When a statement returns boolean value, how to use that value in an if statement

Question

I am a new learner in programming, and I was trying to check whether a word exists in a file. I put the words into a dictionary (the exercise told me to put it in a dict as a key and ignore the value), and then let the user input a word, then I tried to check whether it is in the dictionary.

words_dict = dict()

fhand = open('desktop/codingMaterial/words.txt')
for line in fhand:
    line = line.rstrip()
    t = line.split()
    for word in t:
        words_dict[word] = 0

# check if word exists
v = input('Enter a variable: ')
if v in words_dict == True:
    print('Exist.')
else:
    print('Does not exist.')

I tried to run this, but no matter which word I input, the output was always 'Does not exist.'. I think I did something wrong in the if statement, but I am not sure how to fix it.

1) print the dictionary before you check it 2) try if (v in words_dict) == True: — OneCricketeer
– OneCricketeer, Commented Feb 22, 2020 at 16:52
@cricket_007, why do you give a recommendation without reading question? — Olvin Roght
– Olvin Roght, Commented Feb 22, 2020 at 16:55
Looks like @cricket_007 is a little unpopular on this thread, but I do understand his suggestion to use a set. And I do think it's a good suggestion @MayDing. With a set you don't get duplicate strings in your test data. However, this is a school assignment for dict objects. — Todd
– Todd, Commented Feb 23, 2020 at 0:03

Ch3steR · Accepted Answer · 2020-02-23 04:00:51Z

if v in word_dict == True is evluated as if v in word_dict and word_dict==True. This is called operator chaining.

Check this byte-code using dis module.

import dis
a={'a':1,'b':2,'c':3}

In [53]: dis.dis('"a" in a == True')
  1           0 LOAD_CONST               0 ('a')
              2 LOAD_NAME                0 (a)
              4 DUP_TOP
              6 ROT_THREE
              8 COMPARE_OP               6 (in)
             10 JUMP_IF_FALSE_OR_POP    18
             12 LOAD_CONST               1 (True)
             14 COMPARE_OP               2 (==)
             16 RETURN_VALUE
        >>   18 ROT_TWO
             20 POP_TOP
             22 RETURN_VALUE
In [54]: dis.dis('"a" in a and a==True')
  1           0 LOAD_CONST               0 ('a')
              2 LOAD_NAME                0 (a)
              4 COMPARE_OP               6 (in)
              6 JUMP_IF_FALSE_OR_POP    14
              8 LOAD_NAME                0 (a)
             10 LOAD_CONST               1 (True)
             12 COMPARE_OP               2 (==)
        >>   14 RETURN_VALUE

Both are evaluated in the same way. And in your case word_dict==True is always False as word_dict is a dictionary. So, it would never enter the if block and else block is executed.

if some_bool_expr == True and can be written as if some_bool_expr and if some_bool_expr==False can be written as if not some_bool_expr.

Byte-code Instructions documentation link

LOAD_CONST and LOAD_NAME: push value onto the stack. After line 2 top of the stack is a(not 'a')

DUP_TOP: Duplicates the reference on top of the stack and pushes onto the stack. Now top of the stack is a.

ROT_THREE: Lifts second and third stack item one position up moves top down to position three. Now TOS(top of the stack) is third element (a) and 2nd element (a) is now TOS, 3rd element 'a' is now 2nd element.

COMPARE_OP: Tells the interpreter to pop the two topmost stack elements and perform an membership test(in) between them, pushing the Boolean result back onto the stack. 'a' in a is done and result is pushed onto the stack i.e True. Now stack has TOS as True and duplicate reference from DUP_TOP below it.

JUMP_IF_FALSE_OR_POP: If TOS is false, sets the bytecode counter to target and leaves TOS on the stack. Otherwise (TOS is true), TOS is popped. In our example, TOS is True so. TOS is popped. Now TOS is a.

LOAD_CONST True is pushed onto the stack. COMPARE_OP ==. True==a is done which False. Now TOS is False.

RETURN_VALUE: Returns with TOS to the caller of the function. In our example, TOS is False at this point. False is returned.

POP_TOP: Removes the top-of-stack (TOS) item.

The only difference between both the expressions is that a is evaluate twice in the 2nd one.

Also refer to this answer: https://stackoverflow.com/a/3299724/12416453

I wouldn't say "same way", as dis does show differences and because a is evaluated twice in the long version (which in general can make a difference, as a could've changed between the two evaluations).
This is a good demo. How exactly do you get this output? Nevermind, I figured it out.
This is a good demo for experienced Pythoners. Maybe a little explanation on the effect of dup and rot would help.
@Todd Yeah, I should've at least left a link to dis module documentation.
@Ch3steR, you have, "perform an inequality comparison" in your explanation. I think you meant an "'in' membership test operation".

Kelly Bundy · Accepted Answer · 2020-02-22 20:50:54Z

3

A test of how x in y == z gets evaluated, by using wrappers that print what's going on:

class C:
    def __init__(self, value):
        self.value = value
    def __eq__(self, other):
        print(self, '==', other)
        return True
    def __contains__(self, value):
        print(value, 'in', self)
        return True
    def __str__(self):
        return str(self.value)

C(1) in C(2) == C(3)

Prints 1 in 2 followed by 2 == 3, as it should.

Not 1 in 2 followed by True == 3.

Not 2 == 3 followed by 1 in True.

edited Feb 22, 2020 at 20:50

answered Feb 22, 2020 at 18:16

Kelly Bundy

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3 Comments

Ch3steR Over a year ago

Nice example to show operator chaining. Heap can you tell the differences in the examples I provided. Python docs says expr1 opr1 expr2 opr2 expr3 is evaluated as expr1 opr1 expr2 and expr2 opr2 expr3 right?

Kelly Bundy Over a year ago

@Ch3steR In your particular example, the chain version and the and version have the same meaning, they're just executed a bit differently and probably have a speed difference so small it's hard to measure. In general, they don't mean the same thing. You're only showing the first half of the documentation's sentence. The second half is "except that [expr2] is evaluated only once". A typical counter-example is when expr2 is an expression resulting in a different value when it's evaluated a second time. If it's just a variable, I need to get dirty: repl.it/repls/UnusualWhoppingMap

Ch3steR Over a year ago

Yes, got your point. And I went through the docs. And dis documentation made it clear. Nice example btw ;)

Akhil Nambiar · Accepted Answer · 2020-02-22 16:51:49Z

2

Remove the == True.
It should be just .

if v in words_dict:
    print('Exist.')

answered Feb 22, 2020 at 16:51

Akhil Nambiar

3254 silver badges19 bronze badges

Comments

kederrac · Accepted Answer · 2020-02-22 18:05:10Z

1

change if v in words_dict == True: with if v in words_dict:

your issue is related to Chaining comparison operators, both operators in and == have the same precedence and are evaluated left-to-right, an equivalent to your code is (already pointed by @Ch3steR):

if v in word_dict and word_dict==True

edited Feb 22, 2020 at 18:05

answered Feb 22, 2020 at 16:51

kederrac

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12 Comments

OneCricketeer Over a year ago

It's the same thing, no?

Olvin Roght Over a year ago

@cricket_007, no.

jizhihaoSAMA Over a year ago

@cricket_007 if v in words_dict: maybe == if (v in words_dict) == True:

May Ding Over a year ago

Thank you for your detailed explanation. So in python the '==' has a higher priority than the 'in'? Am I right?

May Ding Over a year ago

what if I want to use the 'False' value? if v not in words_dict: Is this a right statement? I run this and it worked, but I'd like to confirm this with you. Thank you

|

Todd · Accepted Answer · 2020-02-22 22:24:26Z

Verifying that the Python interpreter evaluates the subexpression words_dict == True first. And not v in words_dict (wrong assumption has been made). I see from other posts that this statement is not accurate about what the interpreter actually does. Which is pretty interesting. From other descriptions here, it appears the interp regards the in operator as being at the same level of precedence as ==. In either case, you could group the first expression to force the order of evaluation.

>>> words_dict == True
False
>>> 'hand' in words_dict == True
False
>>> 'hand' in words_dict
True
>>> ('hand' in words_dict) == True
True

If this were an example of production code, one could implement this logic as (a in b) == c if that was the intent of the expression. Or if the expression were truly intended to be a in b == c as discussed, then it could be implemented explicitly as (a in b) and (b == c).

In many coding style specifications I've seen, there's a requirement that developers use parenthesis to group their subexpressions in boolean expressions. Whether or not the author believes he has a solid grasp of order-of-precedence (without parenthesis), other developers appreciate the diligence and no-assumptions approach of using parenthesis, not only as a safeguard to ensure there aren't any incorrect assumptions, but to show the other developers the intended order of precedence with no ambiguity. This is a good case-in-point.

There are tables showing the order of precedence of operators online that list in below ==. However, this one is directly from docs.python.org and accurately shows that in and == are at the same level: https://docs.python.org/3/reference/expressions.html#operator-precedence

Not sure why you mention points. What I mean is that your demo is wrong.
Well.. the demo's not "wrong", the commands and their results are accurately displayed. But the assumption that == takes precedence over in was wrong. From other examples, it looks as if they are at the same order of precedence.
How would you use parenthesis in a in b == c without changing the meaning?
I was trying to change the meaning ;-) Adding parenthesis ensures that the logic is evaluated in the order I intend.

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