topic Re: field calculator in Geoprocessing Questions

field calculator

AndrewWindham — Mon, 15 Jun 2015 21:16:41 GMT

I am using ArcGIS 10.3 basic desktop to interpolate GPS soil points and then create a lime prescription map. I can easily interpolate the points however when I use the field calculator to enter the target lime formula for the desired pHw level I get an error message of 99999 that says a field name was not found or there were unbalanced quotation marks. The formula was produced by a leading university and has been used in numerous types of software so I don't believe it is the problem.

Thanks

Re: field calculator

JeffWard — Mon, 15 Jun 2015 21:20:22 GMT

What does the formula look like?

Are you using python parser of VB parser?

Re: field calculator

AndrewWindham — Mon, 15 Jun 2015 21:28:24 GMT

VB parser. I am not familiar with python code.

Re: field calculator

DarrenWiens2 — Mon, 15 Jun 2015 21:30:06 GMT

^ the image doesn't seem to work. Just copy/paste the expression into your post.

Re: field calculator

AndrewWindham — Mon, 15 Jun 2015 21:35:01 GMT

phw[0,6.24] : ((6.5-(-0.6))2*1.5)

phw[6.24,max] : 0

This is for a ph target of 6.3

Re: field calculator

DarrenWiens2 — Mon, 15 Jun 2015 21:42:16 GMT

This doesn't appear to be either VB Script or Python, your only two options in the field calculator. As I'm not familiar with your subject matter whatsoever, can you please explain, in words, what that formula is supposed to do?

Re: field calculator

SepheFox — Mon, 15 Jun 2015 21:43:04 GMT

S, you have a feature class of points, and you're trying to calculate the value of a field based on the phw field, is that correct?

Re: field calculator

AndrewWindham — Mon, 15 Jun 2015 21:46:25 GMT

No this is a soil equation not python or VB, This is a formula that is intended to raise the ph level of soil matter by applying lime. The formula is applied to areas of low ph with the intent of increasing the ph level to 6.3. Hope this helps.

Re: field calculator

AndrewWindham — Mon, 15 Jun 2015 21:49:27 GMT

I have already calculated the value of the field, now I need to apply the formula to the interpolated points in the attribute table that will raise the soil ph level to 6.3. I have the formula but it will not work in field calculator I guess because it is not a python or vb code. The code is intended to target areas that have a ph level below 6.3 and bring them to that level. The formula is below.

phw[0,6.24] : ((6.5-(-0.6))2*1.5)

phw[6.24,max] : 0

Thanks

Re: field calculator

DarrenWiens2 — Mon, 15 Jun 2015 21:53:59 GMT

No, this still isn't in language I, or the field calculator, will understand. The field calculator works like: for each row in the attribute table, take some value(s) in another column(s), manipulate it somehow, and return some new value into the calculated column for that row.

Is "phw" the name of the column we are getting values from or calculating to? What does your notation phw[0,6.24] mean - something like if phw is between 0 and 6.24, apply this part of the formula?

Re: field calculator

SepheFox — Mon, 15 Jun 2015 21:57:03 GMT

Assuming you want to replace the value of the PHw field with new values, your field calculator should look like this:

Parser: VB Script Expression: phw Code Block: Dim phw If [phw field] > 0 and [phw_field] < 6.24 phw = (((6.5-(-0.6))2*1.5)) else [phw field] > 6.24 phw = 0 end if

Re: field calculator

DarrenWiens2 — Mon, 15 Jun 2015 22:01:35 GMT

From the beginning, this equation has been missing an operator before the 2 in: (((6.5-(-0.6))2*1.5))

Re: field calculator

AndrewWindham — Mon, 15 Jun 2015 22:05:37 GMT

Yes the phw field contains the ph values that we getting the values from. The formula only applies lime to areas that have ph levels of below 6.3.

Re: field calculator

DanPatterson_Retired — Mon, 15 Jun 2015 22:06:03 GMT

soil pH measured in water? (it has been a long time)

phw[0,6.24] : ((6.5-(-0.6))2*1.5) = (7.1)*2**1.5) = 14**1.5 ????

really? something is missing in the formula since the actual value of pH isn't a variable anywhere. Can you go to the original reference and put the formula in without any substitution of terms. Or at least provide a reference (ie USDA, Brady et al or any soils text)

Re: field calculator

SepheFox — Mon, 15 Jun 2015 22:06:20 GMT

Good point. In a math equation, isn't that just seen as multiplication, as in (7.1)3 = 21.9? What I don't understand, is this equation is just a number isn't it? Why use a complicated expression, for what appears to simply be, if phw < 6.4, result = 21.9. Obviously we are missing something.

Re: field calculator

AndrewWindham — Mon, 15 Jun 2015 22:09:14 GMT

Try this one

phw[0,6.24] : ((6.5-(-0.6))2*1.5)

phw[6.24,max] : 0

Re: field calculator

DanPatterson_Retired — Mon, 15 Jun 2015 22:12:35 GMT

the image isnt coming through on the iThingy and what you typed is not a formula, since there are no variables

right now, you take 6.5...subtract -0.6 from it...I presume multiply by 2 raised to the power of 2... that is a number. Just copy and paste that number down the column and you are done. or ... write out the formula that shows where the variables are...I am wildly guessing that it is in the -0.6 term but you have the source

Re: field calculator

SepheFox — Mon, 15 Jun 2015 22:12:43 GMT

Isn't that the same? I don't see the difference.

Re: field calculator

DarrenWiens2 — Mon, 15 Jun 2015 22:13:05 GMT

Since these are all constants, not variables from your table, you may as well just type in the number. Also, you do not have enough mathematical operators in the first part of the equation.

(6.5 minus -0.6) ______ 2 times 0.5

Re: field calculator

DanPatterson_Retired — Mon, 15 Jun 2015 22:15:14 GMT

I think this may a job for ArcObjects