根据上一个/下一个像元值确定像元值 [英] Determining a cell value based on previous/next cell value(s)
问题描述
这是我的上一篇文章的后续文章：根据其他单元格值更改多个单元格值
This is a follow-up from a previous post of mine: Change multiple cell values based on other cell value(s)
这是对象的正常循环看起来像：
this is what a normal cycle of an object looks like:
DateTime limitswitchopen limitswitchclose safetyedgeclose safetyedgeopen photocells traploopext rectloopext moving comment
0 2018-11-12 15:04:46.861 0 1 0 0 0 0 0 close NaN
1 2018-11-12 15:04:57.149 0 1 0 0 0 0 0 close NaN
2 2018-11-12 15:05:05.046 0 1 0 0 0 0 0 close Normaal
3 2018-11-12 15:05:06.859 0 0 0 0 0 0 0 movingToopen Normaal
4 2018-11-12 15:05:10.080 0 0 0 0 0 0 0 movingToopen Normaal
5 2018-11-12 15:05:11.801 1 0 0 0 0 0 0 open Normaal
6 2018-11-12 15:05:13.409 1 0 0 0 0 0 0 open Normaal
7 2018-11-12 15:05:17.142 1 0 0 0 0 1 0 open Normaal
8 2018-11-12 15:05:18.754 1 0 0 0 0 1 0 open Normaal
9 2018-11-12 15:05:19.055 1 0 0 0 0 0 1 open Normaal
10 2018-11-12 15:05:19.763 1 0 0 0 0 0 1 open Normaal
11 2018-11-12 15:05:20.367 1 0 0 0 0 0 0 open Normaal
12 2018-11-12 15:05:21.575 0 0 0 0 0 0 0 movingToclose Normaal
13 2018-11-12 15:05:23.385 0 0 0 0 0 0 0 movingToclose Normaal
14 2018-11-12 15:05:26.505 0 1 0 0 0 0 0 close Normaal
15 2018-11-12 15:05:26.906 0 1 0 0 0 0 0 close NaN
我需要知道对象朝哪个方向移动。所以我所做的是
I need to know in which direction the object is moving. So what i did was
df [’moving] = df [（df.limitswitchclose == 0）& （df.limitswitchopen == 0）]
df ['open'] = df [（dfsamen.limitswitchclose == 0）& （df.limitswitchopen == 1）]
df [’close’] = df [（dfsamen.limitswitchclose == 1）& （df.limitswitchopen == 0）]
我合并为一列，然后发表了上一篇文章。
，然后我使用该代码在显示方向的上方创建数据框。在这种情况下有效。但是当周期中断时，方向会根据打开/关闭期间的中断而改变。
which I merged into one column and then made the previous post. and then i used that code to create dataframe above where the direction is shown. Which works in this case. But when the cycle gets interrupted the direction changes based on an interruption during opening/closing.
DateTime limitswitchopen limitswitchclose safetyedgeclose safetyedgeopen photocells traploopext rectloopext moving comment
41 2018-11-12 15:06:09.931 0 1 0 0 0 0 0 close Fotocellopen
42 2018-11-12 15:06:11.944 0 0 0 0 0 0 0 movingToclose Fotocellopen
43 2018-11-12 15:06:13.756 0 0 0 0 1 0 0 movingToclose Fotocellopen
44 2018-11-12 15:06:15.168 0 0 0 0 0 0 0 movingToclose Fotocellopen
45 2018-11-12 15:06:18.388 0 1 0 0 0 0 0 close Fotocellopen
46 2018-11-12 15:06:20.100 0 0 0 0 0 0 0 movingToopen Fotocellopen
47 2018-11-12 15:06:23.316 0 0 0 0 0 0 0 movingToopen Fotocellopen
48 2018-11-12 15:06:25.730 1 0 0 0 0 0 0 open Fotocellopen
49 2018-11-12 15:06:26.637 1 0 0 0 0 0 0 open Fotocellopen
50 2018-11-12 15:06:27.644 1 0 0 0 0 1 0 open Fotocellopen
51 2018-11-12 15:06:28.550 1 0 0 0 0 1 1 open Fotocellopen
52 2018-11-12 15:06:28.855 1 0 0 0 0 0 1 open Fotocellopen
53 2018-11-12 15:06:29.356 1 0 0 0 0 0 0 open Fotocellopen
54 2018-11-12 15:06:30.563 1 0 0 0 0 0 0 open Fotocellopen
55 2018-11-12 15:06:31.369 0 0 0 0 0 0 0 movingToclose Fotocellopen
56 2018-11-12 15:06:32.575 0 0 0 0 0 0 0 movingToclose Fotocellopen
57 2018-11-12 15:06:35.593 0 1 0 0 0 0 0 close Fotocellopen
在 43 2018-11-12 15：06：13.756 光电池= 1 使对象关闭，然后再次开始打开。
at 43 2018-11-12 15:06:13.756 photocells = 1 this will make the object close and then start to open again.
所以此数据帧应该是：
DateTime limitswitchopen limitswitchclose safetyedgeclose safetyedgeopen photocells traploopext rectloopext moving comment
41 2018-11-12 15:06:09.931 0 1 0 0 0 0 0 close Fotocellopen
42 2018-11-12 15:06:11.944 0 0 0 0 0 0 0 movingToopen Fotocellopen
43 2018-11-12 15:06:13.756 0 0 0 0 1 0 0 movingToopen Fotocellopen
44 2018-11-12 15:06:15.168 0 0 0 0 0 0 0 movingToclose Fotocellopen
45 2018-11-12 15:06:18.388 0 1 0 0 0 0 0 close Fotocellopen
46 2018-11-12 15:06:20.100 0 0 0 0 0 0 0 movingToopen Fotocellopen
47 2018-11-12 15:06:23.316 0 0 0 0 0 0 0 movingToopen Fotocellopen
48 2018-11-12 15:06:25.730 1 0 0 0 0 0 0 open Fotocellopen
49 2018-11-12 15:06:26.637 1 0 0 0 0 0 0 open Fotocellopen
50 2018-11-12 15:06:27.644 1 0 0 0 0 1 0 open Fotocellopen
51 2018-11-12 15:06:28.550 1 0 0 0 0 1 1 open Fotocellopen
52 2018-11-12 15:06:28.855 1 0 0 0 0 0 1 open Fotocellopen
53 2018-11-12 15:06:29.356 1 0 0 0 0 0 0 open Fotocellopen
54 2018-11-12 15:06:30.563 1 0 0 0 0 0 0 open Fotocellopen
55 2018-11-12 15:06:31.369 0 0 0 0 0 0 0 movingToclose Fotocellopen
56 2018-11-12 15:06:32.575 0 0 0 0 0 0 0 movingToclose Fotocellopen
57 2018-11-12 15:06:35.593 0 1 0 0 0 0 0 close Fotocellopen
所以我需要一种确定对象是打开还是关闭的方法。
如果limitswitchclose从 1 变为 0 ，它将始终处于打开状态，并且limitswitchopen从<$ c开始$ c> 1 到 0 ，它将始终关闭。但是基于df中的其他列，它可以改变方向。如果在打开期间 safetyedgeopen = 1 ，它将再次关闭。但是，如果在打开期间 traploopext = 1 ，它将继续打开。
So what i need is a way to determine whether the object is opening or closing.
If limitswitchclose goes from 1 to 0 it will always be opening and if limitswitchopen goes from 1 to 0 it will always be closing. But based on the other columns in df it can change direction. If safetyedgeopen = 1 during opening it will close again. But if traploopext = 1 during opening it will continue opening.
我该如何解决这个问题？
How do i tackle this problem?
（我将继续尝试解决该问题并将其答案发布，如果可以的话，我可以举更多示例说明我希望数据框如何显示，但该帖子越来越long）
(I'll continue trying to solve it and post my answer if it works, I can give more examples of how I want the dataframe to look but the post was getting long)
推荐答案
如果我了解您的主要要求，那就是如何根据上一行的值分配值。我这样做的方法是将过滤条件简单地移动一行。我不会提供所有选项的解决方案，但是您可以适当地扩展过滤条件。
If I understand what you mainly ask, it's "how do I assign a value based on the value of previous row." The way I Would do this is to simply shift the filtering condition by one row. I won't give the solution for all the options, but you can just expand the filtering condition appropriately.
此行代码根据相邻行在光电管中的值为1进行过滤，如果是这种情况，则将当前行 tomoving替换为
this line of code filters by where the neighbouring row would have the value of 1 in photocells, and if that's the case, then replaces the current rows "tomoving" to "toclosing", or "toclosing", to "tomoving"
df.loc [df.shift（1）.photocells == 1，' moving'] = df.loc [df.shift（1）.photocells == 1，'moving']。str.replace（'Toclose'，'To_open'）。str.replace（'Toopen'，'To_close'） .str.replace（'_'，''）
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