'for loop' exit at first passage

[cvp]

@mibradoc this code works but I don't understand yours: why ==8 and the "*3" is not visible

Édit: visually, all between the two "*" becomes italic

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

[cvp]

@mibradoc Sure that all lines in the loop are showed here?

[mibradoc]

Yes, I'am. And loop works fine diretly in shell, but fails in my script, checked with stop points

[stephen]

@mibradoc code prints

>>> 6
>>> 7
>>> 8

as requested. are you wanting this to print that resault 3 times?

>>> 6
>>> 7
>>> 8

>>> 6
>>> 7
>>> 8

>>> 6
>>> 7
>>> 8

[stephen]

@cvp

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