CODEWITHSUNDEEP
pythonpython-3.xfunctionreturn-value
Amanda
Amanda

Reputation: 1

Function returning same value for different arguments

I am very new to Python (and using Python 3). Apologies---I know I must be making a very basic mistake.

Here is the structure of the mistake and then I'll give an example: I wrote a function func(x). To test that it does the "right" things, I plugged in value x1, i.e. print(func(x1)). The output was y1 which was correct. Then I tried print(func(x2)). That again gave me y1 and not the correct y2. And similarly with other values of x. At first I thought it was a mistake with that specific function, but then I had similar issues with other functions. Am I making a mistake common to all the functions?

Below is one example; I've tried this (and other functions) with several different variations and still had the problem. I can give other examples, if that would help. 

def num_to_day(x):
   if 0:
      return "Sunday"
   elif 1:
      return "Monday"
   elif 2:
      return "Tuesday"
   elif 3:
      return "Wednesday"
   elif 4:
      return "Thursday"
   elif 5:
      return "Friday"
   elif 6:
      return "Saturday"
   else:
      return "Not Valid"
print(num_to_day(5))

Upvotes: 0

Views: 3024

Answers (3)

Brad Solomon
Brad Solomon

Reputation: 40878

With your if statements, you need to specify x equals each number--for example, if x == 0:.

Currently, you're testing the truth value of a number itself, with no relation to what x you have specified. To see this, try:

if 1:
    print('Monday')
Monday

You might find it useful to know that in Python, 0 evaluates to False while other integers return True:

print([bool(num) for num in range(7)])
[False, True, True, True, True, True, True]

bool tests the truth value of its argument.

Lastly, one alternative of many would be to lookup your weekday from a dict. .get allows you to specify a value that gets returned if an error is thrown.

def num_to_day(x):
    days =  {
        0 : 'Sunday',
        1 : 'Monday',
        2 : 'Tuesday',
        3 : 'Wednesday',
        4 : 'Thursday',
        5 : 'Friday',
        6 : 'Saturday'
        }
    return days.get(x, 'Not Valid')

num_to_day(0)
Out[54]: 'Sunday'

num_to_day(10)
Out[55]: 'Not Valid'

Upvotes: 4

user3053452
user3053452

Reputation: 675

You're not using your argument inside your function at all.

def num_to_day(x):
   if x == 0: # <-- Now x gets evaluated
      return "Sunday"
   elif x == 1:
      return "Monday"
   elif x == 2:
      return "Tuesday"
   elif x == 3:
      return "Wednesday"
   elif x == 4:
      return "Thursday"
   elif x == 5:
      return "Friday"
   elif x == 6:
      return "Saturday"
   else:
      return "Not Valid"
print(num_to_day(5))

Upvotes: 2

Błotosmętek
Błotosmętek

Reputation: 12927

You're passing an argument to your function and then not using it at all. To make if work, you must actually compare x to all those numbers, simple if 0: is equivalent to if False:, not to if x == 0:.

And anyway, this cascade of elifs is a prime example of how not to code. A much better solution would be to use a list or a dictionary:

def num_to_day(x):
    days_of_week = [ "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday" ]
    return days_of_week[x]

Upvotes: 0

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