passing variable arguments in python

Question

I need to define a function that may or may not take one argument of the 3 arguments defined in a function. However, I get an error message as invalid syntax.

Now, if I make my third argument as variable [value3], I get an error message as 'float' object is not iterable.

Also, I have realized that when all the arguments are passed, it creates a tuple, which is unfavorable.

Could someone help me solve the problem?

def createValues(value1, *value2, value3):
    value = dict()
    value["VALUE1"] = value1
    value["VALUE2"] = value2
    value["VALUE3"] = value3

    print (value["VALUE1"],value["VALUE1"],value["VALUE1"])


createValues(2000,21000001,1)
createValues(2000,,1)

Can you please make examples how your function is supposed to behave when called with 0, 1, 2, 3 arguments? — timgeb, May 23 '17 at 09:00

score 3 · Answer 1 · answered May 23 '17 at 09:03

What you want is default arguments. This allows your function to be called by passing only a part of its parameters, and defaults the other to a pre-defined value.

For instance:

def createValues(value1=12, value2=27, value3=42):
    value = dict()
    value["VALUE1"] = value1
    value["VALUE2"] = value2
    value["VALUE3"] = value3

    print (value["VALUE1"],value["VALUE1"],value["VALUE1"])

will allow you to call your function by either of the following ways:

>>> createValues(1, 2, 3)
1, 2, 3
>>> createValues()
12, 27, 42
>>> createValues(value2=1)
12, 1, 42
>>> createValues(0, 5)
0, 5, 42

Since you seem confused with the * unary operator, I suggest that you read a bit about arguments unpacking (for example, check this post).

Besides, using two commas as in createValues(2000,,1) is in no way a valid syntax in Python.

score 0 · Answer 2 · answered May 23 '17 at 09:07

From my beginners point of view, I see two ways :

Using lists

The arg of createValues become a list so you have as many values as you want.

def createValues(values):
    value = dict()
    #Inserting
    for val in iter(values):
        index = str(values.index(val))
        value["VALUE"+index] = val

    #Printing
    for (key, value) in value.iteritems(): 
        print ("Key : " + key + " | Value : " + value)


createValues([2000,21000001,1])
createValues([2000,1])

Using default value

3 args for your specific case

def createValues(value1, value3, value2 = None):
    value = dict()
    value["VALUE1"] = value1
    if value2 is not None:
        value["VALUE2"] = value2
    value["VALUE3"] = value3

    print (value["VALUE1"],value["VALUE1"],value["VALUE1"])


createValues(2000,21000001,1)
createValues(value1 = 2000, value3 = 1)

score 0 · Answer 3 · answered May 23 '17 at 09:08

You cannot call a function like this:

createValues(2000,,1)

You can however check if the argument is None and work with that.

def createValues(value1, value2, value3):
    value = dict()
    value["VALUE1"] = value1 or 0
    value["VALUE2"] = value2 or 0
    value["VALUE3"] = value3 or 0

    print (value["VALUE1"],value["VALUE1"],value["VALUE1"])

Now, you can use the function like this:

createValues(2000, None, 1)

I have used the default value 0 for any missing argument.

Also, the arguments are not being converted into a tuple. You are packing the arguments into a tuple when printing. Try this instead (note the lack of parenthesis):

print value["VALUE1"], value["VALUE1"], value["VALUE1"]

passing variable arguments in python

3 Answers3