英文:
Calculating the Mode
问题
以下是翻译好的内容：
public double mode() {
if (data == null) {
double mode;
mode = Double.NaN;
return mode; // 如果数组为空，则返回无效值
}
double mode = 0;
double modeCounter = 0;
for (int c = 0; c < data.length; ++c) {
int count = 0;
for (int i = 0; i < data.length; ++i) {
if (data[i] == data[c]) // 修正这里的比较，应为 data[i] == data[c]
++count;
}
if (count > modeCounter) {
modeCounter = count;
mode = data[c];
}
}
if (!(modeCounter > 1)) {
return Double.NaN;
}
return mode;
}
英文:
I am trying to write a method that will calculate the mode of a set of data and return it. If there is no mode or more than one mode (for example: if 4 & 2 returned the most amount of times) I, want the function to return Double.NaN. I was able to get the mode, but I'm having trouble making my code return Double NaN if there is more than one mode.
Here is my code!
public double mode() {
if (data == null) {
double mode;
mode = Double.NaN;
return mode; // returns no value if array is null
}
double mode = 0;
double modeCounter = 0;
for (int c = 0; c < data.length; ++c) {
int count = 0;
for (int i = 0; i < data.length; ++i) {
if (data[i] == data[i])
++count;
}
if (count > modeCounter) {
modeCounter = count;
mode = data[c];
}
}
if (!(modeCounter > 1)) {
return Double.NaN;
}
return mode;
}
答案1
得分: 0
假设您正在使用Java。
我们可以通过简单地创建一个Map，其中键是单独的数据点，值是它们在数据数组中的对应计数，来找到众数（以及您设置的返回Double.NaN的限制）。
使用Stream可以编写可读性强且简洁的代码。
如果您的数组/ArrayList中每个成员的计数都已知，您可以根据您的要求使用此计数来得出结论。
以下是实现上述解释的代码：
public static Double mode(List<Double> data) {
if (data.isEmpty()) {
// 如果输入的ArrayList为空，则返回NaN
return Double.NaN;
}
Map<Double, Integer> count = new HashMap<>();
data.forEach(
i -> {
count.putIfAbsent(i, 1);
count.put(i, count.get(i) + 1);
}
);
int maxCount = count.values().stream().max(Integer::compare).get(); // 这假设数据不为空
if (count.values().stream().filter(i -> i == maxCount).count() > 1) {
// 存在多个众数
return Double.NaN;
}
// 在这里我们可以确保只有一个众数存在
return count
.entrySet()
.stream()
.filter(e -> e.getValue() == maxCount)
.findFirst() // 因为只有一个众数
.get()
.getKey();
}
英文:
I assume that you're using Java.
We can find mode (along with the restrictions that you place to return Double.NaN) by simply creating a Map in which keys are the individual data points and the values are their corresponding count in the data array.
Streams will be superuseful to write readable and succinct code.
If you have count of every member in your array/ArrayList, you can simply use this count to draw conclusions based on your requirement.
See the following code that implements the above explanation:
public static Double mode(List<Double> data) {
if (data.isEmpty()) {
// returning NaN if the input arraylist is empty
return Double.NaN;
}
Map<Double, Integer> count = new HashMap<>();
data.forEach(
i -> {
count.putIfAbsent(i, 1);
count.put(i, count.get(i) + 1);
}
);
int maxCount = count.values().stream().max(Integer::compare).get(); // this assumes that the data is not empty
if (count.values().stream().filter(i -> i == maxCount).count() > 1) {
// more than one mode present
return Double.NaN;
}
// here we'll be sure that only one mode exists
return count
.entrySet()
.stream()
.filter(e -> e.getValue() == maxCount)
.findFirst() // as we've only one mode
.get()
.getKey();
}
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