数学与bigdecimal导致0

时间:2012-12-03 11:26:37

标签: java precision

我正试图用BigDecimal解决问题。我的代码:

BigDecimal tweetcount = new BigDecimal(3344048);
BigDecimal emotionCountBig = new BigDecimal(855937);
BigDecimal emotionCountSentenceBig = new BigDecimal(84988); 

MathContext mc = new MathContext(64);
PMI[cnt] = (emotionCountSentenceBig.divide((tweetcount.multiply(emotionCountBig,mc)),RoundingMode.HALF_UP));

我想做的是:emotionCountSentenceBig/(emotionCountBig*tweetcount)

(值可以更大)

如果我尝试这个,我得到零,这是不可能的。有什么帮助吗?

2 个答案:

答案 0 :(得分:4)

您还需要为分部指定MathContext:

emotionCountSentenceBig.divide(tweetcount.multiply(emotionCountBig, mc), mc);

这给出了预期的结果:

  

2.969226352632111794036880818610913852084810652372969382467557947E-8

现在正如@PeterLawrey正确评论你可以使用双打:

public static void main(String[] args) throws Exception {
    double tweetcount = 3344048;
    double emotionCount = 855937;
    double emotionCountSentence = 84988;

    double result = emotionCountSentence / (tweetcount * emotionCount);

    System.out.println("result = " + result);
}

打印:

  

result = 2.9692263526321117E-8

请注意,如果您使用:

double result = 84988 / (3344048 * 855937);

你实际上正在对整数进行操作(*和/),它将返回0.你可以通过显式使用double来阻止它(例如d):

double result = 84988d / (3344048d * 855937);

答案 1 :(得分:2)

我会使用double

int tweetcount = 3344048;
int emotionCountBig = 855937;
int emotionCountSentenceBig = 84988;

double pmi = emotionCountSentenceBig/((double) tweetcount * emotionCountBig);
System.out.println(pmi);

打印

2.9692263526321117E-8

使用BigDecimal接近答案

2.969226352632111794036880818610913852084810652372969382467557947E-8