Cfrtjryk

I have a list such as:

a = {1, 2, 4, 3};

and I want to get the 'number' of the list. For this example, I want to get this number: 1243. I do not know how to get it easily. For now, my solution is:
a[[1]]*10^3+a[[2]]*10^2+a[[3]]*10^1+a[[4]];
It works. But it is too complex. I want to know a way more convenient.

a = {1, 2, 4, 3};

FromDigits[a]



(* 1243 *)

Other than the built-in FromDigits, we can build our wheel

Fold[{10, 1}.{##} &, a]

Best way would be to use build-in function FromDigits as shown above.

But just in case you'd like a convoluted way to do it, here is another option

a = {1, 2, 4, 3};

ToExpression[StringJoin[ToString[#] & /@ a]]

These are some methods I would not recommend:

lst = {1, 2, 3, 4};



f = (Curry[StringRiffle][""] /* ToExpression);



g = (Through[{

    Identity,

    Length /* Range /* Reverse /* Curry[Plus][-1] /* Curry[Power, 2][10]

}[#]] & /* MapThread[Times] /* Total);



f[lst] (* 1234 *)

g[lst] (* 1234 *)

Here is a method that is faster than FromDigits for the construction of many numbers at once:

a = RandomInteger[{1, 9}, {1000000, 10}];

r1 = FromDigits /@ a; // RepeatedTiming // First

r2 = FromDigits[Transpose[a]]; // RepeatedTiming

r3 = a.(10^Range[9, 0, -1]); // RepeatedTiming // First

r1 == r2 == r3

0.466

0.068

0.040

True

Admittedly, FromDigits[Transpose[a]] is only slower than a.(10^Range[9, 0, -1]) because of Transpose.