Update Cursor skipping last row?





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I have created a python function within field calculator to calculate the time difference between a timestamp (yyyy-MM-dd HH:mm:ss) in a row and the row after (the interval field in seconds). I have used a update cursor with field calculator to do this, however I am receiving null values for both the first and last row.



The first is correctly missed due to the function being reliant on having a value set by the previous row.



I am unsure as to why the last row is being skipped?



The first row of the attribute is formatted as:



First row



The last:



Last row



I have had this issue on several other update cursor functions including those on simple single row condition statements.



Am I missing a statement to close the update cursor loop?



My code is:



import arcpy, time, datetime
from time import strftime
from datetime import timedelta, datetime
from arcpy import da

def FindTime(table,date,interval):

firstRow = True

with arcpy.da.UpdateCursor(table, ["interval", "date"]) as cursor:

for row in cursor:
gap2 = row[1]

if firstRow == True:
gap1 = gap2
firstRow = False
continue
timedelta = gap2 - gap1
row[0] = timedelta.days * 24 * 3600 + timedelta.seconds
gap1 = gap2
cursor.updateRow(row)


*Updated from below comment's solution:



import arcpy
def FindTime(fc, datefield, daydiff_field):

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break









share|improve this question




















  • 1





    There is no row[2] in your first try, you only have two fields. Are you trying to fetch the value from next row? That is not possible

    – BERA
    yesterday













  • Edited pasting to SE typo!

    – Will
    18 hours ago


















4















I have created a python function within field calculator to calculate the time difference between a timestamp (yyyy-MM-dd HH:mm:ss) in a row and the row after (the interval field in seconds). I have used a update cursor with field calculator to do this, however I am receiving null values for both the first and last row.



The first is correctly missed due to the function being reliant on having a value set by the previous row.



I am unsure as to why the last row is being skipped?



The first row of the attribute is formatted as:



First row



The last:



Last row



I have had this issue on several other update cursor functions including those on simple single row condition statements.



Am I missing a statement to close the update cursor loop?



My code is:



import arcpy, time, datetime
from time import strftime
from datetime import timedelta, datetime
from arcpy import da

def FindTime(table,date,interval):

firstRow = True

with arcpy.da.UpdateCursor(table, ["interval", "date"]) as cursor:

for row in cursor:
gap2 = row[1]

if firstRow == True:
gap1 = gap2
firstRow = False
continue
timedelta = gap2 - gap1
row[0] = timedelta.days * 24 * 3600 + timedelta.seconds
gap1 = gap2
cursor.updateRow(row)


*Updated from below comment's solution:



import arcpy
def FindTime(fc, datefield, daydiff_field):

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break









share|improve this question




















  • 1





    There is no row[2] in your first try, you only have two fields. Are you trying to fetch the value from next row? That is not possible

    – BERA
    yesterday













  • Edited pasting to SE typo!

    – Will
    18 hours ago














4












4








4


0






I have created a python function within field calculator to calculate the time difference between a timestamp (yyyy-MM-dd HH:mm:ss) in a row and the row after (the interval field in seconds). I have used a update cursor with field calculator to do this, however I am receiving null values for both the first and last row.



The first is correctly missed due to the function being reliant on having a value set by the previous row.



I am unsure as to why the last row is being skipped?



The first row of the attribute is formatted as:



First row



The last:



Last row



I have had this issue on several other update cursor functions including those on simple single row condition statements.



Am I missing a statement to close the update cursor loop?



My code is:



import arcpy, time, datetime
from time import strftime
from datetime import timedelta, datetime
from arcpy import da

def FindTime(table,date,interval):

firstRow = True

with arcpy.da.UpdateCursor(table, ["interval", "date"]) as cursor:

for row in cursor:
gap2 = row[1]

if firstRow == True:
gap1 = gap2
firstRow = False
continue
timedelta = gap2 - gap1
row[0] = timedelta.days * 24 * 3600 + timedelta.seconds
gap1 = gap2
cursor.updateRow(row)


*Updated from below comment's solution:



import arcpy
def FindTime(fc, datefield, daydiff_field):

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break









share|improve this question
















I have created a python function within field calculator to calculate the time difference between a timestamp (yyyy-MM-dd HH:mm:ss) in a row and the row after (the interval field in seconds). I have used a update cursor with field calculator to do this, however I am receiving null values for both the first and last row.



The first is correctly missed due to the function being reliant on having a value set by the previous row.



I am unsure as to why the last row is being skipped?



The first row of the attribute is formatted as:



First row



The last:



Last row



I have had this issue on several other update cursor functions including those on simple single row condition statements.



Am I missing a statement to close the update cursor loop?



My code is:



import arcpy, time, datetime
from time import strftime
from datetime import timedelta, datetime
from arcpy import da

def FindTime(table,date,interval):

firstRow = True

with arcpy.da.UpdateCursor(table, ["interval", "date"]) as cursor:

for row in cursor:
gap2 = row[1]

if firstRow == True:
gap1 = gap2
firstRow = False
continue
timedelta = gap2 - gap1
row[0] = timedelta.days * 24 * 3600 + timedelta.seconds
gap1 = gap2
cursor.updateRow(row)


*Updated from below comment's solution:



import arcpy
def FindTime(fc, datefield, daydiff_field):

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break






arcpy modelbuilder field-calculator cursor python-parser






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 19 hours ago







Will

















asked yesterday









WillWill

674




674








  • 1





    There is no row[2] in your first try, you only have two fields. Are you trying to fetch the value from next row? That is not possible

    – BERA
    yesterday













  • Edited pasting to SE typo!

    – Will
    18 hours ago














  • 1





    There is no row[2] in your first try, you only have two fields. Are you trying to fetch the value from next row? That is not possible

    – BERA
    yesterday













  • Edited pasting to SE typo!

    – Will
    18 hours ago








1




1





There is no row[2] in your first try, you only have two fields. Are you trying to fetch the value from next row? That is not possible

– BERA
yesterday







There is no row[2] in your first try, you only have two fields. Are you trying to fetch the value from next row? That is not possible

– BERA
yesterday















Edited pasting to SE typo!

– Will
18 hours ago





Edited pasting to SE typo!

– Will
18 hours ago










1 Answer
1






active

oldest

votes


















4














If your field is type date, code below should work. Last line is not calculated since there is no row after. Or do you want to calculate for example second rows diff as second row-first row?



import arcpy

fc = 'somedates'
datefield = 'date123'
daydiff_field = 'seconddiff_long'

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
#next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break


enter image description here






share|improve this answer





















  • 3





    This is why I love using SE, I have never used the iter() function before, had to go away and look it up, like it! Is there a performance boost using that approach or was it for convenience?

    – Hornbydd
    yesterday






  • 3





    Nice! Dont know if there is a performance boost. I use it because it is a simple way of fetching items in a list instead of trying to iterate over a cursor and list at the same time.

    – BERA
    yesterday






  • 2





    If anything, it will actually be slower, because the table is traversed twice instead of once. But for small tables that may not be an issue.

    – Berend
    yesterday






  • 1





    Slower than what approach?

    – BERA
    yesterday






  • 1





    Slower than using just an update cursor, as OP does

    – Berend
    yesterday












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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

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4














If your field is type date, code below should work. Last line is not calculated since there is no row after. Or do you want to calculate for example second rows diff as second row-first row?



import arcpy

fc = 'somedates'
datefield = 'date123'
daydiff_field = 'seconddiff_long'

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
#next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break


enter image description here






share|improve this answer





















  • 3





    This is why I love using SE, I have never used the iter() function before, had to go away and look it up, like it! Is there a performance boost using that approach or was it for convenience?

    – Hornbydd
    yesterday






  • 3





    Nice! Dont know if there is a performance boost. I use it because it is a simple way of fetching items in a list instead of trying to iterate over a cursor and list at the same time.

    – BERA
    yesterday






  • 2





    If anything, it will actually be slower, because the table is traversed twice instead of once. But for small tables that may not be an issue.

    – Berend
    yesterday






  • 1





    Slower than what approach?

    – BERA
    yesterday






  • 1





    Slower than using just an update cursor, as OP does

    – Berend
    yesterday
















4














If your field is type date, code below should work. Last line is not calculated since there is no row after. Or do you want to calculate for example second rows diff as second row-first row?



import arcpy

fc = 'somedates'
datefield = 'date123'
daydiff_field = 'seconddiff_long'

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
#next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break


enter image description here






share|improve this answer





















  • 3





    This is why I love using SE, I have never used the iter() function before, had to go away and look it up, like it! Is there a performance boost using that approach or was it for convenience?

    – Hornbydd
    yesterday






  • 3





    Nice! Dont know if there is a performance boost. I use it because it is a simple way of fetching items in a list instead of trying to iterate over a cursor and list at the same time.

    – BERA
    yesterday






  • 2





    If anything, it will actually be slower, because the table is traversed twice instead of once. But for small tables that may not be an issue.

    – Berend
    yesterday






  • 1





    Slower than what approach?

    – BERA
    yesterday






  • 1





    Slower than using just an update cursor, as OP does

    – Berend
    yesterday














4












4








4







If your field is type date, code below should work. Last line is not calculated since there is no row after. Or do you want to calculate for example second rows diff as second row-first row?



import arcpy

fc = 'somedates'
datefield = 'date123'
daydiff_field = 'seconddiff_long'

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
#next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break


enter image description here






share|improve this answer















If your field is type date, code below should work. Last line is not calculated since there is no row after. Or do you want to calculate for example second rows diff as second row-first row?



import arcpy

fc = 'somedates'
datefield = 'date123'
daydiff_field = 'seconddiff_long'

all_dates = [i[0] for i in arcpy.da.SearchCursor(fc,datefield)]

diff = [(d1-d0).seconds for d0,d1 in zip(all_dates, all_dates[1:])]
givediff = iter(diff)

with arcpy.da.UpdateCursor(fc, daydiff_field) as cursor:
#next(cursor) #uncomment this line if you want to calculate row2 diff as row2-row1, =first row no diff
for row in cursor:
try:
row[0] = next(givediff) #Fetch diffs until list is empty...
cursor.updateRow(row)
except StopIteration: #...then break the cursor
break


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday

























answered yesterday









BERABERA

17.1k62044




17.1k62044








  • 3





    This is why I love using SE, I have never used the iter() function before, had to go away and look it up, like it! Is there a performance boost using that approach or was it for convenience?

    – Hornbydd
    yesterday






  • 3





    Nice! Dont know if there is a performance boost. I use it because it is a simple way of fetching items in a list instead of trying to iterate over a cursor and list at the same time.

    – BERA
    yesterday






  • 2





    If anything, it will actually be slower, because the table is traversed twice instead of once. But for small tables that may not be an issue.

    – Berend
    yesterday






  • 1





    Slower than what approach?

    – BERA
    yesterday






  • 1





    Slower than using just an update cursor, as OP does

    – Berend
    yesterday














  • 3





    This is why I love using SE, I have never used the iter() function before, had to go away and look it up, like it! Is there a performance boost using that approach or was it for convenience?

    – Hornbydd
    yesterday






  • 3





    Nice! Dont know if there is a performance boost. I use it because it is a simple way of fetching items in a list instead of trying to iterate over a cursor and list at the same time.

    – BERA
    yesterday






  • 2





    If anything, it will actually be slower, because the table is traversed twice instead of once. But for small tables that may not be an issue.

    – Berend
    yesterday






  • 1





    Slower than what approach?

    – BERA
    yesterday






  • 1





    Slower than using just an update cursor, as OP does

    – Berend
    yesterday








3




3





This is why I love using SE, I have never used the iter() function before, had to go away and look it up, like it! Is there a performance boost using that approach or was it for convenience?

– Hornbydd
yesterday





This is why I love using SE, I have never used the iter() function before, had to go away and look it up, like it! Is there a performance boost using that approach or was it for convenience?

– Hornbydd
yesterday




3




3





Nice! Dont know if there is a performance boost. I use it because it is a simple way of fetching items in a list instead of trying to iterate over a cursor and list at the same time.

– BERA
yesterday





Nice! Dont know if there is a performance boost. I use it because it is a simple way of fetching items in a list instead of trying to iterate over a cursor and list at the same time.

– BERA
yesterday




2




2





If anything, it will actually be slower, because the table is traversed twice instead of once. But for small tables that may not be an issue.

– Berend
yesterday





If anything, it will actually be slower, because the table is traversed twice instead of once. But for small tables that may not be an issue.

– Berend
yesterday




1




1





Slower than what approach?

– BERA
yesterday





Slower than what approach?

– BERA
yesterday




1




1





Slower than using just an update cursor, as OP does

– Berend
yesterday





Slower than using just an update cursor, as OP does

– Berend
yesterday


















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