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Skip particular repetition in nested foor loop - Python

for k in range(8):
for i in range(2): #number of columns
for j in range(4): #number of row
print(k,j,i)
I want an output like this. no repetition of first for loop
k,j,i
-----
0 0 0
1 1 0
2 2 0
3 3 0
4 0 1
5 1 1
6 2 1
7 3 1
How I will achieve this?

Normally i would do
for i in range(8):
print(i, i%4, i%2)
Output:
0 0 0
1 1 1
2 2 0
3 3 1
4 0 0
5 1 1
6 2 0
7 3 1
But to reproduce your exact output:
for i in range(8):
print(i, i%4, int(i>3))
Output:
0 0 0
1 1 0
2 2 0
3 3 0
4 0 0
5 1 1
6 2 1
7 3 1

You can use if statements to say things like if k == 7 or something along those lines. This will only allow it to loop the first loop before moving on to the second loop.

Other answers have shown similar ways to produce your exact output, but this is another way to do it, and this would still work if you wanted the number of rows to be more than 8 and you wanted i to keep increasing
for i in range(8):
print(i, i%4, i//4)
I hope this is useful, sorry if it isn't

Looking at i and j, you have the cartesian product of {0,1} and {0,1,2,3}. You can compute that with itertools.product(range(2), range(4)), then use enumerate to number them for your k value.
from itertools import product
for k, (i, j) in enumerate(product(range(2), range(4))):
print(k, j, i)
Earlier arguments to product vary more slowly than later arguments.

Why does this Python nested for loop produce the output I get?

I'm very new to learning python, though I understand the basics of the looping, I am unable to understand the method in which output is arrived at.
In particular, how does the mapping of all three for loops happen to give the desired output, as I finding it impossible to understand the logic to be applied, when I try to write the output on paper without referring to IDE.
Code:
n = 4
a = 3
z = 2
for i in range(n):
for j in range(a):
for p in range(z):
print(i, j, p)
Output is:
0 0 0
0 0 1
0 1 0
0 1 1
0 2 0
0 2 1
1 0 0
1 0 1
1 1 0
1 1 1
1 2 0
1 2 1
2 0 0
2 0 1
2 1 0
2 1 1
2 2 0
2 2 1
3 0 0
3 0 1
3 1 0
3 1 1
3 2 0
3 2 1

The first loop iterates four times.
The second loop iterates three times. However since it is embedded inside the first loop, it actually iterates twelve times (4 * 3.)
The third loop iterates two times. However since it is embedded inside the first and second loops, it actually iterates twenty-four times (4 * 3 * 2).

Dynamic For Loops in Python

i understand that to create dynamic for loops, recursive or itertools module in python is the way to go. Lets say I am doing it in recursive.
What I want is
for var1 in range(var1_lowerlimit, var1_upperlimit, var1_stepsize):
for var2 in range(var2_lowerlimit, var2_upperlimit, var2_stepsize):
:
:
# do_whatever()
repeat for n loops where n is the number of variables
What I have now is I have 2 lists
variable_list = [ var1, var2, var3, ... ]
boundaries_list = [ [var1_lowerlimit, var1_upperlimit, var1_stepsize],
[var2_lowerlimit, var2_upperlimit, var2_stepsize], ...]
def dynamic_for_loop(variable_list , boundaries_list, no_of_loops, list_index = 0):
if no_of_loops <= 0:
# do_whatever()
else:
lower_bound = boundaries_list[list_index][0]
upper_bound = boundaries_list[list_index][1]
step_size = boundaries_list[list_index][2]
for index in range(lower_bound, upper_bound, step_size):
list_index += 1
try:
dynamic_for_loop(variable_list , boundaries_list, no_of_loops - 1, list_index)
except:
list_index = 0
dynamic_for_loop(variable_list , boundaries_list, no_of_loops - 1, list_index)
I did a reset on list_index as it gets out of range, but i couldn't get the result I want. Can someone enlighten me what went wrong?

Use the itertools.product() function to generate the values over a variable number of ranges:
for values in product(*(range(*b) for b in boundaries_list)):
# do things with the values tuple, do_whatever(*values) perhaps
Don't try to set a variable number of variables; just iterate over the values tuple or use indexing as needed.
Using * in a call tells Python to take all elements of an iterable and apply them as separate arguments. So each b in your boundaries_list is applied to range() as separate arguments, as if you called range(b[0], b[1], b[2]).
The same applies to the product() call; each range() object the generator expression produces is passed to product() as a separate argument. This way you can pass a dynamic number of range() objects to that call.

Just for fun, I thought that I would implement this using recursion to perhaps demonstrate the pythonic style. Of course, the most pythonic way would be to use the itertools package as demonstrated by Martijn Pieters.
def dynamic_for_loop(boundaries, *vargs):
if not boundaries:
print(*vargs) # or whatever you want to do with the values
else:
bounds = boundaries[0]
for i in range(*bounds):
dynamic_for_loop(boundaries[1:], *(vargs + (i,)))
Now we can use it like so:
In [2]: boundaries = [[0,5,1], [0,3,1], [0,3,1]]
In [3]: dynamic_for_loop(boundaries)
0 0 0
0 0 1
0 0 2
0 1 0
0 1 1
0 1 2
0 2 0
0 2 1
0 2 2
1 0 0
1 0 1
1 0 2
1 1 0
1 1 1
1 1 2
1 2 0
1 2 1
1 2 2
2 0 0
2 0 1
2 0 2
2 1 0
2 1 1
2 1 2
2 2 0
2 2 1
2 2 2
3 0 0
3 0 1
3 0 2
3 1 0
3 1 1
3 1 2
3 2 0
3 2 1
3 2 2
4 0 0
4 0 1
4 0 2
4 1 0
4 1 1
4 1 2
4 2 0
4 2 1
4 2 2

how to make a iterable object like this?

Question 1
how to make an iterable object like this:
0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1 ....
Question 2
if use list(obj) on the object above would this eat up machine's memory? how to prevent it?
Please don't use python2

You can make an infinite generator that counts up and down:
def updown(n):
while True:
for i in range(n):
yield i
for i in range(n - 2, 0, -1):
yield i
uptofive = updown(6)
for i in range(20):
print uptofive.next(),
would output:
0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1
You cannot prevent list(updown(6)) from trying to consume all memory, no. As the doctor would say: "Then don't do that!".
Use .next() calls instead, or use your generator with another statement that limits the number of times you iterate over the generator. The itertools.islice() function would do just that:
import itertools
list(itertools.islice(updown(6), 20))

An alternative is to use itertools.cycle():
from itertools import cycle
def oscillator(start, stop):
# assumes stop >= start
return cycle(range(start, stop+1) + range(stop-1, start, -1))
o = oscillator(0, 5)
for i in range(30):
print o.next(),

Writing Sudoku Solver wih Python

Here is my Sudoku Solver written in python language, When I run this program there seems to be a problem with in Update function and Solve function.
No matter how much time I look over and move the codes around, I seem to have no luck
Can anyone Help me?
import copy
def display (A):
if A:
for i in range (9):
for j in range (9):
if type (A[i][j]) == type ([]): print A[i][j][0],
else: print A[i][j]
print
print
else: print A
def has_conflict(A):
for i in range(9):
for j in range(9):
for (x,y) in get_neighbors(i,j):
if len(A[i][j])==1 and A[i][j]==A[x][y]: return True
return False
def get_neighbors(x,y):
neighbors = []
for i in range(3):
for j in range(3):
a = 3*(x / 3)+i
b = 3*(y / 3)+j
if (x,y) != (a,b):
neighbors += [(a,b)]
for i in range(9):
if (x,y) != (x,i) and (x,i) not in neighbors:
neighbors += [(x,i)]
if (x,y) != (i,y) and (i,y) not in neighbors:
neighbors += [(i,y)]
return neighbors
def update(A,x,y,value):
B = copy.deepcopy(A)
B[x][y] = [value]
for (i,j) in get_neighbors(x,y):
if B[i][j] == B[x][y]:
if len(B[i][j]) > 1: B[i][j].remove(value)
else: return []
if has_conflict(B) == True: return []
else: return B
def solve(A):
for x in range (9):
for y in range(9):
if len(A[x][y]) == 1: return A[x][y]
if len(A[x][y]) > 1:
lst = update(A,x,y,A[x][y])
if len(lst[x][y]) > 1: solve(lst)
if lst == []: return []
if len(lst[x][y]) == 1: return lst
else: return A[x][y]
A=[]
infile = open('puzzle1.txt','r')
for i in range(9):
A += [[]]
for j in range(9):
num = int(infile.read(2))
if num: A[i] += [[num]]
else: A[i] += [[1,2,3,4,5,6,7,8,9]]
for i in range(9):
for j in range(9):
if len(A[i][j])==1: A = update(A, i, j, A[i][j][0])
if A == []: break
if A==[]: break
if A<>[]: A = solve(A)
display(A)
Here are some puzzles:
Puzzle 1
0 0 0 2 6 0 7 0 1
6 8 0 0 7 0 0 9 0
1 9 0 0 0 4 5 0 0
8 2 0 1 0 0 0 4 0
0 0 4 6 0 2 9 0 0
0 5 0 0 0 3 0 2 8
0 0 9 3 0 0 0 7 4
0 4 0 0 5 0 0 3 6
7 0 3 0 1 8 0 0 0
Puzzle 2
1 0 0 4 8 9 0 0 6
7 3 0 0 0 0 0 4 0
0 0 0 0 0 1 2 9 5
0 0 7 1 2 0 6 0 0
5 0 0 7 0 3 0 0 8
0 0 6 0 9 5 7 0 0
9 1 4 6 0 0 0 0 0
0 2 0 0 0 0 0 3 7
8 0 0 5 1 2 0 0 4
Puzzle 3
0 2 0 6 0 8 0 0 0
5 8 0 0 0 9 7 0 0
0 0 0 0 4 0 0 0 0
3 7 0 0 0 0 5 0 0
6 0 0 0 0 0 0 0 4
0 0 8 0 0 0 0 1 3
0 0 0 0 2 0 0 0 0
0 0 9 8 0 0 0 3 6
0 0 0 3 0 6 0 9 0

I would avoid things like "move the codes around". This is called "Programming by Coincidence" (see The Pragmatic Programmer). Programming like this won't make you a better programmer.
Instead, you should take out a paper and pencil, and start thinking how things should work. Then, read the code and carefully write what it actually does. Only when you understand, go back to the computer terminal.

If you want to stabilize your code, then write small test cases for each function which make sure that they work correctly.
In your case, run a puzzle, and determine which field is wrong. Then guess which function might produce the wrong output. Call it with the input to see what it really does. Repeat for every bug you find.
[EDIT] The module unittest is your friend.

I'd like to help you in a way that you can write the actual code, so here is some explanation, and some pseudo-code.
Those sudoku solvers that don't mimic human deduction logic are bruteforce-based. Basically, you'll need a backtrack algorithm. You have your has_conflict method, which checks whether the candidate is ok at first look. Then you write the backtrack algorithm like this:
Solve(s):
Pick a candidate.
Does it have a conflict? If yes, go back, and pick another one.
No more empty cells? Then cool, return True.
Have you run out of candidates? Then it cant be solved, return False.
At this point, it seems ok. So call Solve(s) again, lets see how it works
out with the new candidate.
If Solve returned false, then after all it was a bad candidate. Go
back to picking another one.
If Solve returned True, then you solved the sudoku!
The main idea here is that if your guess was wrong, despite not having a conflict at first look, then a confliction will reveal itself somewhere deeper in the call tree.
The original sudokus have only one solution. You can extend this method to different solutions for sudokus that have them by trying any candidates despite the return value of Solve (but that will be very slow with this approach).
There's a nice trick to find out if a sudoku has more than one solutions. First try the candidates in natural order in every call of solve. Then try them backwards. Then do these two steps again, but this time run the algorithm from the last cell of the sudoku, stepping backwards. If these four solutions are identical, then it has only one solution. Unfortunately I don't have a formal proof, but it seemed to work all the time. I tried to prove it, but I'm not that great with graphs.

Solving sudoku need some bruteforcing method, I dont see those in your codes.
I also tried to do before, but finally I saw norvig's codes, its just working perfect.
and ended up with learning his codes finally.