I am creating a physics calculator and currently working on average velocity. For displacement I would like the user to input a string i.e. "5km north". Then using a method assign only the numeric values in the string to be doubles using Double.parseDouble or something of the like.
Below is my method to pull out numeric values
private double getNumericalValue(String userInput) {
double numericalOutput = 0.0;
for (int i = 0; i < userInput.length(); i++) {
if (userInput.charAt(i) >= 65 && userInput.charAt(i) <= 122) {
numericalOutput = Double.parseDouble(userInput.substring(0, i));
}
}
return numericalOutput;
}
and lastly the getAverageVelocity method
private void getAverageVelocity() {
// formula average velocity = displacement s / change in time t
double avgVelocity = 0.0;
System.out.println("Enter the displacement: [ex. 5 km north]");
String displacement = getStringInput();
System.out.println("Enter the change in time: [ex. 1 hour]");
String changeInTime = getStringInput();
// parse the string and change numerical input into double
double numericalS = getNumericalValue(displacement);
double numericalT = getNumericalValue(changeInTime);
avgVelocity = numericalS / numericalT;
System.out.println("The Average Velocity is: " + avgVelocity);
}
As you can see I figured I could simply compare chars to their ascii value. However, it is including the k or a space in my double output which is throwing this error.
Exception in thread "main" java.lang.NumberFormatException: For input string: "5 k"
at java.base/jdk.internal.math.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:2054)
at java.base/jdk.internal.math.FloatingDecimal.parseDouble(FloatingDecimal.java:110)
at java.base/java.lang.Double.parseDouble(Double.java:556)
at Main.getNumericalValue(Main.java:122)
at Main.getAverageVelocity(Main.java:393)
at Main.kineticsIn1DCalculator(Main.java:375)
at Main.performAction(Main.java:141)
at Main.runner(Main.java:29)
at Main.main(Main.java:22)
Any help and a thorough explanation would be great. Thanks.
What you are actually testing in the if block is whether the character at i is a letter A-Z, a-z and other symbols(i.e. [ ^ ], ...etc) to check if it's a number (i.e. 0-9) the corresponding decimal range is 48-57 check this link
as for the exception it's because of substring(0, 0) will result in an empty string, and when trying to parse it, Double.parseDouble() will throw an exception.
here is a solution you can use:
private double getNumericalValue(String userInput) {
double numericalOutput = 0.0;
for (int i = 0; i < userInput.length(); i++) {
if (userInput.charAt(i) >= 48 && userInput.charAt(i) <= 57) {
numericalOutput = Double.parseDouble(userInput.substring(0, i + 1));
}
}
return numericalOutput;
}
Note: substring(i, j) i is inclusive, j exclusive
Update :
just noticed, in your case the exception happens because you are trying to parse the letter "k" since it passes the condition, resulting in Double.parseDouble() throwing an exception.
You have the error because even if the first number is found the for loop continues. You simply have to add break instruction, then the for loop ends at the first number found.
This is the final code of the method getNumericalValue:
private double getNumericalValue(String userInput) {
double numericalOutput = 0.0;
for (int i = 0; i < userInput.length(); i++) {
if (userInput.charAt(i) >= 65 && userInput.charAt(i) <= 122) {
numericalOutput = Double.parseDouble(userInput.substring(0, i));
break;
}
}
return numericalOutput;
}
You simply should input the line and parse its parts.
Best to introduce a class to hold a value with unit, where "km north" counts as a unit.
Since java 14 there exists the record class.
record ValueWithUnit(double value, String unit) {
public static ValueWithUnit from(String s) {
String v = s.replaceFirst("^(.*?)([ A-z].*)$", "$1");
String u = s.replaceFirst("^(.*?)([ A-z].*)$", "$2");
if (u == v) {
u = "";
}
double x = Double.parseDouble(v); // Replaces getNumericalValue(v);
return new Value(x, u);
}
#Override
public String toString() {
String.format("%f %s", value, unit);
}
}
You could use it directly as:
ValueWithUnit vu = new ValueWithUnit(30.0, "km");
System.out.println(vu);
The code then becomes:
private void getAverageVelocity() {
// formula average velocity = displacement s / change in time t
double avgVelocity = 0.0;
System.out.println("Enter the displacement: [ex. 5 km north]");
ValueWithUnit displacement = ValueWithUnit.from(getStringInput());
System.out.println("Enter the change in time: [ex. 1 hour]");
ValueWithUnit changeInTime = ValueWithUnit.from(getStringInput());
// parse the string and change numerical input into double
double numericalS = displacement.value;
double numericalT = changeInTime.value;
avgVelocity = numericalS / numericalT;
System.out.printf("The Average Velocity is: %f %s%n", avgVelocity,
displacement.unit + "/" + changeInTime.unit);
}
For example when I parse a string "12345678901234567890" to double using Double.parseDouble() it returns the value "12345678901234567000" since it can hold up to 17 digits.
I want to validate this scenario and the user should be allowed to pass only 17 digits. How do I do this?
Example :
1.2345678901234567890 is invalid because it has more than 17 digits total
1.2345E+10 is valid
Tried something like this which can count the digits using split function
String input="12345678901234567E100";
String inputWithoutSign;
int lengthFullNumber;
int lengthFraction;
double v = Double.parseDouble(input);
if(input.startsWith("+") || input.startsWith("-")){
inputWithoutSign = input.split("[-+]",2)[1];
}
else inputWithoutSign = input;
String num = inputWithoutSign.split("[eE]", 2)[0];
if(num.indexOf('.') == -1){
lengthFullNumber = num.length();
lengthFraction = 0;
}else{
String[] splitNum = num.split("\\.", 2);
lengthFullNumber = splitNum[0].length();
lengthFraction = splitNum[1].length();
}
System.out.println("length:"+(lengthFullNumber+lengthFraction));
Presuming I understand your goal of limiting the number of digits, this may help solve the problem.
Test cases
String[] vals = {
"12345678901234567890", "123456789091919191919",
"182828282.18282828", "182828282.182828282", "191929e10",
"192929.22929e10"
};
Try and parse them
for (String v : vals) {
// remove possible decimal point and signs
String test = v.replaceAll("[.+-]", "");
// remove any exponents at end of string
test = test.replace("\\D+.*", "");
if (test.length() > 17) {
System.out.println(v + " has too many digits");
continue;
}
double d = Double.parseDouble(v);
System.out.println(v + " parses to " + d);
}
How to best print 2 float numbers in scientific-notation but with same exponent?
eg: I'd like to print numbers like this:
1.234e-6
11.234e-6
And I would like some function to detect automatically best exponent - that is smaller number always start at first decimal digit and bigger number prints how it must with same exponent.
eg: 0.1 and 100 would print
1.000e-1
1000.000e-1
But even when I ask explicitly for 2 decimal places String.format("%2.3e",11.234e-6) I got 1.123e-5
So far I come up with code bellow. It works as I want. But as you can see it is not exactly short or swift... It would be great if someone would point to some Java native functions which helps me to do it more elegant...
public static String repeat(int count, String with) {
if(count<0){
return "";
}
if(count>1e5){
return "";
}
return new String(new char[count]).replace("\0", with);
}
public static String getFormattedDouble(double num,int posInt,int posFrac){
if(num == 0) return "0"; // correct 0 value to be print only with one 0
String sInt = repeat(posInt,"0");
String sFrac = repeat(posFrac,"0");
String sSing = num<0 ? "" : "+";
DecimalFormat form = new DecimalFormat(sSing+sInt+"."+sFrac+"E0");
String s = form.format(num);
s = s.replace("E","e"); // I really thing capital E looks ugly
return s;
}
public static String[] get2doublesSameExp(double a, double b){
String[] s = new String[2];
int expA;
if(a == 0) expA = 0;
else expA = (int)Math.floor(Math.log10(Math.abs(a)));
int expB;
if(b == 0) expB = 0;
else expB = (int)Math.floor(Math.log10(Math.abs(b)));
double expDif = Math.abs((double)expA-(double)expB);
int fractPos = 3;
if(expDif > 4) fractPos = 1; // too big exponent difference reduce fraction digits
if(expDif > 6){
// too big exponent difference print it normally it will be nicer
s[0] = String.format("%1.3e",a);
s[1] = String.format("%1.3e",b);
return s;
}
int minExp = Math.min(expA,expB) - 1;
s[0] = getFormattedDouble(a, expA - minExp, fractPos );
s[1] = getFormattedDouble(b, expB - minExp, fractPos );
// just text right justification
String rightJust = repeat((int)expDif," ");
int justIdx = expA < expB ? 0 : 1;
s[justIdx] = rightJust + s[justIdx];
return s;
}
String[] s = get2doublesSameExp(1.234e-6,11.234e-6);
I have a Double value xx.yyy and I want to convert to string "xxyyy" or "-xxyy", if the value is negative.
How could I do it?
Regards.
double yourDouble = 61.9155;
String str = String.valueOf(yourDouble).replace(".", "");
Explanation:
String.valueOf(): converts your double to a String
str.replace(s1, s2): returns a new string equals to str where all s1's are replaced by s2's
Update:
The OP had some extra conditions (but I don't know exactly with one):
negative number -> only two decimals.
public static String doubleToSpecialString(double d)
{
if (d >= 0)
{
return String.valueOf(d).replace(".", "");
} else
{
return String.format("%.2f", d).replace(",", "");
}
}
negative number -> one decimal less
public static String doubleToSpecialString(double d)
{
if (d >= 0)
{
return String.valueOf(d).replace(".", "");
} else
{
String str = String.valueOf(d);
int dotIndex = str.indexOf(".");
int decimals = str.length() - dotIndex - 1;
return String.format("%." + (decimals - 1) + "f", d).replace(",", "");
}
}
This answer uses a Decimal Formatter. It assumes that the input number is always strictly of the form (-)xx.yyy.
/**
* Converts a double of the form xx.yyy to xxyyy and -xx.yyy to -xxyy.
* No rounding is performed.
*
* #param number The double to format
* #return The formatted number string
*/
public static String format(double number){
DecimalFormat formatter = new DecimalFormat("#");
formatter.setRoundingMode(RoundingMode.DOWN);
number *= number < 0.0 ? 100 : 1000;
String result = formatter.format(number);
return result;
}
How do you left pad an int with zeros when converting to a String in java?
I'm basically looking to pad out integers up to 9999 with leading zeros (e.g. 1 = 0001).
Use java.lang.String.format(String,Object...) like this:
String.format("%05d", yournumber);
for zero-padding with a length of 5. For hexadecimal output replace the d with an x as in "%05x".
The full formatting options are documented as part of java.util.Formatter.
Let's say you want to print 11 as 011
You could use a formatter: "%03d".
You can use this formatter like this:
int a = 11;
String with3digits = String.format("%03d", a);
System.out.println(with3digits);
Alternatively, some java methods directly support these formatters:
System.out.printf("%03d", a);
If you for any reason use pre 1.5 Java then may try with Apache Commons Lang method
org.apache.commons.lang.StringUtils.leftPad(String str, int size, '0')
Found this example... Will test...
import java.text.DecimalFormat;
class TestingAndQualityAssuranceDepartment
{
public static void main(String [] args)
{
int x=1;
DecimalFormat df = new DecimalFormat("00");
System.out.println(df.format(x));
}
}
Tested this and:
String.format("%05d",number);
Both work, for my purposes I think String.Format is better and more succinct.
Try this one:
import java.text.DecimalFormat;
DecimalFormat df = new DecimalFormat("0000");
String c = df.format(9); // Output: 0009
String a = df.format(99); // Output: 0099
String b = df.format(999); // Output: 0999
If performance is important in your case you could do it yourself with less overhead compared to the String.format function:
/**
* #param in The integer value
* #param fill The number of digits to fill
* #return The given value left padded with the given number of digits
*/
public static String lPadZero(int in, int fill){
boolean negative = false;
int value, len = 0;
if(in >= 0){
value = in;
} else {
negative = true;
value = - in;
in = - in;
len ++;
}
if(value == 0){
len = 1;
} else{
for(; value != 0; len ++){
value /= 10;
}
}
StringBuilder sb = new StringBuilder();
if(negative){
sb.append('-');
}
for(int i = fill; i > len; i--){
sb.append('0');
}
sb.append(in);
return sb.toString();
}
Performance
public static void main(String[] args) {
Random rdm;
long start;
// Using own function
rdm = new Random(0);
start = System.nanoTime();
for(int i = 10000000; i != 0; i--){
lPadZero(rdm.nextInt(20000) - 10000, 4);
}
System.out.println("Own function: " + ((System.nanoTime() - start) / 1000000) + "ms");
// Using String.format
rdm = new Random(0);
start = System.nanoTime();
for(int i = 10000000; i != 0; i--){
String.format("%04d", rdm.nextInt(20000) - 10000);
}
System.out.println("String.format: " + ((System.nanoTime() - start) / 1000000) + "ms");
}
Result
Own function: 1697ms
String.format: 38134ms
Here is how you can format your string without using DecimalFormat.
String.format("%02d", 9)
09
String.format("%03d", 19)
019
String.format("%04d", 119)
0119
You can use Google Guava:
Maven:
<dependency>
<artifactId>guava</artifactId>
<groupId>com.google.guava</groupId>
<version>14.0.1</version>
</dependency>
Sample code:
String paddedString1 = Strings.padStart("7", 3, '0'); //"007"
String paddedString2 = Strings.padStart("2020", 3, '0'); //"2020"
Note:
Guava is very useful library, it also provides lots of features which related to Collections, Caches, Functional idioms, Concurrency, Strings, Primitives, Ranges, IO, Hashing, EventBus, etc
Ref: GuavaExplained
Although many of the above approaches are good, but sometimes we need to format integers as well as floats. We can use this, particularly when we need to pad particular number of zeroes on left as well as right of decimal numbers.
import java.text.NumberFormat;
public class NumberFormatMain {
public static void main(String[] args) {
int intNumber = 25;
float floatNumber = 25.546f;
NumberFormat format=NumberFormat.getInstance();
format.setMaximumIntegerDigits(6);
format.setMaximumFractionDigits(6);
format.setMinimumFractionDigits(6);
format.setMinimumIntegerDigits(6);
System.out.println("Formatted Integer : "+format.format(intNumber).replace(",",""));
System.out.println("Formatted Float : "+format.format(floatNumber).replace(",",""));
}
}
int x = 1;
System.out.format("%05d",x);
if you want to print the formatted text directly onto the screen.
You need to use a Formatter, following code uses NumberFormat
int inputNo = 1;
NumberFormat nf = NumberFormat.getInstance();
nf.setMaximumIntegerDigits(4);
nf.setMinimumIntegerDigits(4);
nf.setGroupingUsed(false);
System.out.println("Formatted Integer : " + nf.format(inputNo));
Output: 0001
Use the class DecimalFormat, like so:
NumberFormat formatter = new DecimalFormat("0000"); //i use 4 Zero but you can also another number
System.out.println("OUTPUT : "+formatter.format(811));
OUTPUT : 0000811
You can add leading 0 to your string like this. Define a string that will be the maximum length of the string that you want. In my case i need a string that will be only 9 char long.
String d = "602939";
d = "000000000".substring(0, (9-d.length())) + d;
System.out.println(d);
Output : 000602939
Check my code that will work for integer and String.
Assume our first number is 2. And we want to add zeros to that so the the length of final string will be 4. For that you can use following code
int number=2;
int requiredLengthAfterPadding=4;
String resultString=Integer.toString(number);
int inputStringLengh=resultString.length();
int diff=requiredLengthAfterPadding-inputStringLengh;
if(inputStringLengh<requiredLengthAfterPadding)
{
resultString=new String(new char[diff]).replace("\0", "0")+number;
}
System.out.println(resultString);
Use this simple extension function
fun Int.padZero(): String {
return if (this < 10) {
"0$this"
} else {
this.toString()
}
}
For Kotlin
fun Calendar.getFullDate(): String {
val mYear = "${this.get(Calendar.YEAR)}-"
val mMonth = if (this.get(Calendar.MONTH) + 1 < 10) {
"0${this.get(Calendar.MONTH) + 1}-"
} else {
"${this.get(Calendar.MONTH)+ 1}-"
}
val mDate = if (this.get(Calendar.DAY_OF_MONTH) < 10) {
"0${this.get(Calendar.DAY_OF_MONTH)}"
} else {
"${this.get(Calendar.DAY_OF_MONTH)}"
}
return mYear + mMonth + mDate
}
and use it as
val date: String = calendar.getFullDate()
Here is another way to pad an integer with zeros on the left. You can increase the number of zeros as per your convenience. Have added a check to return the same value as is in case of negative number or a value greater than or equals to zeros configured. You can further modify as per your requirement.
/**
*
* #author Dinesh.Lomte
*
*/
public class AddLeadingZerosToNum {
/**
*
* #param args
*/
public static void main(String[] args) {
System.out.println(getLeadingZerosToNum(0));
System.out.println(getLeadingZerosToNum(7));
System.out.println(getLeadingZerosToNum(13));
System.out.println(getLeadingZerosToNum(713));
System.out.println(getLeadingZerosToNum(7013));
System.out.println(getLeadingZerosToNum(9999));
}
/**
*
* #param num
* #return
*/
private static String getLeadingZerosToNum(int num) {
// Initializing the string of zeros with required size
String zeros = new String("0000");
// Validating if num value is less then zero or if the length of number
// is greater then zeros configured to return the num value as is
if (num < 0 || String.valueOf(num).length() >= zeros.length()) {
return String.valueOf(num);
}
// Returning zeros in case if value is zero.
if (num == 0) {
return zeros;
}
return new StringBuilder(zeros.substring(0, zeros.length() -
String.valueOf(num).length())).append(
String.valueOf(num)).toString();
}
}
Input
0
7
13
713
7013
9999
Output
0000
0007
0013
7013
9999
No packages needed:
String paddedString = i < 100 ? i < 10 ? "00" + i : "0" + i : "" + i;
This will pad the string to three characters, and it is easy to add a part more for four or five. I know this is not the perfect solution in any way (especially if you want a large padded string), but I like it.