system.tag.readAll COUNTIF

Ignition
1

Ok, first things first, it's been awhile since I last posted, and it appears that my scripting skills have rusted over time.

I have 450 OPC Tags that are all integers. There are only 12 values that they can have. What I need is the equivalent of excel's COUNTIF function. I have been unsuccessful in using the system.tag.readAll scripting function.

Example of Tag Layout

There are 13 folders in the Curing folder:

Tags-> US9_FIS-> Curing->
0000
0100
...
1200

For 0000, there are 44 INT Tags. For 0100 - 1200 there are 34 INT Tags in each folder:

US9_FIS
      Curing
            0000
                0001_DTCode_Right
                0001_DTCode_Left
                0003_DTCode_Right
                0003_DTCode_Left
                .
                .
                .
                0041_DTCode_Right
                0041_DTCode_Left
            0100
                0101_DTCode_Right
                0101_DTCode_Left
                0103_DTCode_Right
                0103_DTCode_Left
                .
                .
                .
                0133_DTCode_Right
                0133_DTCode_Left
            .
            .
            .
            1200
                1201_DTCode_Right
                1201_DTCode_Left
                1203_DTCode_Right
                1203_DTCode_Left
                .
                .
                .
                1233_DTCode_Right
                1233_DTCode_Left

My idea is this, create 12 Memory tags to hold the count from each type of code that can exist. Then, loop through all the tags with a system.tag.readAll and count while looping through the tags. I also imagine that I would have to loop for each code, so 12 times? Any assistance would be greatly appreciated!!

Thank you all in advance.

2

I believe you can do this in one loop by using if statements:

code1Count = 0
code2Count = 0
...
code12Count = 0
tags = ["Tags/US9_FIS/Curing/0000/0001_DTCode_Right","Tags/US9_FIS/Curing/0000/0001_DTCode_Right", etc....]
values = system.tag.readAll(tags)
for x in range(len(tags)):
   value = x.value
   if value==#(Put Code one Value here):
      code1Count += 1
  elif value==#(Put Code two Value here):
      code2Count += 1
...
  else code12Count += 1
         
#Update memory tags with count values
system.tag.write("Tags/code1Count",code1Count)
system.tag.write("Tags/code2Count",code2Count)
...
system.tag.write("Tags/code12Count",code12Count)

I hope this helps

1 Like
3

Along the lines of Brandon’s suggestion, just avoiding the repetitive if statement:

codes = {}
tags = ["Tags/US9_FIS/Curing/0000/0001_DTCode_Right","Tags/US9_FIS/Curing/0000/0001_DTCode_Right", etc....]
values = system.tag.readAll(tags)

for i in range(len(tags)):
	if values[i].value:
		currentCode = tags[i].split('/')[3]
		if codes.get(currentCode) is not None:
			codes[currentCode] += 1
		else:
			codes[currentCode] = 1

Would output a dictionary like this:
{'0000': 1, '0100': 3, '0200': 0} etc

2 Likes
4

Thank you for the responses. Is there a way build the tag list dynamically without having to list all 450 tags? That was also something that I was trying to avoid.

5 
root = "US9_FIS/Curing"
folders = ["%02d00" % i for i in range(13)]
suffix = "_DTCode_"

tags = []

def getpath(folder, i, suffix):
	return '/'.join([root] + [folder] + ["%s%02d%s%s" % (folder[:2], i, suffix, "Left" if i % 2 else "Right")])

for index, folder in enumerate(folders):
	if not int(folder):
		for i in range(44):
			tags.append(getpath(folder, i, suffix))
	else:
		for i in range(34):
			tags.append(getpath(folder, i, suffix))

for tag in tags:
	print tag
2 Likes
6

Paul,

Thank you for the great code. I need to modify it slightly.

Instead of returning, for example, the following:

US9_FIS/Curing/0000/0000_DTCode_Right
US9_FIS/Curing/0000/0001_DTCode_Left
US9_FIS/Curing/0000/0002_DTCode_Right
US9_FIS/Curing/0000/0003_DTCode_Left

I want it to return this:

US9_FIS/Curing/0000/0001_DTCode_Right
US9_FIS/Curing/0000/0001_DTCode_Left
US9_FIS/Curing/0000/0003_DTCode_Right
US9_FIS/Curing/0000/0003_DTCode_Left

I am digging into it myself right now and may come up with a solution on my own (which I would prefer), but any help is appreciated.

I hate feeling like a mooch and a HACK at this, but I only had a loose understanding of Python and Javascript when I first did this 9 years ago. I’ve been out of the game too long and no formal education on this stuff! LOL

Thanks again.

7

To at least point you in the right direction, since my code’s pretty opaque:
The piece you’ll need to modify is the second argument in the string replacement
(folder[:2], i, suffix, "Left" if i % 2 else "Right")])
So, in this case, i, which is just the loop index, needs to be adjusted so that it increases in steps of 2.
I’d recommend looking at the modulo operator that I did to switch between Left and Right; in this case you can use modulo to add/subtract an offset to the loop index and get the return value you want.

1 Like