Thread: Invalid syntax in Python

Can someone tell me what the syntax error in this program is? The program attempts to check a list for duplicates and return True if it has any, False if it has none.

def dupl(L):
----if len(L) <= 1:
--------return False
----elif L[0] in L[1:]:
--------return True
----else:
--------return dupl(L[1:])

The syntax error is marked in bold.

Originally Posted by quiney

Can someone tell me what the syntax error in this program is? The program attempts to check a list for duplicates and return True if it has any, False if it has none.

def dupl(L):
----if len(L) <= 1:
--------return False
----elif L[0] in L[1:]:
--------return True
----else:
--------return dupl(L[1:])

The syntax error is marked in bold.

Works perfectly fine on my computer. Not sure why you would be getting a syntax error.

I'm assuming those hyphens are just in order to preserve indentation on this forum. You can use [code][/code] tags, like so:

Code:

def dupl(L):
    if len(L) <= 1:
        return False
    elif L[0] in L[1:]:
        return True
    else:
        return dupl(L[1:])

L1 = [1,2,3,4,5]
L2 = [1,2,3,2,5]

if dupl(L1):
    print "hello"
if dupl(L2):
    print "goodbye"

Output:

Code:

goodbye

Thanks. The program works fine now. I have one more question. Would something like the following program accomplish what I'm trying to do?

Code:

def dup(L):
	i=1
	dup=False
	if len(L)==0:
		return dup
	else:
		while i<=len(L)-1:
			if sorted(L)[i]==sorted(L)[i-1]:
				dup=True
			else:
				i=i+1
		return dup

When I run this program, for some lists, it returns the correct answer, but for others, the program freezes up.

Originally Posted by quiney

Thanks. The program works fine now. I have one more question. Would something like the following program accomplish what I'm trying to do?

Code:

def dup(L):
	i=1
	dup=False
	if len(L)==0:
		return dup
	else:
		while i<=len(L)-1:
			if sorted(L)[i]==sorted(L)[i-1]:
				dup=True
			else:
				i=i+1
		return dup

When I run this program, for some lists, it returns the correct answer, but for others, the program freezes up.

Probably where you wrote dup=True you meant to write return True. What you have causes an infinite loop in the event of a duplicate.

Or maybe you meant to write a break statement, but might as well just return, it's simpler.

Instead of calling the sorted method so many times, why not just write L2 = sorted(L) near the beginning of the function?

Also after you replace dup=True with return True you'll see that the variable dup is superfluous.

Thanks! So far I have this:

Code:

def dup(L):
	i=1
	L2=sorted(L)
	if len(L)==0:
		return False
	else:
		while i<=len(L)-1:
			if L2[i]==L2[i-1]:
				return True
			else:
				i=i+1

This returns True whenever there is a duplicate, but I still don't know how to get it to return false when there is no duplicate and the list is greater than 0.

Originally Posted by quiney

Thanks! So far I have this:

This returns True whenever there is a duplicate, but I still don't know how to get it to return false when there is no duplicate and the list is greater than 0.

I took the liberty of changing a few things around. Compare, and use whatever aspects of it you like

Code:

def dupl(L):
    if len(L)<2:
        return False
    i=0
    L2=sorted(L)
    while i<len(L)-1:
        if L2[i]==L2[i+1]:
            return True
        i=i+1
    return False

L1 = [1,2,3,4,5]
L2 = [1,2,3,2,5]

if dupl(L1):
    print "hello"
if dupl(L2):
    print "goodbye"

Note in particular that a "return" can render an "else" unnecessary.

Thanks. I also figured out a much shorter way to do it:

Code:

def dupl(L):
	if len(L)==len(set(L)):
		return False
	return True

Originally Posted by quiney

Thanks. I also figured out a much shorter way to do it:

Code:

def dupl(L):
	if len(L)==len(set(L)):
		return False
	return True

Nice! I come from a non-Python background so I have not seen this before, but it makes perfect sense. However, you can save some keystrokes as follows:

Code:

def dupl(L):
    return len(L)!=len(set(L))