根据列表中的最大值计算上限

时间:2022-09-28 18:25:56

I have a list of transactions, integer. I need to create range to display that list in a bar chart. I am trying to compute upper bound of that range based on the maximum value from the list. Eg: If the list is {8, 9 , 91, 45}, then 1. calculate max as 91. 2. calculate range, lower bound always 0 but upper bound can be 100.

我有一个事务列表,整数。我需要创建范围以在条形图中显示该列表。我试图根据列表中的最大值计算该范围的上限。例如:如果列表是{8,9,91,45},则1.将max计算为91. 2.计算范围,下限始终为0但上限可以为100。

Any Ideas how to do that?

任何想法如何做到这一点?

1 个解决方案

#1


1  

Assuming that you're using Java 8, use a java.util.Stream:

假设您使用的是Java 8,请使用java.util.Stream:

List<Integer> list = Arrays.asList(8, 9 , 91, 45); //make your list

int max = list
    .stream() //get a stream for the list
    .mapToInt((i) -> i) //Get an intstream
    .max() //Get the max value
    .getAsInt(); //Unwrap the optional

Since any number greater than or equal to the max is an upper bound, we can just use the max as one:

由于任何大于或等于max的数字都是上限,我们可以使用max作为一个:

int upperBound = max;

If, as @JohnColeman suggested, you need upperBound to be a multiple of 10, just do:

如果像@JohnColeman建议的那样,你需要将upperBound设为10的倍数,那么就这样做:

if (max % 10 == 0) {
    upperBound = max; //if already a multiple of 10, leave as is
} else {
    upperBound = max + (10 - max % 10); //Otherwise, add the amount which will make it a multiple of 10.
}

Since you want the lower bound to be 0, that's easy:

由于您希望下限为0,这很容易:

int lowerBound = 0;

However, if you want the lower bound to be equal to the min, just use another stream:

但是,如果您希望下限等于min,只需使用另一个流:

int min = list
    .stream() //get a stream for the list
    .mapToInt((i) -> i) //Get an intstream
    .min() //Get the min value
    .getAsInt(); //Unwrap the optional

int lowerBound = min;

Again, you can make this a multiple of 10 if that's what you want:

同样,如果这是你想要的,你可以使它成为10的倍数:

if (min % 10 == 0) {
    lowerBound = min; //if already a multiple of 10, leave as is
} else {
    upperBound = min - min % 10; //Subtract it mod 10
}

Further Reading:

#1


1  

Assuming that you're using Java 8, use a java.util.Stream:

假设您使用的是Java 8,请使用java.util.Stream:

List<Integer> list = Arrays.asList(8, 9 , 91, 45); //make your list

int max = list
    .stream() //get a stream for the list
    .mapToInt((i) -> i) //Get an intstream
    .max() //Get the max value
    .getAsInt(); //Unwrap the optional

Since any number greater than or equal to the max is an upper bound, we can just use the max as one:

由于任何大于或等于max的数字都是上限,我们可以使用max作为一个:

int upperBound = max;

If, as @JohnColeman suggested, you need upperBound to be a multiple of 10, just do:

如果像@JohnColeman建议的那样,你需要将upperBound设为10的倍数,那么就这样做:

if (max % 10 == 0) {
    upperBound = max; //if already a multiple of 10, leave as is
} else {
    upperBound = max + (10 - max % 10); //Otherwise, add the amount which will make it a multiple of 10.
}

Since you want the lower bound to be 0, that's easy:

由于您希望下限为0,这很容易:

int lowerBound = 0;

However, if you want the lower bound to be equal to the min, just use another stream:

但是,如果您希望下限等于min,只需使用另一个流:

int min = list
    .stream() //get a stream for the list
    .mapToInt((i) -> i) //Get an intstream
    .min() //Get the min value
    .getAsInt(); //Unwrap the optional

int lowerBound = min;

Again, you can make this a multiple of 10 if that's what you want:

同样,如果这是你想要的,你可以使它成为10的倍数:

if (min % 10 == 0) {
    lowerBound = min; //if already a multiple of 10, leave as is
} else {
    upperBound = min - min % 10; //Subtract it mod 10
}

Further Reading: