獲取最高值和最低值時統一代碼

Question

我的 IDE 給了我重復代碼警告，現在我試圖找出有什么方法可以統一或找到這兩種方法的共同抽象，它們的結構非常相似，但是做相反的事情嗎？ 我自己想不通。 謝謝

public double getLowestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    double lowestRate = Double.MAX_VALUE;
    for(Map<String, Double> ratesByDate : rates.values()){
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if(baseCurrecyRateToTargetCurrency < lowestRate){
            lowestRate = baseCurrecyRateToTargetCurrency;
        }
    }
    return lowestRate;
}

public double getHighestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    double highestRate = Double.MIN_VALUE;
    for(Map<String, Double> ratesByDate : rates.values()){
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if(baseCurrecyRateToTargetCurrency > highestRate){
            highestRate = baseCurrecyRateToTargetCurrency;
        }
    }
    return highestRate;
}

Answer 1

在我看來，那里有很多不必要的代碼。 您可以使用 Streams 以更好的方式實現相同的目標。 它看起來像這樣：

public double getHighestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency, String variant) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    DoubleStream doubleStream = rates.values().stream()
                 .mapToDouble(entry -> entry.get(targetCurrency));
    OptionalDouble result = variant.equals("max") ? doubleStream.max() : doubleStream.min();
    return result.orElse(0);
}

Answer 2

我同意前面的幾個答案，並相信它們可以很好地使用。

就個人而言，我會創建一個統一的方法來處理公共邏輯流，然后有兩個描述性方法調用這個“公共邏輯方法”

因此，我不必將布爾標志作為參數處理（在這種情況下是雙標志），我有機會表達哪個名稱最適合該方法，最后我將所有邏輯流集中在一個方法中。

例子：

public double getLowestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    return commonLogic(targetCurrency, getAllRatesInCurrentMonth(baseCurrency), Double.MAX_VALUE);
}

public double getHighestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    return commonLogic(targetCurrency, getAllRatesInCurrentMonth(baseCurrency), Double.MIN_VALUE);
}

private double commonLogic(String targetCurrency, Map<String, Map<String, Double>> rates, double rate) {
    for(Map<String, Double> ratesByDate : rates.values()){
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if(rate == Double.MIN_VALUE ? baseCurrecyRateToTargetCurrency > rate : baseCurrecyRateToTargetCurrency < rate){
            rate = baseCurrecyRateToTargetCurrency;
        }
    }
    return rate;
}

Answer 3

一個簡單的解決方案如下：你給函數一個速率，它要么是double.MIN_VALUE要么是double.MAX_VALUE並稱之為rate 。 在函數中，您可以檢查 rate 是double.MIN_VALUE還是double.MAX_VALUE 。 或者只是檢查rate是更小還是更大baseCurrecyRateToTargetCurrency 。 我不太確定它是否是一個常數。 這將是一個例子：

public double getCurrencyRateExtremaInTheCurrentMonth(String baseCurrency, String targetCurrency, double rate) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    double resultRate = rate
    for(String key : rates.keySet()){
        Map<String, Double> ratesByDate = rates.get(key);
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if((rate == double.MIN_VALUE && baseCurrecyRateToTargetCurrency > rate) || (rate == double.MAX_VALUE && baseCurrecyRateToTargetCurrency < rate)){
            resultRate = baseCurrecyRateToTargetCurrency;
        }
    }
    return resultRate;
}

我假設baseCurrecyRateToTargetCurrency具有相同的值，因為它在函數中以相同的方式檢索。 那正確嗎？