Using ArcGIS Field Calculator to create classes in new field?

Question

I first classified firstCol into groups and then I want to give the groups with different weighted values in another field, SecondCol. I will need to give different weighted values very often. I find it very time-consuming whenever I modify the weighted values in my script.

---

My column looks like this:

enter image description here

I use the following script to calculate the SecondCol:

def TextValue(gridcode):
if gridcode <= 5:
 return 1
elif 5 < gridcode <= 15:
 return 2
elif 15 < gridcode <= 30:
 return 3
elif 30 < gridcode <= 40:
 return 4
elif 40 < gridcode <= 55:
 return 5
elif 55 < gridcode <= 100:
 return 6
elif 100 < gridcode :
 return 7
else:
 return "N/A"
TextValue( !firstCol! )

As you can see, I give the weighted value as below:

(weighted values , group).
1 = group "firstCol < 5" 
2 = group "firstCol = 5 - 15" 
3 = group "firstCol = 15 - 30" 
4 = group "firstCol = 30 - 40" 
5 = group "firstCol = 40 - 55" 
6 = group "firstCol = 55 - 100" 
7 = group "firstCol > 100"

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I wonder if there is a better way to give new weighted values more efficiently. Like, just give a new column in the beginning of the script.

61 = group "firstCol < 5" 
8 = group "firstCol = 5 - 15" 
88 = group "firstCol = 15 - 30" 
70 = group "firstCol = 30 - 40" 
52 = group "firstCol = 40 - 55" 
2 = group "firstCol = 55 - 100" 
6 = group "firstCol > 100"
def TextValue( gridcode):
if ...
...
else...
TextValue( !firstCol! )

Then, the script will use the new values, instead of me modifying it in the TextValue.

Answer 1

Try this out:

def Get_Class(depth):
 breaks=[0.5,1,2,3,5,10,100000]
 n=len(breaks)
 for i in range(n):
 if breaks[i] >= depth:
 m=i+1; break
 return m
'===============================================
Get_Class( !AREA_HA! )

It will look cooler with enumerate though

• NewHere: Thanks, FelixIP! Your script works for sequence number very well. But what if I need to give non-sequence numbers? like change 1, 2, 3, 4, 5, 6 into 12, 42, 11, 68, 54, 10?
• FelixIP: Create second list and pick from it using [m]
• FelixIP: Your comment and question are mismatch. One thing reclass using interval (Question) and very different is reclass using lookup (Comment)
• NewHere: Sorry for my misleading description. I will edit my question.
• Learner: @ FelixIP Does breaks[i] >= depth: considers the first label(i.e. 5) since then it needs <5?

Answer 2

If you need MORE control over the class/label name and interval WISELY and want to use ANY type of equality check then a way is to use dictionary that keeps order (i.e. OrderedDict) and eval check as below where you have good command in interval for class and label for a class.To edit class (i.e. 1,2,3 etc.) and interval (i.e. '0<x<=5','5<x<=15','15<x<=30' etc.) just edit

{1:'0<x<=5',2:'5<x<=15',:'15<x<=30',4:'30<x<=40',5:'40<x<=55',6:'55<x<=100',7:'100<x'}`

See full implementation

from collections import OrderedDict
#Setup class name and interal definition for that class/label- you can change here as you want
lookup_table = OrderedDict({1:'0<x<=5', 2:'5<x<=15', 3:'15<x<=30',4:'30<x<=40',5:'40<x<=55',6:'55<x<=100',7:'100<x'})
#Define function that determines the class/label
def labeller(val):
 try:
 #find the True value for check that is used to determine class/label
 chks = [eval(v.replace('x',str(val))) for v in lookup_table.values()]#It is just a true/false list
 if True in chks and chks.count(True)==1:
 v= lookup_table[chks.index(True)+1]
 for ky,vl in lookup_table.iteritems():
 if vl == v:
 return str(ky)
 except:
 return 'N/A'

And run it-

labeller(!firstCol!)

N.B. I used OrderedDict for performance - you can use normal dictionary too.