How to write inline if statement for print?

I need to print some stuff only when a boolean variable is set to True. So, after looking at this, I tried with a simple example:

>>> a = 100
>>> b = True
>>> print a if b
  File "<stdin>", line 1
    print a if b
             ^
SyntaxError: invalid syntax  

Same thing if I write print a if b==True.

What am I missing here?

Python does not have a trailing if statement.

There are two kinds of if in Python:

1. if statement:

   if condition: statement
   if condition:
       block
   

2. if expression (introduced in Python 2.5)

   expression_if_true if condition else expression_if_false
   

And note, that both print a and b = a are statements. Only the a part is an expression. So if you write

print a if b else 0

it means

print (a if b else 0)

and similarly when you write

x = a if b else 0

it means

x = (a if b else 0)

Now what would it print/assign if there was no else clause? The print/assignment is still there.

And note, that if you don't want it to be there, you can always write the regular if statement on a single line, though it's less readable and there is really no reason to avoid the two-line variant.

Inline if-else EXPRESSION must always contain else clause, e.g:

a = 1 if b else 0

If you want to leave your 'a' variable value unchanged - assing old 'a' value (else is still required by syntax demands):

a = 1 if b else a

This piece of code leaves a unchanged when b turns to be False.

The 'else' statement is mandatory. You can do stuff like this :

>>> b = True
>>> a = 1 if b else None
>>> a
1
>>> b = False
>>> a = 1 if b else None
>>> a
>>> 

EDIT:

Or, depending of your needs, you may try:

>>> if b: print(a)

If you don't want to from __future__ import print_function you can do the following:

a = 100
b = True
print a if b else "",  # Note the comma!
print "see no new line"

Which prints:

100 see no new line

If you're not aversed to from __future__ import print_function or are using python 3 or later:

from __future__ import print_function
a = False
b = 100
print(b if a else "", end = "")

Adding the else is the only change you need to make to make your code syntactically correct, you need the else for the conditional expression (the "in line if else blocks")

The reason I didn't use None or 0 like others in the thread have used, is because using None/0 would cause the program to print None or print 0 in the cases where b is False.

If you want to read about this topic I've included a link to the release notes for the patch that this feature was added to Python.

The 'pattern' above is very similar to the pattern shown in PEP 308:

> This syntax may seem strange and backwards; why does the condition go > in the middle of the expression, and not in the front as in C's c ? x > : y? The decision was checked by applying the new syntax to the > modules in the standard library and seeing how the resulting code > read. In many cases where a conditional expression is used, one value > seems to be the 'common case' and one value is an 'exceptional case', > used only on rarer occasions when the condition isn't met. The > conditional syntax makes this pattern a bit more obvious: > > contents = ((doc + '\n') if doc else '')

So I think overall this is a reasonable way of approching it but you can't argue with the simplicity of:

if logging: print data

Since 2.5 you can use equivalent of C’s ”?:” ternary conditional operator and the syntax is:

[on_true] if [expression] else [on_false]

So your example is fine, but you've to simply add else, like:

print a if b else ''

This can be done with string formatting. It works with the % notation as well as .format() and f-strings (new to 3.6)

print '%s' % (a if b else "")

or

print '{}'.format(a if b else "")

or

print(f'{a if b else ""}')

You can write an inline ternary operator like so:

sure = True

# inline operator
is_true = 'yes' if sure else 'no'

# print the outcome
print(is_true)

You can use:

print (1==2 and "only if condition true" or "in case condition is false")

Just as well you can keep going like:

print (1==2 and "aa" or ((2==3) and "bb" or "cc"))

Real world example:

>>> print ("%d item%s found." % (count, (count!=1 and 's' or '')))
1 item found.
>>> count = 2
>>> print ("%d item%s found." % (count, (count!=1 and 's' or '')))
2 items found.

For your case this works:

a = b or 0

Edit: How does this work?

In the question

b = True

So evaluating

b or 0

results in

True

which is assigned to a.

If b == False?, b or 0 would evaluate to the second operand 0 which would be assigned to a.

Try this . It might help you

a=100
b=True

if b:
   print a

You're simply overcomplicating.

if b:
   print a

You always need an else in an inline if:

a = 1 if b else 0

But an easier way to do it would be a = int(b).

hmmm, you can do it with a list comprehension. This would only make sense if you had a real range.. but it does do the job:

print([a for i in range(0,1) if b])

or using just those two variables:

print([a for a in range(a,a+1) if b])

Well why don't you simply write:

if b:
    print a
else:
    print 'b is false'