绘制 x 轴点的更有效方法？
我一直在从事一个需要用价格结果填充条形图的项目。该图表显示给定价格范围内的商品数量。例如，如果在亚马逊上有 9 个商品的价格范围为 0-10 美元，则 x 轴将显示 0-10 美元，y 轴将填充值 9。
我的条形图有 8 个条形，全部带有类似的价格范围。 $0-$10、$10-$20、$20-$30...等等。
我的问题是：定义这些单独点的最佳方式是什么？这些商品之间没有共同的价格范围，因此x 轴不能是静态数字。它们必须在结果范围内动态计算。
因此，目前我正在按如下方式创建 x 轴点：
我采用最低结果： @numbers[0] = results[0];
我取最高的结果： @numbers[8] = results[-1];
然后我找到两者的中位数： @数字[4] = (@数字[0]+@数字[8])/2;
然后我再重复该过程 6 次
@numbers[2] = (@numbers[0]+@numbers[4])/2; @数字[6] = (@数字[4]+@数字[8])/2; @数字[1] = (@数字[0]+@数字[2])/2; @数字[3] = (@数字[2]+@数字[4])/2; @数字[5] = (@数字[4]+@数字[6])/2; @数字[7] = (@数字[6]+@数字[8])/2;
这给了我需要的结果，但它看起来非常重复，我想有更好的方法。
我尝试创建一个循环，但无法以更简洁的方式编写它。
有没有更快的方法来做到这一点，或者也许有更多类似于 DRY 的方法？
I have been working on a project that requires a bar graph to be populated with price results. The chart displays the number of items within a given price range. For instance, if on amazon there are 9 items within the price range of $0-$10 the x-axis would display $0-$10 and the y-axis would be populated with a value of 9.
My bar graph has 8 bars, all with similar price ranges. $0-$10, $10-$20, $20-$30... etc.
My question is this: What is the best way to define those individual points? There is no common price range between these items, so the x-axis cannot be static numbers. They must be dynamically calculated within the range of results.
As such, currently I am creating the x-axis points as follows:
I take the lowest result:@numbers[0] = results[0];
And I take the highest result:@numbers[8] = results[-1];
Then I find the median of the two:
@numbers[4] = (@numbers[0]+@numbers[8])/2;
I then repeat the process 6 more times
@numbers[2] = (@numbers[0]+@numbers[4])/2; @numbers[6] = (@numbers[4]+@numbers[8])/2; @numbers[1] = (@numbers[0]+@numbers[2])/2; @numbers[3] = (@numbers[2]+@numbers[4])/2; @numbers[5] = (@numbers[4]+@numbers[6])/2; @numbers[7] = (@numbers[6]+@numbers[8])/2;
This gives me the results I need, but it seems awfully repetitive and I would imagine there is a better way.
I tried creating a loop, but I could not write it in a less verbose manner.
Is there a quicker way to do this, or perhaps something more along the lines of DRY?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
您的垃圾箱大小始终相同吗？在您的示例中，所有共享范围=10。如果是这样，那么你可以这样做：
bin n 的位置将是numbers[0] 的 x 轴位置加上 n 乘以 binspacing。
Are your bins always of equal size? In your example, all share range=10. If so, then you could do:
and the position of bin n would be the x-axis position of numbers[0] plus n times the binspacing.