Position of the first nonzero number

You can specify number of object to return in Position. For example,

list = RandomInteger[{-5, 5}, 100000];
Position[list, _?(# != 0 &), 1, 1] // AbsoluteTiming

{0.003050, {{1}}}

Position[list, _?(# != 0 &)] // First // AbsoluteTiming

{0.138709, {1}}

SeedRandom[1];
LengthWhile[RandomInteger[{-5, 5}, 100000], # == 0 &] + 1
(* 1 *)

We can use SparseArray. It computes all the positions, but it's a bit faster than Position, esp. on packed arrays, since Position unpacks a packed array.

First @ SparseArray[packedarray]["NonzeroPositions"]

or

First @ SparseArray[Developer`ToPackedArray @ unpackedlist]["NonzeroPositions"]

Timings on ten random lists:

SeedRandom[1];
n = 10;
list = RandomChoice[{0.9, 0.1} -> {0, 1}, {n, 100000}];
packedlist = Developer`ToPackedArray @ list;

Table[
  First @ SparseArray[Developer`ToPackedArray @ list[[i]]]["NonzeroPositions"],
  {i, n}] // AbsoluteTiming
Table[
  Position[list[[i]], _?(# != 0 &), 1, 1],
  {i, n}] // AbsoluteTiming

Table[
  First @ SparseArray[packedlist[[i]]]["NonzeroPositions"],
  {i, n}] // AbsoluteTiming
Table[
  Position[packedlist[[i]], _?(# != 0 &), 1, 1],
  {i, n}] // AbsoluteTiming

{0.007372, {{15}, {28}, {4}, {20}, {13}, {2}, {8}, {7}, {42}, {9}}}
{0.009183, {{{15}}, {{28}}, {{4}}, {{20}}, {{13}}, {{2}}, {{8}}, {{7}}, {{42}}, {{9}}}}

{0.005098, {{15}, {28}, {4}, {20}, {13}, {2}, {8}, {7}, {42}, {9}}}
{0.024295, {{{15}}, {{28}}, {{4}}, {{20}}, {{13}}, {{2}}, {{8}}, {{7}}, {{42}}, {{9}}}}

---

Compiled

It's rather straightforward in an imperative paradigm:

firstnzp = Compile[{{list, _Integer, 1}},
  Do[If[list[[i]] != 0, Return[i]], {i, Length@list}],
  RuntimeOptions -> "Speed"
  ]

And fast:

Table[firstnzp @ list[[i]], {i, n}] // AbsoluteTiming

{0.004164, {15, 28, 4, 20, 13, 2, 8, 7, 42, 9}}