我正在使用count函数来知道count是否大于0。 但是要花费超过5分钟才能使特定色谱柱的大小达到40,000,000个。
下面是我的代码垃圾。
specficManufacturerdetailsSource = source.filter(col("ManufacturerSource").equalTo(individualManufacturerName));
specficManufacturerdetailsTarget = target.filter(col("ManufacturerTarget").equalTo(individualManufacturerName));
manufacturerSourceCount=specficManufacturerdetailsSource.count();
manufacturerTargetCount=specficManufacturerdetailsTarget.count();
System.out.println("Size of specfic manufacturer source ML :"+manufacturerSourceCount+"Size of specfic manufacturer target"+manufacturerTargetCount);
if(manufacturerSourceCount > 0 && manufacturerTargetCount > 0 ){
}
答案 0 :(得分:0)
根据您提到的上述要求,您不需要计算。
您可以使用findFirst()
代替计数,如果您发现任何值
manufacturerSourceCount.isPresent()
则表示count > 0