Use and Meaning of "In" in an If Statement

Use and meaning of in in an if statement?

It depends on what next is.

If it's a string (as in your example), then in checks for substrings.

>>> "in" in "indigo"
True
>>> "in" in "violet"
False
>>> "0" in "10"
True
>>> "1" in "10"
True

If it's a different kind of iterable (list, tuple, set, dictionary...), then in checks for membership.

>>> "in" in ["in", "out"]
True
>>> "in" in ["indigo", "violet"]
False

In a dictionary, membership is seen as "being one of the keys":

>>> "in" in {"in": "out"}
True
>>> "in" in {"out": "in"}
False

Python's equivalent of && (logical-and) in an if-statement

Use and instead of &&.

What does if condition mean, when the condition is not a statement?

It checks if x is "truthy", which means that it shouldn't be :

• None
• False
• Zero
• Empty

As an example :

if 1 and True and "string" and ["list"] and {'key' : 'value'} and\
    not None and not 0 and not False and not "" and not [] and not {}:
  print "I understood truthiness!"
# => I understood truthiness!

Meaning of conditional statements in if[] then statement in makefile

man test

(And your question is about shell, not make).

Meaning of if statement without == operator

If you run it in MATLAB when fl = 0, nothing happens, showing that 0 means false. If you change fl to anything but 0 then the code executes and would print out whatever code is in the if, for the case of a = 5 it would print a = 6

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

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

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