While loop in a function with input()

Question

So I'm trying to get this function to return what I write only if it is equal to a character in an array. If not I want it to keep looping.

def pick():
    picks = input('Type one letter. ')
    choice = {'q', 'w', 'e', 'r', 't', 'y'}
    for x in choice:
        while x in choice != picks:
            picks = input('Pick again. ')
        else:
            return x
pick()

I am just getting really confused with this.

Example:

Type one letter. z

Pick again. q

then it will return q to the function to be used in another function.

Also It has to only continue to the next function if this one is right (returns the proper character). The other function while compare this functions answer to its own. Will it stop other functions from "starting" while this one keeps looping if it is not right?

Can you explain what are you expecting while x in choice != picks: to do? — Tom Dalton
– Tom Dalton, Commented Jan 18, 2016 at 0:31
If you were calling this a lot of times, it might be worth defining choice outside the pick function. — Paul Rooney
– Paul Rooney, Commented Jan 18, 2016 at 0:44

Suever · Accepted Answer · 2016-01-18 00:34:49Z

3

It is unnecessary to loop through choices every time. Also I'm not quite sure what you're trying to do with your while condition, but something like this should do the trick. The motif x in y returns a boolean which indicates if x is a member of y.

def pick():

    picks = input('Type one letter. ')
    choice = {'q', 'w', 'e', 'r', 't', 'y'}

    while picks not in choice:
        picks = input('Pick again. ')

    return picks

pick()

Also I would probably do a little bit to cleanup the user's input.

while picks.lower().rstrip() not in choice:

answered Jan 18, 2016 at 0:34

Suever

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3 Comments

Will Over a year ago

While your answer is greatly improved over OP's original code, it doesn't run: NameError: name 'q' is not defined. This is primarily because of the way input() is being used instead of raw_input().

Suever Over a year ago

@Will it's a python 3.x question in which the behavior of input() is different. Oh the joys of 2.x vs 3.x

Will Over a year ago

Oh, fun! Fixed my answer and included Python 2 and 3 examples then :) Sorry, upvoted!

Will · Accepted Answer · 2016-01-18 00:42:45Z

1

Try (Python 3):

def pick():
    choices = {'q', 'w', 'e', 'r', 't', 'y'}

    picked = input('Type one letter: ')

    while picked not in choices:
        picked = input('Pick again: ')

    return picked

pick()

For Python 2:

def pick():
    choices = {'q', 'w', 'e', 'r', 't', 'y'}

    picked = raw_input('Type one letter: ')

    while picked not in choices:
        picked = raw_input('Pick again: ')

    return picked

pick()

You don't need to iterate through choices every time as in checks against every member of the set.

When you use input() in Python 2, Python will try to interpret the input as a variable, but raw_input() will return a string.

And, yes, this function will block until valid input is received, stopping subsequent functions from executing.

edited Jan 18, 2016 at 0:42

answered Jan 18, 2016 at 0:39

Will

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