Certain conditions evaluate differently then expected in Python

Question

Currently creating a piece that takes values a given value in an array and adds 32

My IDE (PyCharm) recommended that I remove a redundancy in my code by doing the following

if ascii_key[i] >= 65 and ascii_key[i] <= 90:
    ascii_key[i] = ascii_key[i] + 32

to

if 65 <= ascii_key[i] <= 90:
    ascii_key[i] = ascii_key[i] + 32

Both solutions work for me, but I'm curious as to why this arrangement of conditions breaks

if ascii_key[i] <= 65 >= 90:
    ascii_key[i] = ascii_key[i] + 32

I'd appreciate an explanation in pseudo-code if convenient, I'm still a bit novice

`a < b < c` is just shorthand notation for `(a < b) and (b < c)`. Writing `ascii_key[i] <= 65 >= 90` is the same as `(ascii_key[i] <= 65) and (65 >= 90)`, which is probably not what you want. — jfaccioni, Feb 13 '20 at 20:59
See [chained comparison](https://docs.python.org/3/reference/expressions.html#comparisons) — Patrick Artner, Feb 13 '20 at 21:01
[How do chained comparisons in Python actually work?](https://stackoverflow.com/q/28754726/1324033) — Sayse, Feb 13 '20 at 21:01

score 3 · Accepted Answer · answered Feb 13 '20 at 21:00

In the third code provided by you.

ascii_key[i] <= 65 >= 90 this is always evaluated to False.

ascii_key[i] <= 65 >= 90 is evaluated as ascii_key[i] <= 65 and 65 >= 90. 65>=90 is False.

Byte-code evaluation of a<b<c.

In [17]: dis.dis('a<b<c')
  1           0 LOAD_NAME                0 (a)
              2 LOAD_NAME                1 (b)
              4 DUP_TOP
              6 ROT_THREE
              8 COMPARE_OP               0 (<)
             10 JUMP_IF_FALSE_OR_POP    18
             12 LOAD_NAME                2 (c)
             14 COMPARE_OP               0 (<)
             16 RETURN_VALUE
        >>   18 ROT_TWO
             20 POP_TOP
             22 RETURN_VALUE

Byte-code evaluation of a<b and b<c.

In [18]: dis.dis('a<b and b<c')
  1           0 LOAD_NAME                0 (a)
              2 LOAD_NAME                1 (b)
              4 COMPARE_OP               0 (<)
              6 JUMP_IF_FALSE_OR_POP    14
              8 LOAD_NAME                1 (b)
             10 LOAD_NAME                2 (c)
             12 COMPARE_OP               0 (<)
        >>   14 RETURN_VALUE

You can observe observe a<b<c is equivalent to a<b and b<c.

score 0 · Answer 2 · answered Feb 13 '20 at 21:08

it is due to how the conditions are resolved. if ascii_key[i] >= 65 and ascii_key[i] <= 90 and if 65 <= ascii_key[i] <= 90 are equivalent. While if ascii_key[i] <= 65 >= 90 gets resolved to this: if ascii_key[i] <= 65 and 65 >= 90 65 >= 90 is never true.

First the condition 65 <= ascii_key[i] and then the other condition ascii_key[i] <= 90 are evaluated and chained with an AND.

Cheers

Certain conditions evaluate differently then expected in Python

2 Answers2