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Splitting string ID code into various parts
Splitting string in Python parser of ArcGIS Field Calculator?Select polygons contained inside a polygon and assign IDHow to place points along a line in a specific offset using python / arcpy?Help with formatting ArcGIS text elements with PythonAssigning a vector to a field in feature class (UpdateCursor)Splitting string column into 2 columns in Python?Find the first occurrence of any letter in an alphanumeric stringHow can I convert Bing's “quadtree” tile addresses to ZXY tile addresses in Python?Address Prefix Strip Using PythonOverlay two linestring objects in geopandas, accounting for the attributes
I have a series of identification codes that I need to split out. The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)]
.
An example of some codes includes S22-201
, TT100-12
, and V6-1B
. Often there is no subdistrict, and all points fall within the same larger district (so no As or Cs or whatever at the end of the string.
I can do parts of it, like splitting at the hyphen.
!Original_ID!.split('-')[0]
and then extracting the district
!Split_ID![1:3]
But it seems like two steps for this are unnecessary, and only works when I know the specific number of characters in the string, which isn't realistic for a large data set.
I'd like to be able to grab each piece at once:
- letters on the left of the hyphen
- numbers on the left of the hyphen
- numbers on the right of the hyphen
- letters (if any) on the right of the hyphen.
I'd need the numeric fields to be integers (or I guess possibly floats in some rare cases maybe).
python arcmap field-calculator
New contributor
add a comment |
I have a series of identification codes that I need to split out. The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)]
.
An example of some codes includes S22-201
, TT100-12
, and V6-1B
. Often there is no subdistrict, and all points fall within the same larger district (so no As or Cs or whatever at the end of the string.
I can do parts of it, like splitting at the hyphen.
!Original_ID!.split('-')[0]
and then extracting the district
!Split_ID![1:3]
But it seems like two steps for this are unnecessary, and only works when I know the specific number of characters in the string, which isn't realistic for a large data set.
I'd like to be able to grab each piece at once:
- letters on the left of the hyphen
- numbers on the left of the hyphen
- numbers on the right of the hyphen
- letters (if any) on the right of the hyphen.
I'd need the numeric fields to be integers (or I guess possibly floats in some rare cases maybe).
python arcmap field-calculator
New contributor
add a comment |
I have a series of identification codes that I need to split out. The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)]
.
An example of some codes includes S22-201
, TT100-12
, and V6-1B
. Often there is no subdistrict, and all points fall within the same larger district (so no As or Cs or whatever at the end of the string.
I can do parts of it, like splitting at the hyphen.
!Original_ID!.split('-')[0]
and then extracting the district
!Split_ID![1:3]
But it seems like two steps for this are unnecessary, and only works when I know the specific number of characters in the string, which isn't realistic for a large data set.
I'd like to be able to grab each piece at once:
- letters on the left of the hyphen
- numbers on the left of the hyphen
- numbers on the right of the hyphen
- letters (if any) on the right of the hyphen.
I'd need the numeric fields to be integers (or I guess possibly floats in some rare cases maybe).
python arcmap field-calculator
New contributor
I have a series of identification codes that I need to split out. The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)]
.
An example of some codes includes S22-201
, TT100-12
, and V6-1B
. Often there is no subdistrict, and all points fall within the same larger district (so no As or Cs or whatever at the end of the string.
I can do parts of it, like splitting at the hyphen.
!Original_ID!.split('-')[0]
and then extracting the district
!Split_ID![1:3]
But it seems like two steps for this are unnecessary, and only works when I know the specific number of characters in the string, which isn't realistic for a large data set.
I'd like to be able to grab each piece at once:
- letters on the left of the hyphen
- numbers on the left of the hyphen
- numbers on the right of the hyphen
- letters (if any) on the right of the hyphen.
I'd need the numeric fields to be integers (or I guess possibly floats in some rare cases maybe).
python arcmap field-calculator
python arcmap field-calculator
New contributor
New contributor
edited 13 mins ago
Taras
2,2342727
2,2342727
New contributor
asked 5 hours ago
vce500vce500
62
62
New contributor
New contributor
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2 Answers
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oldest
votes
You're not going to be able to calculate two fields in one go.. though you can split it up into two calcs. I would do this with an update cursor:
with arcpy.da.UpdateCursor(YourFeatureClass,['Original_ID','District','Split_ID']) as uCur:
for sRow in uCur:
OrigID = sRow[0].split('-')[0] # first element in the Original_ID
charRng = range(len(OrigID)) # a range to iterate over
Chars = ''
Numbers = ''
for Idx in charRng:
if OrigID[Idx].isnumeric():
Numbers += OrigID[Idx]
else:
chars += OrigID[Idx]
sRow[1] = float(Numbers)
sRow[2] = Chars
uCur.updateRow(sRow)
This shows how to break up a string into numbers and not numbers and put the values into a row, it should give you some ideas where to start from.
add a comment |
Assuming you have four fields, region, district, place and subdistrict already added and you want to use the field calculator to populate them. You would have to run the calculator four times using an expression like:
Code Block
import re
def parse(s):
"""The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)].
An example of a some codes include S22-201, TT100-12, and V6-1B.
Often there is no subdistrict, and all points fall within the same larger district
(so no As or Cs or whatever at the end of the string)."""
letters = re.findall(r'[a-z A-Z]+', s)
numbers = re.findall(r'[0-9]+', s)
region = letters[0]
district, place = [int(n) for n in numbers]
try:
subdistrict = letters[1]
except IndexError:
subdistrict = None
return region, district, place, subdistrict
Then for the region field, use:
parse(!Original_ID!)[0]
For district:
parse(!Original_ID!)[1]
For place:
parse(!Original_ID!)[2]
For subdistrict:
parse(!Original_ID!)[3]
However, I would use the update cursor approach suggested by Michael Stimson so you could update all four fields in one hit. Use the following in the python window of ArcMap/ArcGIS Pro:
import re
def parse(s):
etc... from code block above
with arcpy.da.UpdateCursor(YourFeatureClass, ['Original_ID','Region', 'District', 'Place', 'Subdistrict']) as rows:
for row in rows:
region, district, place, subdistrict = parse(row[0])
row = [row[0], region, district, place, subdistrict]
rows.updateRow(row)
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You're not going to be able to calculate two fields in one go.. though you can split it up into two calcs. I would do this with an update cursor:
with arcpy.da.UpdateCursor(YourFeatureClass,['Original_ID','District','Split_ID']) as uCur:
for sRow in uCur:
OrigID = sRow[0].split('-')[0] # first element in the Original_ID
charRng = range(len(OrigID)) # a range to iterate over
Chars = ''
Numbers = ''
for Idx in charRng:
if OrigID[Idx].isnumeric():
Numbers += OrigID[Idx]
else:
chars += OrigID[Idx]
sRow[1] = float(Numbers)
sRow[2] = Chars
uCur.updateRow(sRow)
This shows how to break up a string into numbers and not numbers and put the values into a row, it should give you some ideas where to start from.
add a comment |
You're not going to be able to calculate two fields in one go.. though you can split it up into two calcs. I would do this with an update cursor:
with arcpy.da.UpdateCursor(YourFeatureClass,['Original_ID','District','Split_ID']) as uCur:
for sRow in uCur:
OrigID = sRow[0].split('-')[0] # first element in the Original_ID
charRng = range(len(OrigID)) # a range to iterate over
Chars = ''
Numbers = ''
for Idx in charRng:
if OrigID[Idx].isnumeric():
Numbers += OrigID[Idx]
else:
chars += OrigID[Idx]
sRow[1] = float(Numbers)
sRow[2] = Chars
uCur.updateRow(sRow)
This shows how to break up a string into numbers and not numbers and put the values into a row, it should give you some ideas where to start from.
add a comment |
You're not going to be able to calculate two fields in one go.. though you can split it up into two calcs. I would do this with an update cursor:
with arcpy.da.UpdateCursor(YourFeatureClass,['Original_ID','District','Split_ID']) as uCur:
for sRow in uCur:
OrigID = sRow[0].split('-')[0] # first element in the Original_ID
charRng = range(len(OrigID)) # a range to iterate over
Chars = ''
Numbers = ''
for Idx in charRng:
if OrigID[Idx].isnumeric():
Numbers += OrigID[Idx]
else:
chars += OrigID[Idx]
sRow[1] = float(Numbers)
sRow[2] = Chars
uCur.updateRow(sRow)
This shows how to break up a string into numbers and not numbers and put the values into a row, it should give you some ideas where to start from.
You're not going to be able to calculate two fields in one go.. though you can split it up into two calcs. I would do this with an update cursor:
with arcpy.da.UpdateCursor(YourFeatureClass,['Original_ID','District','Split_ID']) as uCur:
for sRow in uCur:
OrigID = sRow[0].split('-')[0] # first element in the Original_ID
charRng = range(len(OrigID)) # a range to iterate over
Chars = ''
Numbers = ''
for Idx in charRng:
if OrigID[Idx].isnumeric():
Numbers += OrigID[Idx]
else:
chars += OrigID[Idx]
sRow[1] = float(Numbers)
sRow[2] = Chars
uCur.updateRow(sRow)
This shows how to break up a string into numbers and not numbers and put the values into a row, it should give you some ideas where to start from.
answered 4 hours ago
Michael StimsonMichael Stimson
21.6k22360
21.6k22360
add a comment |
add a comment |
Assuming you have four fields, region, district, place and subdistrict already added and you want to use the field calculator to populate them. You would have to run the calculator four times using an expression like:
Code Block
import re
def parse(s):
"""The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)].
An example of a some codes include S22-201, TT100-12, and V6-1B.
Often there is no subdistrict, and all points fall within the same larger district
(so no As or Cs or whatever at the end of the string)."""
letters = re.findall(r'[a-z A-Z]+', s)
numbers = re.findall(r'[0-9]+', s)
region = letters[0]
district, place = [int(n) for n in numbers]
try:
subdistrict = letters[1]
except IndexError:
subdistrict = None
return region, district, place, subdistrict
Then for the region field, use:
parse(!Original_ID!)[0]
For district:
parse(!Original_ID!)[1]
For place:
parse(!Original_ID!)[2]
For subdistrict:
parse(!Original_ID!)[3]
However, I would use the update cursor approach suggested by Michael Stimson so you could update all four fields in one hit. Use the following in the python window of ArcMap/ArcGIS Pro:
import re
def parse(s):
etc... from code block above
with arcpy.da.UpdateCursor(YourFeatureClass, ['Original_ID','Region', 'District', 'Place', 'Subdistrict']) as rows:
for row in rows:
region, district, place, subdistrict = parse(row[0])
row = [row[0], region, district, place, subdistrict]
rows.updateRow(row)
add a comment |
Assuming you have four fields, region, district, place and subdistrict already added and you want to use the field calculator to populate them. You would have to run the calculator four times using an expression like:
Code Block
import re
def parse(s):
"""The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)].
An example of a some codes include S22-201, TT100-12, and V6-1B.
Often there is no subdistrict, and all points fall within the same larger district
(so no As or Cs or whatever at the end of the string)."""
letters = re.findall(r'[a-z A-Z]+', s)
numbers = re.findall(r'[0-9]+', s)
region = letters[0]
district, place = [int(n) for n in numbers]
try:
subdistrict = letters[1]
except IndexError:
subdistrict = None
return region, district, place, subdistrict
Then for the region field, use:
parse(!Original_ID!)[0]
For district:
parse(!Original_ID!)[1]
For place:
parse(!Original_ID!)[2]
For subdistrict:
parse(!Original_ID!)[3]
However, I would use the update cursor approach suggested by Michael Stimson so you could update all four fields in one hit. Use the following in the python window of ArcMap/ArcGIS Pro:
import re
def parse(s):
etc... from code block above
with arcpy.da.UpdateCursor(YourFeatureClass, ['Original_ID','Region', 'District', 'Place', 'Subdistrict']) as rows:
for row in rows:
region, district, place, subdistrict = parse(row[0])
row = [row[0], region, district, place, subdistrict]
rows.updateRow(row)
add a comment |
Assuming you have four fields, region, district, place and subdistrict already added and you want to use the field calculator to populate them. You would have to run the calculator four times using an expression like:
Code Block
import re
def parse(s):
"""The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)].
An example of a some codes include S22-201, TT100-12, and V6-1B.
Often there is no subdistrict, and all points fall within the same larger district
(so no As or Cs or whatever at the end of the string)."""
letters = re.findall(r'[a-z A-Z]+', s)
numbers = re.findall(r'[0-9]+', s)
region = letters[0]
district, place = [int(n) for n in numbers]
try:
subdistrict = letters[1]
except IndexError:
subdistrict = None
return region, district, place, subdistrict
Then for the region field, use:
parse(!Original_ID!)[0]
For district:
parse(!Original_ID!)[1]
For place:
parse(!Original_ID!)[2]
For subdistrict:
parse(!Original_ID!)[3]
However, I would use the update cursor approach suggested by Michael Stimson so you could update all four fields in one hit. Use the following in the python window of ArcMap/ArcGIS Pro:
import re
def parse(s):
etc... from code block above
with arcpy.da.UpdateCursor(YourFeatureClass, ['Original_ID','Region', 'District', 'Place', 'Subdistrict']) as rows:
for row in rows:
region, district, place, subdistrict = parse(row[0])
row = [row[0], region, district, place, subdistrict]
rows.updateRow(row)
Assuming you have four fields, region, district, place and subdistrict already added and you want to use the field calculator to populate them. You would have to run the calculator four times using an expression like:
Code Block
import re
def parse(s):
"""The format of these codes is [region(letter)][district(number)] - [place(number)][subdistrict(letter)].
An example of a some codes include S22-201, TT100-12, and V6-1B.
Often there is no subdistrict, and all points fall within the same larger district
(so no As or Cs or whatever at the end of the string)."""
letters = re.findall(r'[a-z A-Z]+', s)
numbers = re.findall(r'[0-9]+', s)
region = letters[0]
district, place = [int(n) for n in numbers]
try:
subdistrict = letters[1]
except IndexError:
subdistrict = None
return region, district, place, subdistrict
Then for the region field, use:
parse(!Original_ID!)[0]
For district:
parse(!Original_ID!)[1]
For place:
parse(!Original_ID!)[2]
For subdistrict:
parse(!Original_ID!)[3]
However, I would use the update cursor approach suggested by Michael Stimson so you could update all four fields in one hit. Use the following in the python window of ArcMap/ArcGIS Pro:
import re
def parse(s):
etc... from code block above
with arcpy.da.UpdateCursor(YourFeatureClass, ['Original_ID','Region', 'District', 'Place', 'Subdistrict']) as rows:
for row in rows:
region, district, place, subdistrict = parse(row[0])
row = [row[0], region, district, place, subdistrict]
rows.updateRow(row)
edited 2 hours ago
answered 2 hours ago
user2856user2856
30.3k258105
30.3k258105
add a comment |
add a comment |
vce500 is a new contributor. Be nice, and check out our Code of Conduct.
vce500 is a new contributor. Be nice, and check out our Code of Conduct.
vce500 is a new contributor. Be nice, and check out our Code of Conduct.
vce500 is a new contributor. Be nice, and check out our Code of Conduct.
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