You can use slicing:

for item in some_list[2:]:
    # do stuff

This will start at the third element and iterate to the end.

Index is an object, and default index starts from 0:

>>> result.index
Int64Index([0, 1, 2], dtype=int64)

You can shift this index by 1 with

>>> result.index += 1 
>>> result.index
Int64Index([1, 2, 3], dtype=int64)

Python (and therefore sage) lists are always numbered from 0, and there isn't a way to change that.

Looking at CPython's source, in http://hg.python.org/cpython/file/70274d53c1dd/Objects/listobject.c on line 449:

static PyObject *
list_item(PyListObject *a, Py_ssize_t i)
{
    if (i < 0 || i >= Py_SIZE(a)) {
        if (indexerr == NULL) {
            indexerr = PyString_FromString(
                "list index out of range");
            if (indexerr == NULL)
                return NULL;
        }
        PyErr_SetObject(PyExc_IndexError, indexerr);
        return NULL;
    }
    Py_INCREF(a->ob_item[i]);
    return a->ob_item[i];
}

The item lookup delegates straight to the underlying C array, and C arrays are always zero-based. So Python lists are always zero-based as well.

You don't have that choice. Python indexing starts at 0, and is not configurable.

You can just subtract 1 from your indices when indexing:

array.insert(i - 1, element)  # but better just use array.append(element)
print(i, array[i - 1])

or (more wasteful), start your list with a dummy value at index 0:

array = [None]

at which point the next index used will be 1.

Just assign directly a new index array:

df.index = np.arange(1, len(df)+1)

Or if the index is already 0 based, just:

df.index += 1

Example:

In [151]:

df = pd.DataFrame({'a': np.random.randn(5)})
df
Out[151]:
          a
0  0.443638
1  0.037882
2 -0.210275
3 -0.344092
4  0.997045
In [152]:

df.index = np.arange(1, len(df)+1)
df
Out[152]:
          a
1  0.443638
2  0.037882
3 -0.210275
4 -0.344092
5  0.997045

TIMINGS

For some reason I can't take timings on reset_index but the following are timings on a 100,000 row df:

In [160]:

%timeit df.index = df.index + 1
The slowest run took 6.45 times longer than the fastest. This could mean that an intermediate result is being cached 
10000 loops, best of 3: 107 µs per loop


In [161]:

%timeit df.index = np.arange(1, len(df)+1)
10000 loops, best of 3: 154 µs per loop

So without the timing for reset_index I can't say definitively, however it looks like just adding 1 to each index value will be faster if the index is already 0 based