[python] 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?

You're simply overcomplicating.

if b:
   print a

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 ''

Well why don't you simply write:

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

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.

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)

Try this . It might help you

a=100
b=True

if b:
   print a

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.

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 ""}')

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.

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).

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

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])