Indexing list of lists

[Hilde Roth]

So you want l[seq] to be a shorter way for current

[l for i in seq]

Which is pythonic as it is explicit, easier to read if the code is not
written by you yesterday

But consistency is pythonic, too ;-) and, as I pointed in another message,
if you accept l[sequence_abstraction], why not l[actual_sequence]??

Hilde

 

[David C. Fox]

This is a red herring.

No, it's not. It is a hint that, despite the similarity of notation
between Matrix[j] and NestedList[j], there is something
fundamentally different between the two. See below for another example.

What do you expect a[;1] to return if a = [[], [1, 2, 3], [4], 5]?

Whatever error python returns if you ask, e.g., for (1 2 3)[4].

Hilde

Okay, here's a better example which I thought of just after I posted my
previous reply:

Given

x = [[1, 2], [3, 4]]

the statement

x[0] = [5, 6]

will result in a nested list x = [ [5, 6], [3, 4] ]. If you think of x
as a multi-dimensional array represented by a nested list, then I've
just replaced a row of the original array

1 2
3 4

with a new row, yielding

5 6
3 4

If we had an x[;0] notation, then you would expect to be able to do the
same thing to replace a column:

x[;0] = [7, 8]

Unfortunately, there is no pre-existing list representing the first
column of x, so x[;0] has to return a new list [1, 3], and assigning to
that new list has no affect on x.

Again, my point is that nested lists are a fundamentally different
structure than multi-dimensional arrays. For simple things like
x[j], you can use a nested list to represent a multi-dimensional
array, but if you actually want to manipulate a two-dimensional array
like a matrix, you are better off using a class designed for that
purpose, like the ones defined in Numeric Python.

David