I have a doubt to confirm. Let there is a list

  f = {{l}, {c}, {d}, {e}, {1, 2, 3}, {3, 4, 5}}

And I expect it to be traversed as a normal list which I can do by,

Table[f[[i, k]], {i, Length[f]}, {k, Length[f[[i]]]}]

But if the list is

a = {b, c, d, e, {1, 2, 3}, {3, 4, 5}}

Than I get the following output

   {{}, {}, {}, {}, {1, 2, 3}, {3, 4, 5}}

Is it must that there shall be list of sublists and no single element can be kept without sublist or I am accessing it wrong way. Doesn't Mathematica automatically consider it to be sublist for obvious reason.

  • $\begingroup$ ...and you've just seen that Part[] does not do anything too useful to symbols, as opposed to what it does for lists and list-like objects. $\endgroup$ Commented Jun 3, 2013 at 18:00

1 Answer 1


In simple form:

f = {a, b};

gives you


as you expect. But what is the length of f[[1]]?


Which explains why your Table operates the way it does: the length of any symbol is 0. In fact, the length of any indivisible object (like a number or a symbol or a string) is 0, as is the length of any expression for which AtomQ is True.


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