How to write inline if statement for print?

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 

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) 

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 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. 

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 '' 

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

Answer

If your print statement must print an empty line when the expression is false, the correct syntax is:

print(a if b else '') 

The reason is you're using the conditional expression which has two mandatory clauses, one when b is true preceding if, one when b is false following else.

Both clauses are themselves expressions. The conditional expression is also called the ternary operator, making it clear it operates on three elements, a condition and two expressions. In your code the else part is missing.

However I guess your idea was not to print an empty line when the expression was false, but to do nothing. In that case you have to use print within a conditional statement, in order to condition its execution to the result of the evaluation of b.

Details: Expression vs. statement

The conditional statement can be used without the else part:

• The if statement is a compound statement with further instructions to execute depending on the result of the condition evaluation.

• It is not required to have an else clause where the appropriate additional instructions are provided. Without it, when the condition is false no further instructions are executed after the test.

• The conditional expression is an expression. Any expression must be convertible to a final value, regardless of the subsequent use of this value by subsequent statements (here the print statement).

• Python makes no assumption about what should be the value of the expression when the condition is false.

You could have used an if statement this way:

if b: print(a) 

Note the difference:

• There is no instructions executed by the if statement when the condition is false, nothing is printed because print is not called.

• In a print statement with a conditional expression print(a if b else ''), print is always executed. What it prints is the if conditional expression. This expression is evaluated prior to executing the print statement. So print outputs an empty line when the condition is false.

Note your other attempt print(a if b==True) is just equivalent to the first one.

Since b==True is syntactically equivalent to b, I guess Python interpreter just replaces the former by the latter before execution. Your second attempt likely comes down to a repeat of the first one.

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'