Accessing user input taken in while loop (Python)

Question

I want user to enter random numbers. I run a while loop until the user enters 0.

while input()!=0:
    print "Your number is:...??"

My question is: Are there any special variables (like $_ in Perl) using which I can access the user input? For example (in the above case) I want to print what the user has entered.

wim · Accepted Answer · 2013-02-25 06:34:21Z

2

You need to assign the result to a variable:

s = None
while s != 0:
  s = int(input("Enter number: "))
  print("Your number is: {}".format(s))

This is a python3 example. For python 2.x you should be using raw_input instead.

answered Feb 25, 2013 at 6:34

wim

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1 Comment

Holmes Over a year ago

Hmm.. It seems we can not run this loop without creating a variable. Thank you.

Jon Clements · Accepted Answer · 2013-02-25 06:36:03Z

1

An alternative method is to use iter and provide a callable:

for num in iter(lambda: int(raw_input('Your number is: ')), 0):
    print 'You entered', num

answered Feb 25, 2013 at 6:36

Jon Clements

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4 Comments

Octipi Over a year ago

This is nice. According to the docs, the value of the callable will be returned unless it is the sentinel which will raise StopIteration.

Holmes Over a year ago

@Jon This is nice method. Thank you. But I never clearly understand what lambda exactly does. Any help is appreciated.

Jon Clements Over a year ago

@Holmes lambda creates a callable... You can think of it as a less flexible def... So the above could define def somefunc(): raw_input('Your number is: ') and instead of the lambda, use somefunc.... - see docs.python.org/2/reference/expressions.html#lambda

Jon Clements Over a year ago

@Holmes In fairness, I should point out although I prefer this method - it's not an oft-used syntax of the language, so not all Python programmers will immediately grok it - while everyone will grok the while and break system...

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