Java - Convert double to int array

up vote
10
down vote

favorite

I have a double

double pi = 3.1415;

I want to convert this to a int array

int piArray = 3,1,4,1,5;

I came up with this

double pi = 3.1415;
String piString = Double.toString(pi).replace(".", "");
int piArray = new int[piString.length()];
for (int i = 0; i <= piString.length()-1; i++)
 piArray[i] = piString.charAt(i) - '0';

It's working but I don't like this solution because I think a lot of conversions between datatypes can lead to errors. Is my code even complete or do I need to check for something else?

And how would you approach this problem? Wasn't able to find a similar problem here on stackoverflow for Java.

asked 35 mins ago

Ian Fako

282112

You might be interested in base 2 digits, a double as array of 0s and 1s. The Double class has conversion of the bits. Above you could have the power-of-ten in the array: -4 in your example. Doubles are themselves problematic, as the toString could well give "3.14149999998"
â€“Â Joop Eggen
25 mins ago

add a commentÂ |Â

up vote
10
down vote

favorite

I have a double

double pi = 3.1415;

I want to convert this to a int array

int piArray = 3,1,4,1,5;

I came up with this

double pi = 3.1415;
String piString = Double.toString(pi).replace(".", "");
int piArray = new int[piString.length()];
for (int i = 0; i <= piString.length()-1; i++)
 piArray[i] = piString.charAt(i) - '0';

It's working but I don't like this solution because I think a lot of conversions between datatypes can lead to errors. Is my code even complete or do I need to check for something else?

And how would you approach this problem? Wasn't able to find a similar problem here on stackoverflow for Java.

asked 35 mins ago

Ian Fako

282112

You might be interested in base 2 digits, a double as array of 0s and 1s. The Double class has conversion of the bits. Above you could have the power-of-ten in the array: -4 in your example. Doubles are themselves problematic, as the toString could well give "3.14149999998"
â€“Â Joop Eggen
25 mins ago

add a commentÂ |Â

up vote
10
down vote

favorite

I have a double

double pi = 3.1415;

I want to convert this to a int array

int piArray = 3,1,4,1,5;

I came up with this

double pi = 3.1415;
String piString = Double.toString(pi).replace(".", "");
int piArray = new int[piString.length()];
for (int i = 0; i <= piString.length()-1; i++)
 piArray[i] = piString.charAt(i) - '0';

It's working but I don't like this solution because I think a lot of conversions between datatypes can lead to errors. Is my code even complete or do I need to check for something else?

And how would you approach this problem? Wasn't able to find a similar problem here on stackoverflow for Java.

asked 35 mins ago

Ian Fako

282112

I have a double

double pi = 3.1415;

I want to convert this to a int array

int piArray = 3,1,4,1,5;

I came up with this

double pi = 3.1415;
String piString = Double.toString(pi).replace(".", "");
int piArray = new int[piString.length()];
for (int i = 0; i <= piString.length()-1; i++)
 piArray[i] = piString.charAt(i) - '0';

It's working but I don't like this solution because I think a lot of conversions between datatypes can lead to errors. Is my code even complete or do I need to check for something else?

And how would you approach this problem? Wasn't able to find a similar problem here on stackoverflow for Java.

java arrays type-conversion int double

asked 35 mins ago

Ian Fako

282112

asked 35 mins ago

Ian Fako

282112

asked 35 mins ago

Ian Fako

282112

asked 35 mins ago

Ian Fako

282112

asked 35 mins ago

Ian Fako

282112

You might be interested in base 2 digits, a double as array of 0s and 1s. The Double class has conversion of the bits. Above you could have the power-of-ten in the array: -4 in your example. Doubles are themselves problematic, as the toString could well give "3.14149999998"
â€“Â Joop Eggen
25 mins ago

add a commentÂ |Â

You might be interested in base 2 digits, a double as array of 0s and 1s. The Double class has conversion of the bits. Above you could have the power-of-ten in the array: -4 in your example. Doubles are themselves problematic, as the toString could well give "3.14149999998"
â€“Â Joop Eggen
25 mins ago

You might be interested in base 2 digits, a double as array of 0s and 1s. The Double class has conversion of the bits. Above you could have the power-of-ten in the array: -4 in your example. Doubles are themselves problematic, as the toString could well give "3.14149999998"
â€“Â Joop Eggen
25 mins ago

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
6
down vote

I don't know stright way but I think It is simpler:

int piArray = String.valueOf(pi).replaceAll("\D", "").chars().map(Character::getNumericValue).toArray();

edited 20 mins ago

answered 28 mins ago

Rahim Dastar

419110

1

Came up with the same solution, you were a bit earlier :)
â€“Â Glains
27 mins ago

4

I have used replaceAll("\D", "") tough to filter any non digits.
â€“Â Glains
25 mins ago

Glains post your solution or edit, as i think it's the more robust solution.
â€“Â fl0w
23 mins ago

Thanks @Glains, post edited
â€“Â Rahim Dastar
20 mins ago

@RahimDastar since a double has a single ., use replaceFirst instead of replaceAll, otherwise 1+
â€“Â Eugene
11 mins ago

Â |Â
show 3 more comments

up vote
1
down vote

int result = Stream.of(pi)
 .map(String::valueOf)
 .flatMap(x -> Arrays.stream(x.split("\.|")))
 .filter(x -> !x.isEmpty())
 .mapToInt(Integer::valueOf)
 .toArray();

Or a safer approach with java-9:

 int result = new Scanner(String.valueOf(pi))
 .findAll(Pattern.compile("\d"))
 .map(MatchResult::group)
 .mapToInt(Integer::valueOf)
 .toArray();

edited 3 mins ago

answered 26 mins ago

Eugene

59.8k982141

Can you explain why it is safer, would be interested in that.
â€“Â Glains
59 secs ago

@Glains safer that my first approach, which does not take care of 1.13E+21...
â€“Â Eugene
23 secs ago

add a commentÂ |Â

up vote
0
down vote

You can use in java 8

int piArray = piString.chars().map(Character::getNumericValue).toArray();

answered 18 mins ago

Sveteek

215

add a commentÂ |Â

up vote
0
down vote

this would work also

int piArray = Double.toString(pi).replace(".", "").chars().map(c -> c-'0').toArray();

answered 3 mins ago

The Scientific Method

568310

add a commentÂ |Â

up vote
-1
down vote

public class Main 
 public static void main(String args) 
 Double pi = 3.14159;
 char chars = pi.toString().toCharArray();
 System.out.println(chars.length);

Use char array to handle it.
The decimal point, handle it yourself.

answered 18 mins ago

CHONG CHEN

New contributor

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
6
down vote

I don't know stright way but I think It is simpler:

int piArray = String.valueOf(pi).replaceAll("\D", "").chars().map(Character::getNumericValue).toArray();

edited 20 mins ago

answered 28 mins ago

Rahim Dastar

419110

1

Came up with the same solution, you were a bit earlier :)
â€“Â Glains
27 mins ago

4

I have used replaceAll("\D", "") tough to filter any non digits.
â€“Â Glains
25 mins ago

Glains post your solution or edit, as i think it's the more robust solution.
â€“Â fl0w
23 mins ago

Thanks @Glains, post edited
â€“Â Rahim Dastar
20 mins ago

@RahimDastar since a double has a single ., use replaceFirst instead of replaceAll, otherwise 1+
â€“Â Eugene
11 mins ago

Â |Â
show 3 more comments

up vote
6
down vote

I don't know stright way but I think It is simpler:

int piArray = String.valueOf(pi).replaceAll("\D", "").chars().map(Character::getNumericValue).toArray();

edited 20 mins ago

answered 28 mins ago

Rahim Dastar

419110

1

Came up with the same solution, you were a bit earlier :)
â€“Â Glains
27 mins ago

4

I have used replaceAll("\D", "") tough to filter any non digits.
â€“Â Glains
25 mins ago

Glains post your solution or edit, as i think it's the more robust solution.
â€“Â fl0w
23 mins ago

Thanks @Glains, post edited
â€“Â Rahim Dastar
20 mins ago

@RahimDastar since a double has a single ., use replaceFirst instead of replaceAll, otherwise 1+
â€“Â Eugene
11 mins ago

Â |Â
show 3 more comments

up vote
6
down vote

I don't know stright way but I think It is simpler:

int piArray = String.valueOf(pi).replaceAll("\D", "").chars().map(Character::getNumericValue).toArray();

edited 20 mins ago

answered 28 mins ago

Rahim Dastar

419110

I don't know stright way but I think It is simpler:

int piArray = String.valueOf(pi).replaceAll("\D", "").chars().map(Character::getNumericValue).toArray();

edited 20 mins ago

answered 28 mins ago

Rahim Dastar

419110

edited 20 mins ago

answered 28 mins ago

Rahim Dastar

419110

answered 28 mins ago

Rahim Dastar

419110

answered 28 mins ago

Rahim Dastar

419110

1

Came up with the same solution, you were a bit earlier :)
â€“Â Glains
27 mins ago

4

I have used replaceAll("\D", "") tough to filter any non digits.
â€“Â Glains
25 mins ago

Glains post your solution or edit, as i think it's the more robust solution.
â€“Â fl0w
23 mins ago

Thanks @Glains, post edited
â€“Â Rahim Dastar
20 mins ago

@RahimDastar since a double has a single ., use replaceFirst instead of replaceAll, otherwise 1+
â€“Â Eugene
11 mins ago

Â |Â
show 3 more comments

1

Came up with the same solution, you were a bit earlier :)
â€“Â Glains
27 mins ago

4

I have used replaceAll("\D", "") tough to filter any non digits.
â€“Â Glains
25 mins ago

Glains post your solution or edit, as i think it's the more robust solution.
â€“Â fl0w
23 mins ago

Thanks @Glains, post edited
â€“Â Rahim Dastar
20 mins ago

@RahimDastar since a double has a single ., use replaceFirst instead of replaceAll, otherwise 1+
â€“Â Eugene
11 mins ago

Came up with the same solution, you were a bit earlier :)
â€“Â Glains
27 mins ago

I have used replaceAll("\D", "") tough to filter any non digits.
â€“Â Glains
25 mins ago

Glains post your solution or edit, as i think it's the more robust solution.
â€“Â fl0w
23 mins ago

Thanks @Glains, post edited
â€“Â Rahim Dastar
20 mins ago

@RahimDastar since a double has a single ., use replaceFirst instead of replaceAll, otherwise 1+
â€“Â Eugene
11 mins ago

Â |Â
show 3 more comments

up vote
1
down vote

int result = Stream.of(pi)
 .map(String::valueOf)
 .flatMap(x -> Arrays.stream(x.split("\.|")))
 .filter(x -> !x.isEmpty())
 .mapToInt(Integer::valueOf)
 .toArray();

Or a safer approach with java-9:

 int result = new Scanner(String.valueOf(pi))
 .findAll(Pattern.compile("\d"))
 .map(MatchResult::group)
 .mapToInt(Integer::valueOf)
 .toArray();

edited 3 mins ago

answered 26 mins ago

Eugene

59.8k982141

Can you explain why it is safer, would be interested in that.
â€“Â Glains
59 secs ago

@Glains safer that my first approach, which does not take care of 1.13E+21...
â€“Â Eugene
23 secs ago

add a commentÂ |Â

up vote
1
down vote

int result = Stream.of(pi)
 .map(String::valueOf)
 .flatMap(x -> Arrays.stream(x.split("\.|")))
 .filter(x -> !x.isEmpty())
 .mapToInt(Integer::valueOf)
 .toArray();

Or a safer approach with java-9:

 int result = new Scanner(String.valueOf(pi))
 .findAll(Pattern.compile("\d"))
 .map(MatchResult::group)
 .mapToInt(Integer::valueOf)
 .toArray();

edited 3 mins ago

answered 26 mins ago

Eugene

59.8k982141

Can you explain why it is safer, would be interested in that.
â€“Â Glains
59 secs ago

@Glains safer that my first approach, which does not take care of 1.13E+21...
â€“Â Eugene
23 secs ago

add a commentÂ |Â

up vote
1
down vote

int result = Stream.of(pi)
 .map(String::valueOf)
 .flatMap(x -> Arrays.stream(x.split("\.|")))
 .filter(x -> !x.isEmpty())
 .mapToInt(Integer::valueOf)
 .toArray();

Or a safer approach with java-9:

 int result = new Scanner(String.valueOf(pi))
 .findAll(Pattern.compile("\d"))
 .map(MatchResult::group)
 .mapToInt(Integer::valueOf)
 .toArray();

edited 3 mins ago

answered 26 mins ago

Eugene

59.8k982141

int result = Stream.of(pi)
 .map(String::valueOf)
 .flatMap(x -> Arrays.stream(x.split("\.|")))
 .filter(x -> !x.isEmpty())
 .mapToInt(Integer::valueOf)
 .toArray();

Or a safer approach with java-9:

 int result = new Scanner(String.valueOf(pi))
 .findAll(Pattern.compile("\d"))
 .map(MatchResult::group)
 .mapToInt(Integer::valueOf)
 .toArray();

edited 3 mins ago

answered 26 mins ago

Eugene

59.8k982141

edited 3 mins ago

answered 26 mins ago

Eugene

59.8k982141

answered 26 mins ago

Eugene

59.8k982141

answered 26 mins ago

Eugene

59.8k982141

Can you explain why it is safer, would be interested in that.
â€“Â Glains
59 secs ago

@Glains safer that my first approach, which does not take care of 1.13E+21...
â€“Â Eugene
23 secs ago

add a commentÂ |Â

Can you explain why it is safer, would be interested in that.
â€“Â Glains
59 secs ago

@Glains safer that my first approach, which does not take care of 1.13E+21...
â€“Â Eugene
23 secs ago

Can you explain why it is safer, would be interested in that.
â€“Â Glains
59 secs ago

@Glains safer that my first approach, which does not take care of 1.13E+21...
â€“Â Eugene
23 secs ago

add a commentÂ |Â

up vote
0
down vote

You can use in java 8

int piArray = piString.chars().map(Character::getNumericValue).toArray();

answered 18 mins ago

Sveteek

215

add a commentÂ |Â

up vote
0
down vote

You can use in java 8

int piArray = piString.chars().map(Character::getNumericValue).toArray();

answered 18 mins ago

Sveteek

215

add a commentÂ |Â

up vote
0
down vote

You can use in java 8

int piArray = piString.chars().map(Character::getNumericValue).toArray();

answered 18 mins ago

Sveteek

215

You can use in java 8

int piArray = piString.chars().map(Character::getNumericValue).toArray();

answered 18 mins ago

Sveteek

215

answered 18 mins ago

Sveteek

215

answered 18 mins ago

Sveteek

215

answered 18 mins ago

Sveteek

215

add a commentÂ |Â

up vote
0
down vote

this would work also

int piArray = Double.toString(pi).replace(".", "").chars().map(c -> c-'0').toArray();

answered 3 mins ago

The Scientific Method

568310

add a commentÂ |Â

up vote
0
down vote

this would work also

int piArray = Double.toString(pi).replace(".", "").chars().map(c -> c-'0').toArray();

answered 3 mins ago

The Scientific Method

568310

add a commentÂ |Â

up vote
0
down vote

this would work also

int piArray = Double.toString(pi).replace(".", "").chars().map(c -> c-'0').toArray();

answered 3 mins ago

The Scientific Method

568310

this would work also

int piArray = Double.toString(pi).replace(".", "").chars().map(c -> c-'0').toArray();

answered 3 mins ago

The Scientific Method

568310

answered 3 mins ago

The Scientific Method

568310

answered 3 mins ago

The Scientific Method

568310

answered 3 mins ago

The Scientific Method

568310

add a commentÂ |Â

up vote
-1
down vote

public class Main 
 public static void main(String args) 
 Double pi = 3.14159;
 char chars = pi.toString().toCharArray();
 System.out.println(chars.length);

Use char array to handle it.
The decimal point, handle it yourself.

answered 18 mins ago

CHONG CHEN

New contributor

add a commentÂ |Â

up vote
-1
down vote

public class Main 
 public static void main(String args) 
 Double pi = 3.14159;
 char chars = pi.toString().toCharArray();
 System.out.println(chars.length);

Use char array to handle it.
The decimal point, handle it yourself.

answered 18 mins ago

CHONG CHEN

New contributor

add a commentÂ |Â

up vote
-1
down vote

public class Main 
 public static void main(String args) 
 Double pi = 3.14159;
 char chars = pi.toString().toCharArray();
 System.out.println(chars.length);

Use char array to handle it.
The decimal point, handle it yourself.

answered 18 mins ago

CHONG CHEN

New contributor

public class Main 
 public static void main(String args) 
 Double pi = 3.14159;
 char chars = pi.toString().toCharArray();
 System.out.println(chars.length);

Use char array to handle it.
The decimal point, handle it yourself.

answered 18 mins ago

CHONG CHEN

New contributor

answered 18 mins ago

CHONG CHEN

New contributor

answered 18 mins ago

CHONG CHEN

answered 18 mins ago

CHONG CHEN

New contributor

CHONG CHEN is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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