iOS 将大数转换为小数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/18267211/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
iOS convert large numbers to smaller format
提问by Kyle Begeman
How can I convert all numbers that are more than 3 digits down to a 4 digit or less number?
如何将所有超过 3 位的数字转换为 4 位或更少的数字？
This is exactly what I mean:
这正是我的意思：
10345 = 10.3k
10012 = 10k
123546 = 123.5k
4384324 = 4.3m
Rounding is not entirely important, but an added plus.
四舍五入并不完全重要，而是一个额外的好处。
I have looked into NSNumberFormatter but have not found the proper solution, and I have yet to find a proper solution here on SO. Any help is greatly appreciated, thanks!
我已经研究了 NSNumberFormatter 但还没有找到合适的解决方案，我还没有在 SO 上找到合适的解决方案。非常感谢任何帮助，谢谢！
采纳答案by Kyle Begeman
Here are two methods I have come up with that work together to produce the desired effect. This will also automatically round up. This will also specify how many numbers total will be visible by passing the int dec.
以下是我提出的两种方法，它们可以协同工作以产生所需的效果。这也将自动四舍五入。这还将通过传递 int dec 来指定可见的数字总数。
Also, in the float to string method, you can change the @"%.1f"to @"%.2f", @"%.3f", etc to tell it how many visible decimals to show after the decimal point.
此外，在浮到字符串的方法，您可以更改@"%.1f"到@"%.2f"，@"%.3f"等来告诉它有多少可见小数，小数点后显示。
For Example:
52935 ---> 53K
52724 ---> 53.7K
-(NSString *)abbreviateNumber:(int)num withDecimal:(int)dec {
NSString *abbrevNum;
float number = (float)num;
NSArray *abbrev = @[@"K", @"M", @"B"];
for (int i = abbrev.count - 1; i >= 0; i--) {
// Convert array index to "1000", "1000000", etc
int size = pow(10,(i+1)*3);
if(size <= number) {
// Here, we multiply by decPlaces, round, and then divide by decPlaces.
// This gives us nice rounding to a particular decimal place.
number = round(number*dec/size)/dec;
NSString *numberString = [self floatToString:number];
// Add the letter for the abbreviation
abbrevNum = [NSString stringWithFormat:@"%@%@", numberString, [abbrev objectAtIndex:i]];
NSLog(@"%@", abbrevNum);
}
}
return abbrevNum;
}
- (NSString *) floatToString:(float) val {
NSString *ret = [NSString stringWithFormat:@"%.1f", val];
unichar c = [ret characterAtIndex:[ret length] - 1];
while (c == 48 || c == 46) { // 0 or .
ret = [ret substringToIndex:[ret length] - 1];
c = [ret characterAtIndex:[ret length] - 1];
}
return ret;
}
Hope this helps anyone else out who needs it!
希望这可以帮助其他需要它的人！
回答by Luca Iaco
-(NSString*) suffixNumber:(NSNumber*)number
{
if (!number)
return @"";
long long num = [number longLongValue];
int s = ( (num < 0) ? -1 : (num > 0) ? 1 : 0 );
NSString* sign = (s == -1 ? @"-" : @"" );
num = llabs(num);
if (num < 1000)
return [NSString stringWithFormat:@"%@%lld",sign,num];
int exp = (int) (log10l(num) / 3.f); //log10l(1000));
NSArray* units = @[@"K",@"M",@"G",@"T",@"P",@"E"];
return [NSString stringWithFormat:@"%@%.1f%@",sign, (num / pow(1000, exp)), [units objectAtIndex:(exp-1)]];
}
sample usage
示例用法
NSLog(@"%@",[self suffixNumber:@100]); // 100
NSLog(@"%@",[self suffixNumber:@1000]); // 1.0K
NSLog(@"%@",[self suffixNumber:@1500]); // 1.5K
NSLog(@"%@",[self suffixNumber:@24000]); // 24.0K
NSLog(@"%@",[self suffixNumber:@99900]); // 99.9K
NSLog(@"%@",[self suffixNumber:@99999]); // 100.0K
NSLog(@"%@",[self suffixNumber:@109999]); // 110.0K
NSLog(@"%@",[self suffixNumber:@5109999]); // 5.1M
NSLog(@"%@",[self suffixNumber:@8465445223]); // 8.5G
NSLog(@"%@",[self suffixNumber:[NSNumber numberWithInt:-120]]); // -120
NSLog(@"%@",[self suffixNumber:[NSNumber numberWithLong:-5000000]]); // -5.0M
NSLog(@"%@",[self suffixNumber:[NSNumber numberWithDouble:-3.5f]]); // -3
NSLog(@"%@",[self suffixNumber:[NSNumber numberWithDouble:-4000.63f]]); // -4.0K
[Update]
[更新]
Swift version below:
Swift 版本如下：
func suffixNumber(number:NSNumber) -> NSString {
var num:Double = number.doubleValue;
let sign = ((num < 0) ? "-" : "" );
num = fabs(num);
if (num < 1000.0){
return "\(sign)\(num)";
}
let exp:Int = Int(log10(num) / 3.0 ); //log10(1000));
let units:[String] = ["K","M","G","T","P","E"];
let roundedNum:Double = round(10 * num / pow(1000.0,Double(exp))) / 10;
return "\(sign)\(roundedNum)\(units[exp-1])";
}
sample usage
示例用法
print(self.suffixNumber(NSNumber(long: 100))); // 100.0
print(self.suffixNumber(NSNumber(long: 1000))); // 1.0K
print(self.suffixNumber(NSNumber(long: 1500))); // 1.5K
print(self.suffixNumber(NSNumber(long: 24000))); // 24.0K
print(self.suffixNumber(NSNumber(longLong: 99900))); // 99.9K
print(self.suffixNumber(NSNumber(longLong: 99999))); // 100.0K
print(self.suffixNumber(NSNumber(longLong: 109999))); // 110.0K
print(self.suffixNumber(NSNumber(longLong: 5109999))); // 5.1K
print(self.suffixNumber(NSNumber(longLong: 8465445223))); // 8.5G
print(self.suffixNumber(NSNumber(long: -120))); // -120.0
print(self.suffixNumber(NSNumber(longLong: -5000000))); // -5.0M
print(self.suffixNumber(NSNumber(float: -3.5))); // -3.5
print(self.suffixNumber(NSNumber(float: -4000.63))); // -4.0K
Hope it helps
希望能帮助到你
回答by gbitaudeau
Here my version ! Thanks to previous answers. The goals of this version is :
这是我的版本！感谢之前的回答。这个版本的目标是：
- Have better threshold control because small number details are more important that very big number details
- Use as much as possible
NSNumberFormatterto avoid location problems (like comma instead of dot in french) - Avoid ".0" and well rounding numbers, which can be customize using
NSNumberFormatterRoundingMode
- 有更好的阈值控制，因为小数字细节比大数字细节更重要
- 尽可能
NSNumberFormatter使用避免位置问题（如法语中的逗号而不是点） - 避免“.0”和四舍五入的数字，可以使用自定义
NSNumberFormatterRoundingMode
You can use all wonderful NSNumberFormatteroptions to fulfill your needs, see NSNumberFormatter Class Reference
您可以使用所有出色的NSNumberFormatter选项来满足您的需求，请参阅NSNumberFormatter 类参考
The code (gist):
代码（要点）：
extension Int {
func formatUsingAbbrevation () -> String {
let numFormatter = NSNumberFormatter()
typealias Abbrevation = (threshold:Double, divisor:Double, suffix:String)
let abbreviations:[Abbrevation] = [(0, 1, ""),
(1000.0, 1000.0, "K"),
(100_000.0, 1_000_000.0, "M"),
(100_000_000.0, 1_000_000_000.0, "B")]
// you can add more !
let startValue = Double (abs(self))
let abbreviation:Abbrevation = {
var prevAbbreviation = abbreviations[0]
for tmpAbbreviation in abbreviations {
if (startValue < tmpAbbreviation.threshold) {
break
}
prevAbbreviation = tmpAbbreviation
}
return prevAbbreviation
} ()
let value = Double(self) / abbreviation.divisor
numFormatter.positiveSuffix = abbreviation.suffix
numFormatter.negativeSuffix = abbreviation.suffix
numFormatter.allowsFloats = true
numFormatter.minimumIntegerDigits = 1
numFormatter.minimumFractionDigits = 0
numFormatter.maximumFractionDigits = 1
return numFormatter.stringFromNumber(NSNumber (double:value))!
}
}
let testValue:[Int] = [598, -999, 1000, -1284, 9940, 9980, 39900, 99880, 399880, 999898, 999999, 1456384, 12383474]
testValue.forEach() {
print ("Value : \(Value : 598 -> 598
Value : -999 -> -999
Value : 1000 -> 1K
Value : -1284 -> -1.3K
Value : 9940 -> 9.9K
Value : 9980 -> 10K
Value : 39900 -> 39.9K
Value : 99880 -> 99.9K
Value : 399880 -> 0.4M
Value : 999898 -> 1M
Value : 999999 -> 1M
Value : 1456384 -> 1.5M
Value : 12383474 -> 12.4M
) -> \(Results:
[self abbreviateNumber:987] ---> 987
[self abbreviateNumber:1200] ---> 1.2K
[self abbreviateNumber:12000] ----> 12K
[self abbreviateNumber:120000] ----> 120K
[self abbreviateNumber:1200000] ---> 1.2M
[self abbreviateNumber:1340] ---> 1.3K
[self abbreviateNumber:132456] ----> 132.5K
-(NSString *)abbreviateNumber:(int)num {
NSString *abbrevNum;
float number = (float)num;
//Prevent numbers smaller than 1000 to return NULL
if (num >= 1000) {
NSArray *abbrev = @[@"K", @"M", @"B"];
for (int i = abbrev.count - 1; i >= 0; i--) {
// Convert array index to "1000", "1000000", etc
int size = pow(10,(i+1)*3);
if(size <= number) {
// Removed the round and dec to make sure small numbers are included like: 1.1K instead of 1K
number = number/size;
NSString *numberString = [self floatToString:number];
// Add the letter for the abbreviation
abbrevNum = [NSString stringWithFormat:@"%@%@", numberString, [abbrev objectAtIndex:i]];
}
}
} else {
// Numbers like: 999 returns 999 instead of NULL
abbrevNum = [NSString stringWithFormat:@"%d", (int)number];
}
return abbrevNum;
}
- (NSString *) floatToString:(float) val {
NSString *ret = [NSString stringWithFormat:@"%.1f", val];
unichar c = [ret characterAtIndex:[ret length] - 1];
while (c == 48) { // 0
ret = [ret substringToIndex:[ret length] - 1];
c = [ret characterAtIndex:[ret length] - 1];
//After finding the "." we know that everything left is the decimal number, so get a substring excluding the "."
if(c == 46) { // .
ret = [ret substringToIndex:[ret length] - 1];
}
}
return ret;
}
.formatUsingAbbrevation ())")
}
Result :
结果 ：
extension Int {
var abbreviated: String {
let abbrev = "KMBTPE"
return abbrev.characters.enumerated().reversed().reduce(nil as String?) { accum, tuple in
let factor = Double(self) / pow(10, Double(tuple.0 + 1) * 3)
let format = (factor.truncatingRemainder(dividingBy: 1) == 0 ? "%.0f%@" : "%.1f%@")
return accum ?? (factor > 1 ? String(format: format, factor, String(tuple.1)) : nil)
} ?? String(self)
}
}
回答by Ricardo Corrêa de Souza
I had the same issue and ended up using Kyle's approach but unfortunately it breaks when numbers like 120000are used, showing 12kinstead of 120Kand I needed to show small numbers like: 1.1K instead of rounding down to 1K.
我遇到了同样的问题并最终使用了 Kyle 的方法，但不幸的是，当使用120000 之类的数字时它会中断，显示12k而不是120K，我需要显示小数字，例如：1.1K 而不是四舍五入到 1K。
So here's my edit from Kyle's original idea:
所以这是我对凯尔最初想法的编辑：
NSNumberFormatter *numFormatter = [[NSNumberFormatter alloc] init];
[numFormatter setPositiveFormat:@"0M"];
[numFormatter setMultiplier:[NSNumber numberWithDouble:0.000001]];
I Hope this can help you guys.
我希望这可以帮助你们。
回答by AbdulMomen ?????????
Flávio J Vieira Caetano's answerconverted to Swift 3.0
Flávio J Vieira Caetano 的答案转换为 Swift 3.0
NSString *formattedNumber = [numFormatter stringFromNumber:[NSNumber numberWithInteger:4000000]]; //@"4M"
回答by Taidg Murphy
I ran into a similar issue trying to format y-axis values in Shinobi Charts. It required using a NSNumberFormatter, so I eventually came up with this
我在尝试格式化 Shinobi 图表中的 y 轴值时遇到了类似的问题。它需要使用 NSNumberFormatter，所以我最终想出了这个
/*
With "onlyShowDecimalPlaceForNumbersUnder" = 10:
Original number: 598 - Result: 598
Original number: 1000 - Result: 1.0K
Original number: 1284 - Result: 1.3K
Original number: 9980 - Result: 10K
Original number: 39900 - Result: 40K
Original number: 99880 - Result: 100K
Original number: 999898 - Result: 1.0M
Original number: 999999 - Result: 1.0M
Original number: 1456384 - Result: 1.5M
Original number: 12383474 - Result: 12M
*/
- (NSString *)suffixNumber:(NSNumber *)number
{
if (!number)
return @"";
long long num = [number longLongValue];
if (num < 1000)
return [NSString stringWithFormat:@"%lld",num];
int exp = (int) (log(num) / log(1000));
NSArray * units = @[@"K",@"M",@"G",@"T",@"P",@"E"];
int onlyShowDecimalPlaceForNumbersUnder = 10; // Either 10, 100, or 1000 (i.e. 10 means 12.2K would change to 12K, 100 means 120.3K would change to 120K, 1000 means 120.3K stays as is)
NSString *roundedNumStr = [NSString stringWithFormat:@"%.1f", (num / pow(1000, exp))];
int roundedNum = [roundedNumStr integerValue];
if (roundedNum >= onlyShowDecimalPlaceForNumbersUnder) {
roundedNumStr = [NSString stringWithFormat:@"%.0f", (num / pow(1000, exp))];
roundedNum = [roundedNumStr integerValue];
}
if (roundedNum >= 1000) { // This fixes a number like 999,999 from displaying as 1000K by changing it to 1.0M
exp++;
roundedNumStr = [NSString stringWithFormat:@"%.1f", (num / pow(1000, exp))];
}
NSString *result = [NSString stringWithFormat:@"%@%@", roundedNumStr, [units objectAtIndex:(exp-1)]];
NSLog(@"Original number: %@ - Result: %@", number, result);
return result;
}
To get a formatted value
获取格式化的值
extension Int {
var abbreviated: String {
let abbrev = "KMBTPE"
return abbrev.characters
.enumerated()
.reversed()
.reduce(nil as String?) { accum, tuple in
let factor = Double(self) / pow(10, Double(tuple.0 + 1) * 3)
let format = (factor - floor(factor) == 0 ? "%.0f%@" : "%.1f%@")
return accum ?? (factor >= 1 ? String(format: format, factor, String(tuple.1)) : nil)
} ?? String(self)
}
}
This solution does not having rounding included, but if you (or anyone else) just needs something simple, this could work. If you need by the thousand instead of by million, you change the "M" to a "K" in the setPostiveFormat method, and change the NSNumber value in the multiplier to 0.001 .
此解决方案不包含舍入，但如果您（或其他任何人）只需要一些简单的东西，这可以工作。如果您需要千而不是百万，则在 setPostiveFormat 方法中将“M”更改为“K”，并将乘数中的 NSNumber 值更改为 0.001 。
回答by beebcon
After trying a couple of these solutions, Luca laco appears to have it closest, but I've made some amendments to his method in order to have more control over how many digits will appear (i.e. if you want 120.3K to be shorter, you can limit it to 120K). Additionally, I've added an extra step that ensures a number like 999,999 doesn't appear as 1000.0K, rather 1.0M.
在尝试了几个这些解决方案后，Luca laco 似乎最接近它，但我对他的方法进行了一些修改，以便更好地控制将出现的位数（即，如果您希望 120.3K 更短，您可以将其限制为 120K）。此外，我添加了一个额外的步骤，以确保像 999,999 这样的数字不会显示为 1000.0K，而是 1.0M。
func abbreviateNumber(num: NSNumber) -> NSString {
var ret: NSString = ""
let abbrve: [String] = ["K", "M", "B"]
var floatNum = num.floatValue
if floatNum > 1000 {
for i in 0..<abbrve.count {
let size = pow(10.0, (Float(i) + 1.0) * 3.0)
println("\(size) \(floatNum)")
if (size <= floatNum) {
let num = floatNum / size
let str = floatToString(num)
ret = NSString(format: "%@%@", str, abbrve[i])
}
}
} else {
ret = NSString(format: "%d", Int(floatNum))
}
return ret
}
func floatToString(val: Float) -> NSString {
var ret = NSString(format: "%.1f", val)
var c = ret.characterAtIndex(ret.length - 1)
while c == 48 {
ret = ret.substringToIndex(ret.length - 1)
c = ret.characterAtIndex(ret.length - 1)
if (c == 46) {
ret = ret.substringToIndex(ret.length - 1)
}
}
return ret
}
abbreviateNumber(123)
abbreviateNumber(12503)
abbreviateNumber(12934203)
abbreviateNumber(12234200003)
abbreviateNumber(92234203)
abbreviateNumber(9223.3)
回答by Flávio J Vieira Caetano
I know there are already lots of answers and different ways, but this is how I solved it with a more functional approach:
我知道已经有很多答案和不同的方法，但这就是我用更实用的方法解决它的方法：
-(NSString*) suffixNumber:(NSNumber*)number
double value = [number doubleValue];
NSUInteger index = 0;
NSArray *suffixArray = @[@"", @"K", @"M", @"B", @"T", @"P", @"E"];
while ((value/1000) >= 1){
value = value/1000;
index++;
}
//3 line of code below for round doubles to 1 digit
NSNumberFormatter *fmt = [[NSNumberFormatter alloc] init];
[fmt setMaximumFractionDigits:1];
NSString *valueWith1Digit = [fmt stringFromNumber:[NSNumber numberWithFloat:value]];
NSString *svalue = [NSString stringWithFormat:@"%@%@",valueWith1Digit, [suffixArray objectAtIndex:index]];
return svalue;
}
回答by coderek
Swift version
迅捷版
Direct translation from Objective-C version
从 Objective-C 版本直接翻译
NSLog(@"%@",[self suffixNumber:@100]); // 100
NSLog(@"%@",[self suffixNumber:@1000]); // 1K
NSLog(@"%@",[self suffixNumber:@10345]); // 10.3K
NSLog(@"%@",[self suffixNumber:@10012]); // 10K
NSLog(@"%@",[self suffixNumber:@123456]); // 123.5K
NSLog(@"%@",[self suffixNumber:@4384324]); // 4.4M
NSLog(@"%@",[self suffixNumber:@10000000]) // 10M
回答by Phan Van Linh
You can use this simple function, the idea is easy to understand
可以使用这个简单的函数，思路很容易理解
##代码##Test
测试
##代码##