Python: is index() buggy at all?

Question

I'm working through this thing on pyschools and it has me mystified. Here's the code:

def convertVector(numbers):
    totes = []
    for i in numbers:
        if i!= 0:
            totes.append((numbers.index(i),i))
    return dict((totes))

Its supposed to take a 'sparse vector' as input (ex: [1, 0, 1 , 0, 2, 0, 1, 0, 0, 1, 0]) and return a dict mapping non-zero entries to their index. so a dict with 0:1, 2:1, etc where x is the non zero item in the list and y is its index.

So for the example number it wants this: {0: 1, 9: 1, 2: 1, 4: 2, 6: 1} but instead gives me this: {0: 1, 4: 2} (before its turned to a dict it looks like this: [(0, 1), (0, 1), (4, 2), (0, 1), (0, 1)]

My plan is for i to iterate through numbers, create a tuple of that number and its index, and then turn that into a dict. The code seems straightforward, I'm at a loss. It just looks to me like numbers.index(i) is not returning the index, but instead returning some other, unsuspected number.

Is my understanding of index() defective? Are there known index issues? Any ideas?

Answer 1

index() only returns the first:

>>> a = [1,2,3,3]
>>> help(a.index)
Help on built-in function index:

index(...)
    L.index(value, [start, [stop]]) -> integer -- return first index of value.
    Raises ValueError if the value is not present.

If you want both the number and the index, you can take advantage of enumerate:

>>> for i, n in enumerate([10,5,30]):
...     print i,n
... 
0 10
1 5
2 30

and modify your code appropriately:

def convertVector(numbers):
    totes = []
    for i, number in enumerate(numbers):
        if number != 0:
            totes.append((i, number))
    return dict((totes))

which produces

>>> convertVector([1, 0, 1 , 0, 2, 0, 1, 0, 0, 1, 0])
{0: 1, 9: 1, 2: 1, 4: 2, 6: 1}

[Although, as someone pointed out though I can't find it now, it'd be easier to write totes = {} and assign to it directly using totes[i] = number than go via a list.]

Answer 2

What you're trying to do, it could be done in one line:

>>> dict((index,num) for index,num in enumerate(numbers) if num != 0)
{0: 1, 2: 1, 4: 2, 6: 1, 9: 1}