When i = 2
1) (2%2 + 1) == (1 if 2%2 == 0 else 2)
       Evaluation of LE w/ RE (ternary operand (ternary's have 'assignment order' since they return a value))
       1 == 1 -> True
2) Similar logic to 1, but i is now mutated
3) Evaluation of ternary operand w/ another mutation on i, in this case
     if the condition is true it will return a boolean otherwise it returns the val 2
       the evaluation of the ternary condition results in the return val of 2, where 1%2 /= 0 when i = 1
       -->   Can be rewritten as follows for clarity (i%2 + 1 == 1) if (i%2 == 0) else 2 

For more info, ref : https://www.geeksforgeeks.org/ternary-operator-in-python/

rabbott wrote:

Can you explain the following?

>>> i = 2

>>> (i%2 + 1) == (1 if i%2==0 else 2)
True
>>> i = 0

>>> (i%2 + 1) == (1 if i%2==0 else 2)
True

>>> i = 1

>>> i%2 + 1 == 1 if i%2==0 else 2
2

 

1. LHS: (0+1) = 1 i.e True since it’s truthy; RHS=(1) since i%2==0 is true = 1 i.e. True since it’s Truthy; therefore LHS = RHS

2. Since 0%2 is 0; therefore same goes as above

3. Pure magic (or a possible bug?) of precedence. Mod is evaluated first, which returns 1, then + which returns 2, then == which returns False i.e. 0 as well, and finally ternary, which anyway retuns false case i.e. 2.

Edit: I think this has more to it since I just learned if else has higher precedence. Have to look at it.

Edit 2: My bad. This table lists the precedence in least to most binding order which made me confused. So ignore all edits: 

https://docs.python.org/3/reference/expressions.html#operator-precedence

jungsoolim wrote:

I tested few cases. Based on the experiments, the if-then-else function has higher precedence than mod, add, and comparison.

If-then-else is evaluated first. Then %, +, and finally ==.

(Maybe?)

i=1
i%2 + 1 == 1 if i%2==0 else 2
2
i%2 + 1 == 1 if i%2==0 else 100
100
i%2 + 1 == 1 if i%2 !=0 else 100
False
i%2 + 1 == 1 if i%2 == 1 else 100
False
i%2 + 1 == 1 if i%3 == 77 else 77
77
i%2 + 1 == 1 if i%2 == 77 else 77
77
i%2 + 1 == 1 if i%2 != 77 else 77
False
 

Don't you think it gave these outputs because the ternary is evaluated at the end?

Or, it is equivalent with

(i%2 + 1 == 1, 2)[i%2 ==0]
False
(i%2 + 1 == 1, 2)[i%2 !=0]
2

The first part is for true condition, and the second part is for false condition.

(i%2 + 1 == 1, 77)[i%2 !=0]
77

 

 

This is so clever! I really liked this one. And yes, you’re right.

rabbott wrote:

Can you explain the following?

>>> i = 2

>>> (i%2 + 1) == (1 if i%2==0 else 2)
True
>>> i = 0

>>> (i%2 + 1) == (1 if i%2==0 else 2)
True

>>> i = 1

>>> i%2 + 1 == 1 if i%2==0 else 2
2

 

Nice discussion!

As Jay pointed out, the key is the precedence of the operators. (See the table he linked to.)  The == operator binds more tightly than the if-else ternary operator. Notice that the third example (deliberately) doesn't have the parentheses the first two have. If they were put in, the result would be True as in the other examples. 

>>> (i%2 + 1) == (1 if i%2==0 else 2)
True

They will always produce the same result because:

i % 2 will either return 0 or 1. (no values other than 0/1)

When i % 2  returns 0:

"i%2+1" evaluates to 1
and 
"1 if i%2==0 else 2" satisfies the if check and returns 1.

In a similar way,
When i % 2  returns 1:

"i%2+1" evaluates to 2
and 
"1 if i%2==0 else 2" does not satisfy the if check and returns 2

Not sure, If I have understood correctly.

In order to transform, we can use Bitwise AND as shown below:

a = (i & 1) + 1

print(a)

Terrific! Just what I was hoping for.

Another way of putting the next to last step is as follows.

A tautological way to write i%2 is

0 if i%2 == 0 

i%2 =                              = 0 if i%2 == 0 else 1

1 if i%2 == 1