谷歌折线图如何不插入缺失的数据 [英] Google line charts how to NOT interpolate missing data

问题描述

我有一个包含一系列事件的数据库表，这些事件具有发生时的日期。现在我想显示每个月的事件数量并以折线图形式显示。
我通过php获取数据并使用data.addrow在循环中生成js代码

现在这个工作正常，但如果有一个月没有事件不会被添加为一行，并且该行只会插入缺失的月份。
因此，例如1月5日事件，2月0日和3月5日事件，将创建一个从一月到三月的水平线，使其看起来像在那里2月份的5个事件。

我怎样才能使行在空月份下降到零？

我是否需要用空填充空行/月？这看起来并不实用。

解决方案

如果您希望该行下降到零，不同的方式。 Google使用 <$为您提供了两种处理空值（空值）的方法c $ c> interpolateNulls 选项：

1. True 会猜测（图表中的当前行为）


2. False 会在空数据点的行中留下空缺（不是您想要的）

要让所有空值显示为零，您需要使用for循环检查数据，并将任何空值更改为0.类似下面的代码。

  for（i = 0; i< data.getNumberOfRows（）; i ++）{ 
 if（data.getValue（1，i）== null）
 data.setValue（1，i）= 0 
} 
  pre> 
 
 
上面的代码没有经过测试，我不知道您的数据是如何设置的，所以我没有花太多时间测试它（我没有想法有多少列，等你的数据哈s，您可以根据需要进行调整和测试）。 
I have a database table with a series of events, these events have dates for when they happen. Now I would like to display the number of events for each month and show them in a line chart.
I get the data via php and generate the js-code in a loop using data.addrow

Now this works fine, but if there is a month with no events it will not be added as a row and the line will simply interpolate the missing month.
So for example january 5 events, february 0 and march 5 events, would create a horizontal line going from january to march making it look like there where 5 events in february.

How can i make the line go down to zero for a "empty" month?

Do I need to fill in empty rows/months with null? This would not seem very practical.
解决方案
If you want the line to go down to zero, you are interpolating, just in a different way. Google gives you two ways to deal with empty values (nulls) using the interpolateNulls option:

True will guess what value belongs in between (the current behavior in your chart)
False will leave a break in the line for empty data points (not what you want)
To get all null values to show as zero, you would need to go through your data with a for loop and change any null value to 0. Something like the below code.
for (i = 0; i < data.getNumberOfRows(); i++) {
  if ( data.getValue(1, i) == null )
    data.setValue(1, i) = 0
}
The above code isn't tested, and I don't know how your data is set up so I didn't spend much time testing it (I have no idea how many columns, etc. your data has, you can adjust and test as needed).

                        
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