[python] What is the correct syntax for 'else if'?

I'm a new Python programmer who is making the leap from 2.6.4 to 3.1.1. Everything has gone fine until I tried to use the 'else if' statement. The interpreter gives me a syntax error after the 'if' in 'else if' for a reason I can't seem to figure out.

def function(a):
    if a == '1':
        print ('1a')
    else if a == '2'
        print ('2a')
    else print ('3a')

function(input('input:'))

I'm probably missing something very simple; however, I haven't been able to find the answer on my own.

def function(a):
    if a == '1':
        print ('1a')
    elif a == '2':
        print ('2a')
    else:
        print ('3a')

def function(a):
    if a == '1':
        print ('1a')
    else if a == '2'
        print ('2a')
    else print ('3a')

Should be corrected to:

def function(a):
    if a == '1':
        print('1a')
    elif a == '2':
        print('2a')
    else:
        print('3a')

As you can see, else if should be changed to elif, there should be colons after '2' and else, there should be a new line after the else statement, and close the space between print and the parentheses.

since olden times, the correct syntax for if/else if in Python is elif. By the way, you can use dictionary if you have alot of if/else.eg

d={"1":"1a","2":"2a"}
if not a in d: print("3a")
else: print (d[a])

For msw, example of executing functions using dictionary.

def print_one(arg=None):
    print "one"

def print_two(num):
    print "two %s" % num

execfunctions = { 1 : (print_one, ['**arg'] ) , 2 : (print_two , ['**arg'] )}
try:
    execfunctions[1][0]()
except KeyError,e:
    print "Invalid option: ",e

try:
    execfunctions[2][0]("test")
except KeyError,e:
    print "Invalid option: ",e
else:
    sys.exit()

Here is a little refactoring of your function (it does not use "else" or "elif"):

def function(a):
    if a not in (1, 2):
        a = 3
    print(str(a) + "a")

@ghostdog74: Python 3 requires parentheses for "print".

Do you mean elif?