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Code:
for i in range(1000):
print(i) if i%10==0 else pass
Error:
File "<ipython-input-117-6f18883a9539>", line 2
print(i) if i%10==0 else pass
^
SyntaxError: invalid syntax
Why isn't 'pass' working here?
795
Answer 526897a4abf821c5f4002967 If you require code to run only when the statement returns true (and do nothing else if false) then an else statement is not needed. On the other hand if you need a code to execute “A” when true and “B” when false, then you can use the if / else statement.
If you want your last else to do nothing, you can use the continue keyword inside it. For any conditional you can define the conditional to do nothing by defining an empty code block {} for it or by simply ommit the case.
Answer: use _ = 0 .
Simply remove else { return false } .
This is not a good way of doing this, if you see this problem the structure of your code might not be good for your desires, but this will helps you:
print(i) if i%10==0 else None
This is not a direct answer to your question, but I would like suggest a different approach.
First pick the elements you want to print, then print them. Thus you'll not need empty branching.
your_list = [i for i in range(100) if i%10]
# or filter(lambda e: e%10 == 0, range(100))
for number in your_list:
print number
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