work the way you are thinking they do. — sberry
– sberry, Commented Jun 12, 2013 at 5:13

sberry · Accepted Answer · 2013-06-12 05:17:06Z

3

Instead of if

pType == 1 or 2

try

if pType in (1, 2)

or

if pType == 1 or pType == 2.

Comparators in Python don't work the way you are thinking they do. What you are doing actually evaluates to

if ((pType == 1) or (4))

Which, since 4 is truthy, is always True.

answered Jun 12, 2013 at 5:17

sberry

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1 Comment

Moonraker Over a year ago

Thank you. I feel so sheepish! I remember thinking that isn't how I'm used to IF working and thought I'd try it new, but somehow never thought to use a more traditional approach. I appreciate your concise and helpful answer.

thomasd · Accepted Answer · 2013-06-12 05:15:49Z

2

The expression pType == 1 or 2 is the logical union of the two expressions pType == 1, which could be True or False, and 2, which is always True to Python. Thus, the expression pType == 1 or 2 is always True regardless of the value of pType.

answered Jun 12, 2013 at 5:15

thomasd

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