'for loop' exit at first passage

stephen

i do want to add that the original post is saying


where box_altex == 8:
    for num_box in range ( (box_altex//3*3) , (box_altex//*33)+3) :
        print (num_box)

of course that throws error so im assuming that its ( (box_altex//3*3) , (box_altex//3*3)+3)

im guessing a few things:

im thinking @mibradoc is tryingbto run a while do while loop possibly since it seems likenhe wants it to run more than once?
the where box_altex == 8: was meant box_altex = 8 and this states the loop should run once.

i tried many implementations and other than obvious i get the 123 type ressult. so im not sure how his code in script would alter the outcome. unless example givin differs from actuall..

sorry @mibradoc

JonB

@stephen he said where, not while...

stephen

@JonB

yes sir i was just giving my input on "if" it was a typo. similar to his missplacement of the asterisk. this is why i used i think and ?. when helping i try to be as thorough as i can..

cvp

@JonB & @stephen I think that "where box_altex == 8" is not part of the script, only to say that this variable has the value 8. The script is between the 2 "..." lines

cvp

@mibradoc You say that the code jumps to the line after the loop. I think that we need the full code, if it is possible

mibradoc

here is full code :
https://www.dropbox.com/sh/puqht9zcmjfrzsh/AAA1_zJt1Kt0YFU3avW2dHeja?dl=0

stephen

UPDATED

@mibradoc output:


grille à résoudre ? libre

[[1 4 0 0 7 0 9 2 0]
 [3 0 0 0 0 2 8 5 4]
 [0 0 0 0 5 0 6 0 7]
 [0 0 0 2 0 0 0 7 0]
 [9 0 6 0 0 0 4 0 1]
 [0 5 0 0 0 6 0 0 0]
 [0 0 9 0 6 0 0 0 0]
 [5 3 0 7 0 0 0 0 8]
 [0 8 0 0 2 0 0 3 9]]
50
(1, 1) [3, 4] 2
0
1
2
(3, 2) [3, 5] 0
0
1
2
(5, 4) [3, 5] 3
3
4
5
(7, 5) [0, 2] 4
3
4
5
(9, 3) [4, 5] 5
3
4
5
(3, 4) [3, 4] 3
3
4
5
(8, 0) [3, 5] 2
0
1
2
(8, 0) [3, 5] 2
0
1
2

[[1 4 3 0 7 0 9 2 3]
 [3 0 0 0 0 2 8 5 4]
 [0 0 0 0 5 0 6 1 7]
 [0 0 5 2 3 5 3 7 0]
 [9 7 6 0 0 7 4 0 1]
 [0 5 0 0 0 6 0 9 0]
 [0 0 9 0 6 0 0 0 0]
 [5 3 0 7 0 0 0 0 8]
 [0 8 0 0 2 0 0 3 9]]
40
3

cvp

@mibradoc run gives an error

grille à résoudre ? libre

[[1 4 0 0 7 0 9 2 0]
 [3 0 0 0 0 2 8 5 4]
 [0 0 0 0 5 0 6 0 7]
 [0 0 0 2 0 0 0 7 0]
 [9 0 6 0 0 0 4 0 1]
 [0 5 0 0 0 6 0 0 0]
 [0 0 9 0 6 0 0 0 0]
 [5 3 0 7 0 0 0 0 8]
 [0 8 0 0 2 0 0 3 9]]
50
(1, 1) [3, 4] 2
0
Traceback (most recent call last):
  File "/private/var/mobile/Containers/Shared/AppGroup/668A7D98-7216-47ED-917D-AA0B6173167E/Pythonista3/Documents/aa.py", line 147, in <module>
    for colonne in range((numero_carre_sans_chiffre % 3)*3,(numero_carre_sans_chiffre % 3)*3 + 3) :
NameError: name 'numero_carre_sans_chiffre' is not defined

cvp

@stephen did you change code?

cvp

@mibradoc you say
"The for loop is line # 141
using breakpoint you 'll see the jump to line 147 after the first print "

but line 147 is still in the loop, due to its indentation

stephen

@cvp yes im sorry, didnt realize it didnt paste in

I CHANGED:

for colonne in range((numero_carre % 3)*3,(numero_carre % 3)*3 + 3) :
#colonnes_possibles=[] ⇒ colonnes_possibles=[]

not sure what expected outcome isnexpected tho

for numero_carre in range ( (numcarre_altex // 3) * 3,((numcarre_altex//3)*3) + 3 ) :
				print (numero_carre) # BUG !!  sortie de la boucle for après le 1er passage !! 
				#if numero_carre != numcarre_altex and key[0] not in carrint(numero_carre) :
					#numero_carre_sans_chiffre = numero_carre
					
				colonnes_possibles=[]
				for colonne in range((numero_carre % 3)*3,(numero_carre % 3)*3 + 3) :
						if puzzle[ ligne_a_analyser , colonne ] == 0 and key[0] not in puzzle[:9,colonne]:
							colonnes_possibles.append(colonne)
				if len(colonnes_possibles)	== 1 :
					fn_remplir_case ( ligne_a_analyser , colonnes_possibles[0] , key[0] )
						#print()
						#print (puzzle)
						#print ()

cvp

@stephen understood, we will wait on réaction about my last post.

stephen

@cvp this feels more like what hes probably looking for? comented out print (numero_carre)
on line 143

grille à résoudre ? libre

[[1 4 0 0 7 0 9 2 0]
 [3 0 0 0 0 2 8 5 4]
 [0 0 0 0 5 0 6 0 7]
 [0 0 0 2 0 0 0 7 0]
 [9 0 6 0 0 0 4 0 1]
 [0 5 0 0 0 6 0 0 0]
 [0 0 9 0 6 0 0 0 0]
 [5 3 0 7 0 0 0 0 8]
 [0 8 0 0 2 0 0 3 9]]
50
(1, 1) [3, 4] 2
(3, 2) [3, 5] 0
(5, 4) [3, 5] 3
(7, 5) [0, 2] 4
(9, 3) [4, 5] 5
(3, 4) [3, 4] 3
(8, 0) [3, 5] 2
(8, 0) [3, 5] 2

[[1 4 3 0 7 0 9 2 3]
 [3 0 0 0 0 2 8 5 4]
 [0 0 0 0 5 0 6 1 7]
 [0 0 5 2 3 5 3 7 0]
 [9 7 6 0 0 7 4 0 1]
 [0 5 0 0 0 6 0 9 0]
 [0 0 9 0 6 0 0 0 0]
 [5 3 0 7 0 0 0 0 8]
 [0 8 0 0 2 0 0 3 9]]
40
3

cvp

@stephen I don't know, up to @mibradoc to tell us more

mibradoc

That 's correct, line 147 is still in the loop. But I 've commented lines 143 and 144 because the nameError and insert line 142 in order to debug. I don't need this print line in my script.
If you uncomment 143 and 144 and if the loop works fine, the name error would dissapear.
Wich version of Pythonista do you use? Mine is 3.3

cvp

@mibradoc Pythonista 3.3 too
You need also to uncomment 146

cvp

This

			for numero_carre in range ( (numcarre_altex // 3) * 3,((numcarre_altex//3)*3) + 3 ) :
				print ('x:',numero_carre) # BUG !!  sortie de la boucle for après le 1er passage !! 
				if numero_carre != numcarre_altex and key[0] not in carrint(numero_carre) :
					numero_carre_sans_chiffre = numero_carre
					
				colonnes_possibles=[]
				for colonne in range((numero_carre_sans_chiffre % 3)*3,(numero_carre_sans_chiffre % 3)*3 + 3) :

also crashes with

grille à résoudre ? libre

[[1 4 0 0 7 0 9 2 0]
 [3 0 0 0 0 2 8 5 4]
 [0 0 0 0 5 0 6 0 7]
 [0 0 0 2 0 0 0 7 0]
 [9 0 6 0 0 0 4 0 1]
 [0 5 0 0 0 6 0 0 0]
 [0 0 9 0 6 0 0 0 0]
 [5 3 0 7 0 0 0 0 8]
 [0 8 0 0 2 0 0 3 9]]
50
(1, 1) [3, 4] 2
x: 0
Traceback (most recent call last):
  File "/private/var/mobile/Containers/Shared/AppGroup/668A7D98-7216-47ED-917D-AA0B6173167E/Pythonista3/Documents/aa.py", line 147, in <module>
    for colonne in range((numero_carre_sans_chiffre % 3)*3,(numero_carre_sans_chiffre % 3)*3 + 3) :
NameError: name 'numero_carre_sans_chiffre' is not defined

cvp

Your loop on colonne uses numero_carre_sans_chiffre which is only defined in a if.

Are sure you don't need to indent more this loop?

				if numero_carre != numcarre_altex and key[0] not in carrint(numero_carre) :
					numero_carre_sans_chiffre = numero_carre
					
					colonnes_possibles=[]
					for colonne in range((numero_carre_sans_chiffre % 3)*3,(numero_carre_sans_chiffre % 3)*3 + 3) :

mibradoc

@cvp you're right, but the bug remains on my iPad
It's strange that the same code runs without error on your device.
Thanks a lot to you and to @stephen

cvp

@mibradoc As we have made some changes in your code to let it run without name error, I'm not sure we use exactly the same code.
I also have an iPad (Mini 4) and I don't think comes from that.
Could you post a code which runs without name error on your device but which stops after one loop