英文:
Iterating list of map for writing only its content
问题
以下为您翻译的内容：
每个参与者都会为其他人投票一些数值，我想要按照代理的名称来记录这些信息。
通过以下代码，
cj.makingDefault().stream().forEach(a -> out.print(a.entrySet()));
我得到了一个打印出来的Map列表，内容如下：
[ari=[0.03, 0.03, 0.03, 0.03, 0.03, 0.03, 0.03],hyo=[0.31, 0.31, 0.317, 0.31..., ji=[...], yoo=[...], mi=[...]..]
我该如何只打印代理的数值，用换行符分隔开呢？
期望的结果应该是这样的：
0.3, 0.3, 0.3, ...0.30.31, 0.31, 0.31, ...0.31,...
如果有7个参与者，那么会是一个7x7的矩阵。
非常感谢您提前的帮助～
编辑）
作为一个临时解决方案，我修改了“cj.makingDefault()”函数，使其仅返回Map<String, List
cj.makingDefault().entrySet().stream().forEach(a -> out.print(a.getValue() + "\n"));
然后我得到了如下结果：
[0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865][0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144][0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273][0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558][0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038][0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696][0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337]
那么，我该如何消除这些方括号呢？这样就完美了。
英文:
Each participant is voting some numerical values to each others, and I want to write this information by agent's name.
By the following code,
cj.makingDefault().stream().forEach(a->out.print(a.entrySet()));
I get a List of Map printed which is,
[ari=[0.03, 0.03, 0.03, 0.03, 0.03, 0.03, 0.03],hyo=[0.31, 0.31, 0.317, 0.31..., ji=[...], yoo=[...], mi=[...]..]
How can I print out only the numberic value by agents, separated by newline?
Desired results would be like
0.3, 0.3, 0.3, ...0.30.31, 0.31, 0.31, ...0.31,...
If there would be 7 participants, 7 by 7 matrix.
Thank you in advance for your help~
edit)
As a temporary solution, I modified "cj.makingDefault()" function to return only Map<String, List<Double>> and changed code to this.
cj.makingDefault().entrySet().stream().forEach(a->out.print(a.getValue()+"\n"));
and I got
[0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865, 0.0056009042654552865][0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144, 0.31583809304895144][0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273, 0.9924090841607273][0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558, 0.8093898651356558][0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038, 0.015200181208559038][0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696, 0.03947825506938696][0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337, 0.39173806385104337]
Then how can I eliminate the suqare braket of this..? it would be perfect..
答案1
得分: 1
你所看到的是集合的默认toString实现，其中包括方括号。
你需要自己打印列表中的元素。
cj.makingDefault().stream().forEach(a -> print(a.values()));private void print(Collection<Double> values) {System.out.println(values.stream().map(String::valueOf).collect(Collectors.joining(",")));}
英文:
What you are seeing is the default toString implementation of a collection which will include the square brackets.
You have to print the elements in the list yourself.
cj.makingDefault().stream().forEach(a -> print(a.values()));private void print(Collection<Double> values) {System.out.println(values.stream().map(String::valueOf).collect(Collectors.joining(",")));}
答案2
得分: 1
你必须首先迭代地图中的值，这些值是列表，然后迭代列表中的项目，以覆盖标准的toString输出。
Map<String, List<Double>> map = ...String result = map.values().stream().map(s -> s.stream().map(String::valueOf).collect(Collectors.joining(", "))).collect(Collectors.joining("\n"));
英文:
You must first iterate the map values, that are lists, and then the items in those lists to override the standard toString output.
Map<String, List<Double>> map = ...String result = map.values().stream().map(s -> s.stream().map(String::valueOf).collect(Collectors.joining(", "))).collect(Collectors.joining("\n"));
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