给定一个double，我希望将它四舍五入到小数点后的给定精度点数，类似于PHP的round()函数。
我能在Dart文档中找到的最接近的东西是double.toStringAsPrecision()，但这不是我所需要的，因为它包括了精度总分中小数点前的数字。
例如，使用toStringAsPrecision(3):
0.123456789 rounds to 0.123
9.123456789 rounds to 9.12
98.123456789 rounds to 98.1
987.123456789 rounds to 987
9876.123456789 rounds to 9.88e+3
随着数字大小的增加，小数点后的精度也相应降低。
当前回答
如果你不想要任何小数，而结果的小数都是0，这样做是可行的:
String fixedDecimals(double d, int decimals, {bool removeZeroDecimals = true}){
double mod = pow(10.0, decimals);
double result = ((d * mod).round().toDouble() / mod);
if( removeZeroDecimals && result - (result.truncate()) == 0.0 ) decimals = 0;
return result.toStringAsFixed(decimals);
}
如果输入是9.004并且你想要2个小数，这将简单地输出9而不是9.00。
其他回答
我在double上做了这个扩展
import 'dart:math';
extension DoubleExtension on double {
/// rounds the double to a specific decimal place
double roundedPrecision(int places) {
double mod = pow(10.0, places) as double;
return ((this * mod).round().toDouble() / mod);
}
/// good for string output because it can remove trailing zeros
/// and sometimes periods. Or optionally display the exact number of trailing
/// zeros
String roundedPrecisionToString(
int places, {
bool trailingZeros = false,
}) {
double mod = pow(10.0, places) as double;
double round = ((this * mod).round().toDouble() / mod);
String doubleToString =
trailingZeros ? round.toStringAsFixed(places) : round.toString();
if (!trailingZeros) {
RegExp trailingZeros = RegExp(r'^[0-9]+.0+$');
if (trailingZeros.hasMatch(doubleToString)) {
doubleToString = doubleToString.split('.')[0];
}
}
return doubleToString;
}
String toStringNoTrailingZeros() {
String doubleToString = toString();
RegExp trailingZeros = RegExp(r'^[0-9]+.0+$');
if (trailingZeros.hasMatch(doubleToString)) {
doubleToString = doubleToString.split('.')[0];
}
return doubleToString;
}
}
这是通过的测试。
import 'package:flutter_test/flutter_test.dart';
import 'package:project_name/utils/double_extension.dart';
void main() {
group("rounded precision", () {
test("rounding to 0 place results in an int", () {
double num = 5.1234;
double num2 = 5.8234;
expect(num.roundedPrecision(0), 5);
expect(num2.roundedPrecision(0), 6);
});
test("rounding to 1 place rounds correctly to 1 place", () {
double num = 5.12;
double num2 = 5.15;
expect(num.roundedPrecision(1), 5.1);
expect(num2.roundedPrecision(1), 5.2);
});
test(
"rounding a number to a precision that is more accurate than the origional",
() {
double num = 5;
expect(num.roundedPrecision(5), 5);
});
});
group("rounded precision returns the correct string", () {
test("rounding to 0 place results in an int", () {
double num = 5.1234;
double num2 = 5.8234;
expect(num.roundedPrecisionToString(0), "5");
expect(num2.roundedPrecisionToString(0), "6");
});
test("rounding to 1 place rounds correct", () {
double num = 5.12;
double num2 = 5.15;
expect(num.roundedPrecisionToString(1), "5.1");
expect(num2.roundedPrecisionToString(1), "5.2");
});
test("rounding to 2 places rounds correct", () {
double num = 5.123;
double num2 = 5.156;
expect(num.roundedPrecisionToString(2), "5.12");
expect(num2.roundedPrecisionToString(2), "5.16");
});
test("cut off all trailing zeros (and periods)", () {
double num = 5;
double num2 = 5.03000;
expect(num.roundedPrecisionToString(5), "5");
expect(num2.roundedPrecisionToString(5), "5.03");
});
});
}
double value = 2.8032739273;
String formattedValue = value.toStringAsFixed(3);
请参阅num.toStringAsFixed()的文档。
字符串磨损
返回this的小数字符串表示形式。
在计算字符串表示形式之前，将此转换为double。
如果this的绝对值大于或等于10^21，则该方法返回由this. tostringasexponential()计算的指数表示。
例子:
1000000000000000000000.toStringAsExponential(3); // 1.000e+21
否则，结果是与小数点后的fractionDigits位数最接近的字符串表示形式。如果fractionDigits等于0，则小数点将被省略。
参数fractionDigits必须是满足以下条件的整数:0 <= fractionDigits <= 20。
例子:
1.toStringAsFixed(3); // 1.000
(4321.12345678).toStringAsFixed(3); // 4321.123
(4321.12345678).toStringAsFixed(5); // 4321.12346
123456789012345678901.toStringAsFixed(3); // 123456789012345683968.000
1000000000000000000000.toStringAsFixed(3); // 1e+21
5.25.toStringAsFixed(0); // 5
如果你不想要任何小数，而结果的小数都是0，这样做是可行的:
String fixedDecimals(double d, int decimals, {bool removeZeroDecimals = true}){
double mod = pow(10.0, decimals);
double result = ((d * mod).round().toDouble() / mod);
if( removeZeroDecimals && result - (result.truncate()) == 0.0 ) decimals = 0;
return result.toStringAsFixed(decimals);
}
如果输入是9.004并且你想要2个小数，这将简单地输出9而不是9.00。
void main() {
int decimals = 2;
int fac = pow(10, decimals);
double d = 1.234567889;
d = (d * fac).round() / fac;
print("d: $d");
}
打印: 1.23
