How to switch between a binary number as String and a as a list of digits?

Question

I was trying with String Functions and I viewing any posts and have no results

I want to know

1.- From

cromo33 = Table[RandomInteger[], {i, 1, 33}]
cromo33

that gives {1,0,1,1,1,0,1,......,1,0,0} , alist of 33 elements

How can you obtain "1011101.....100" ?

2.- From

IntegerString[RandomInteger[8], 2] // InputForm

that gives 101 (a little add-on... ¿how can I force this to obtain 4 digits in any random number?)

How can you obtain a list {1,0,1} ?

Answer 1

There are built-in functions to play:

NumberForm[FromDigits@RandomInteger[1, 5], 6, NumberPadding -> "0"]

"0011101"

---

Or very straightforward approach:

1

RandomInteger[1, 5]
PadLeft[%, 7]
ToString /@ %
StringJoin@% 

{0, 1, 0, 1, 0}
{0, 0, 0, 1, 0, 1, 0}
{"0", "0", "0", "1", "0", "1", "0"}
"0001010"

2

IntegerString[RandomInteger[8], 2]
Characters@%
PadLeft[%, 4]
ToExpression@%

"101"
{"1", "0", "1"}
{0, "1", "0", "1"}
{0, 1, 0, 1}

Answer 2

Just some other functions to consider:

• IntegerDigits an take a base argument, e.g. IntegerDigits[5,2] yields {1,0,1}
• FromDigits is the inverse: FromDigits[{1,0,1},2] yields 5
• If the issue is for display purpose and not specifically need for strings then BaseForm will display binary numbers with subscript of 2, i.e.BaseForm[5,2] yields: $101_2$
• so for first part if display is aim you can use (perhaps somewhat perversely) BaseForm[ToExpression@Table[RandomInteger[], {i, 1, 33}]]

Answer 3

Let:

SeedRandom[11]

cromo33 = RandomInteger[1, 33];

To produce your string I would use Row and ToString:

ToString @ Row @ cromo33

"001101111011110111010011001000101"

For the second case IntegerString takes a third parameter that specifies length:

SeedRandom[15]
string = IntegerString[RandomInteger[8], 2, 4]

"0101"

To convert this back into to a digit list I propose ToCharacterCode:

ToCharacterCode[string] - 48

{0, 1, 0, 1}