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How can I change the order of 2d list in python

I am working on cut up method in python. Where I am breaking the text of text file in columns. I want to change the order of the elements, meaning if text is sbreaking into column A, B and C, I want to now display the text in column A, C and B. My program is as following
import sys
def first(aList):
for row in colList:
for item in row:
print(item, end=" ")
print()
ncolumns = int(input("Enter Number of Columns:"))
file = open("alice.txt", "r")
rowL= []
colList= []
print(" ")
print(" ")
print("++++++++++++++++++++++++++++++++++++++")
while True:
line = file.readline()
if not line:
break
numElements = len(line.rstrip())
_block= numElements//ncolumns
block = _block
start=0
rowL =[]
for count in range(0,(ncolumns)):
columnChars = ""
for index in range(start,block):
columnChars += line[index]
rowL.append(columnChars)
start = block
block = block + _block
if (block < numElements):
if((block + _block)>numElements):
block = numElements
colList.append(rowL)
file.close()
first(colList)

Just create a new list indexing the elements from the old list.
That is to say,
old_list = [1, 2, 3]
new_list = [old_list[0], old_list[2], old_list[1]]
which would give new_list as [0, 3, 2].

You use slice assignment here.
lst= [0,1,2]
lst[1:]=lst[1:][::-1]
print(lst)
# [0, 2, 1]
Take a look at How slicing as assignment works if you don't get it.

write a program to insert an element in an array at a specific position don't use any function in python

i want to add any specific number in any specific position. so when i come up with solution i get index out of range or list out of range(where the list and array in python are growing dynamically).
from array import *
a = int(input("how many number you want to inout in an arery"))
ary = array('i',[])
for i in range(a):
d = int(input("enter number")) # suppose i take [2,5,4,3,8]
ary.append(d)
pos = int(input("Enter position number")) #i take position is 4
val = int(input("enter value number")) # and value is 10
c = a-1
while(c >= pos-1):
ary[c+1] = ary[c]
c-=1
ary[pos-1]= val
print(ary)
[2,5,4,10,3,8]

What about slicing? Inspired from how you 'd do this with a list. The following adds the value 100 in position 2 (indexing from 0 and not 1)
>>> a = array('i',[1,2,3,4,5])
>>> value = 100
>>> pos = 2 # 0-based
>>> a = array('i', a[:pos]) + array('i', [value]) + array('i', a[pos:])
>>> a
array('i', [1, 2, 100, 3, 4, 5])
Note that I am not familiar with arrays but from what I read they seem to behave like lists. As such, the above way to insert might not be the most efficient since it is creating 3 new arrays and concatenates them into a

The reason why you get this error message is that your array's size increases with 1, because you add a new value, but you forget to actually add that position to your array. So, to solve your problem, you need to add a position, then do the corresponding movements.
The algorithm is this:
you append a position to the array;
you add the last number of the previous array to the new one;
you cycle through the array, from the end to the position, and move the numbers one position towards the end;
when you reach the position, you insert the number read from the keyboard.
In Python, the code is:
from array import *
a = int(input("how many numbers you want to input in an array? "))
ary = array('i', [])
for i in range(a):
d = int(input("enter number")) # suppose i take [2,5,4,3,8]
ary.append(d)
pos = int(input("Enter position number")) # i take position is 4
val = int(input("enter value number")) # and value is 10
ary.append(ary[len(ary) - 1])
x = len(ary) - 1
while x >= pos:
ary[x] = ary[x - 1]
x -= 1
ary[pos - 1] = val
print(ary)
Another thing: ary is the entire array. ary[2] is the 3rd element from the array. You can append to an array, but not to a position of it. Hence, an assignment like:
ary[2].append(whatever)
is wrong.

one line for loop for 2d list wih variable incrementation and 2d list search

i am new to the python one liner loops.
i want the user to enter the data into a 2d list while alerting him about the index of data they will enter.
my code is:
flag=0
x=[[int(input("enter the "+ str(flag)+ "number:")) flag+=1 for x in range(0,col)]for y in range(0,row)]
the above code shows syntax error. how should i increment the flag value alongside the one line for loop.
the next thing is that i want to search a 2d list for integer value in it
x=[[0,3,2],[1,1,1],[9,2,2]]
for i,j in enumerate(x):
try:
y=j.index(1)
except ValueError:
continue
print(i,y)
the above code works fine for a single item searching but i want the index of every 1 in the list.whereas the above code only returns the index of the first "1" it encounters i.e "1 0"
thank you in advance.

Your problem is that a list comprehension has to yield a value, and you are trying to execute the flag +=1 inside of it.
It would be clearer if you did it like so:
>>> flag = 0
>>> row = 1
>>> col = 3
>>> x=[[int(input("enter the number [{}][{}]:".format(y, x))) for x in range(0,col)] for y in range(0,row)]
enter the number [0][0]:1
enter the number [0][1]:2
enter the number [0][2]:3
>>> x
[[1, 2, 3]]
Alternatively, to achieve exactly what you wanted, you can do something like this:
>>> y=[[int(input("enter the {} number:".format(x+y*row))) for x in range(0,col)]for y in range(0,row)]
enter the 0 number:1
enter the 1 number:2
enter the 2 number:3
>>> y
[[1, 2, 3]]
Update for the second part of the question:
You can use more list comprehensions for that:
>>> y=[[int(input("enter the {} number:".format(x+y*row))) for x in range(0,col)]for y in range(0,row)]
enter the 0 number:0
enter the 1 number:1
enter the 2 number:1
>>> y
[[0, 1, 1]]
>>> for arr in y:
... print [index for (index, val) in enumerate(arr) if val == 1]
...
[1, 2]

Convert string to integers

I'm just starting out with programming, and I know I'm still missing some of the basics, but I'm trying to work this out. I have a list of 3 and 4 digit numbers that I've brought in from a text file, and I'm trying to get a sum of these numbers. So far all I can get python to do is perform a sum of each individual number, so if the first number in the list is 427, it's printing 13, rather than adding 427 + 504 + 219, etc.
This is what I have:
myList = []
inFile = open('E:/GIS/GTECH 731/NYCElementarySchools.txt', 'r')
for row in inFile:
col = row.split('\t')
if col[1]=='BK':
myList = (col[3])
intList = [int(x) for x in myList]
print sum(intList)
Additionally, when i have it print length, it gives me a list of 3's and 4's, telling me the length of each number, not giving me the total count of numbers.
I must be missing something fundamental, but I don't know what it is! Any suggestions are appreciated! Thanks!

This:
myList = (col[3])
will set myList to a str, not a list, which would be a representation of a number. Thus:
intList = [int(x) for x in myList]
would convert the digits to numbers. You want int(myList) to convert the whole string to a number.
You can keep a running total (initialised to 0) and do total += int(myList) to total all of the numbers. Then after the loop you can print the result.

In your code:
col = row.split('\t')
if col[1]=='BK':
myList = (col[3])
intList = [int(x) for x in myList]
print sum(intList)
'col = row.split('\t')' makes a list which is divided by TAB.
If the line which is read from the file, looks like this:
# \t is TAB
SOMETHING\tBK\t1\t2\t3
The col structure is:
col[0] = SOMETHING
col[1] = BK
col[2] = 1
col[3] = 2
col[4] = 3
So, if you intend sum col[3] to col[...] then use col[3:] = list of col[3], col[4]
Thus, if you want to accumulate sum result you need another variable.
myList = []
inFile = open('E:/GIS/GTECH 731/NYCElementarySchools.txt', 'r')
sumList = []
for row in inFile:
row_total = 0
col = row.split('\t')
if col[1]=='BK':
intList = [int(x) for x in col[3:]]
row_sum = sum(intList)
# row_sum = map(lambda x: int(x), col[3:])
print 'row total: %d' % (row_sum)
sumList.append(row_sum)
print 'total: %d' % (sum(sumList))

Probably you want to use a slice to assign myList
myList = col[3:]
this will return a list everything from the 3rd column on.

Python sudoku checker

I'm working on an assignment for my CIS class in python. We have to code a Sudoku checker. In a 9x9 board we obviously have to check each row, col and 3x3 square for duplicates. I'm a little stuck on the idea of how to check the numbers by a 3x3 square. Below is my code for checking each row and col, if someone could help me a little with an outline or an approach just something for checking each 3x3 square that would be amazing!
self.columns = [ ]
for col in range(9):
col_tiles = [ ]
self.columns.append(col_tiles)
for row in range(9):
col_tiles.append(self.tiles[row][col])
self.squares = [ ]
for col in range(1, 10, 3):
for row in range(1, 10, 3):
square_tiles = [ ]
self.squares.append(square_tiles)
for x in range(3):
for y in range(3):
square_tiles.append(self.tiles[x][y])

This assumes you have the freedom to read the data and structure how you want. We want a set of unique values 1-9 for each row/column/3x3 grid, so one way is to either use a set or a list comparison (we'll use set here to make it cleaner). If we create a set equal to the numbers from 1 to 9, we have a point against which we can compare all of our other groups. Assume a structure like this (from here):
In [1]: r1 = [9,3,2,5,4,8,1,7,6]
In [2]: r2 = [1,8,7,9,2,6,5,4,3]
In [3]: r3 = [5,4,6,3,7,1,2,8,9]
# Continues....
Where each row represents a full row of data. Now lets create a section of that data that represents the first three rows, pull out one grid and compare the contents to our set:
In [4]: sec1 = [r1, r2, r3]
In [5]: nums = set(range(1, 10))
In [6]: nums == set(n for row in sec1 for n in row[:3])
Out[6]: True
This iterates over the first three rows and returns the first three elements in each of those rows. To get a better visual, here is the equivalent for-loop code to make it a bit easier to decipher:
result = set()
for row in sec1:
for n in row[:3]:
result.add(n)
Since our set of numbers includes everything from 1-9, we know it is valid. To move to the second, we range the row[:3] to row[3:6] (and row[6:9] after that). You'll then need to handle this for the next two sections as well. I'll leave it to you as to how to wrap this in a more dynamic structure (note the multiples of three), but hopefully this will get you started :)

Whenever you're having trouble coming up with an algorithm, just ask yourself: "How would I solve this manually, if the only way I could be given the problem was by a computer".
In other words, if I asked you to check the top left 3x3 grid, your eyes would just go to the top left corner and add up numbers. But if I said, check the top left 3x3 grid, and didn't actually give you the board, you'd say, "OK, give me the top left 3x3 grid".
And I'd say "How?"
And you'd say, "Imagine the tiles are numbered by rows and columns. I want the tiles in spots (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), and (2,2)"
Does that help?

Here's what I would do: create 3 dictionaries, one for the rows, one for the columns, and one for the 3x3 squares.
while you loop through each element in the sudoku puzzle, keep track of your row and column (trivial), and use if statements to keep track of which 3x3 square you're in (a bit more involved)
then just send each element to the corresponding row, column, and 3x3 square dictionary, and compare at the end.
This way you only need to inspect each element once.
EDIT: also, set will probably be useful

This function will do. "sample" gives the randomness, so every time you run this you will get a different one.
from random import sample
def generate_sudoku_checker():
random_list = sample([1,2,3,4,5,6,7,8,9],9)
random_list = random_list + random_list[:9]
for i in range(3):
for j in range(3):
print(random_list[i+j*3:i+j*3+9])

Here is my solution to this :
Funtion:
def check_sudoku(lis):
n = len(lis)
digit = 1 #start from one
while (digit<=n):
i=0
while i<n: # go through each row and column
row_count=0
column_count=0
j=0
while j < n: # for each entry in the row / column
if lis[i][j] == digit: # check row count
row_count = row_count+1
if lis[j][i]== digit :
column_count = column_count+1
j=j+1
if row_count !=1 or column_count!=1:
return False
i=i+1 # next row/column
digit = digit+1 #next digit
return True

Late to the party but this worked for me:
def insert_sudoku():
puzzle = []
for i in range(9):
print("You've entered", len(puzzle), "rows so far")
row = input("Enter a row")
if len(row) < 9:
print("Not enough numbers on this row")
return insert_sudoku()
elif len(row) > 9:
print("Too many numbers. Try again!")
return insert_sudoku()
try:
row = [int(dig) for dig in row]
puzzle.append(row)
except:
print("Whoops, looks like you didn't enter only numbers somewhere. Try again!")
return insert_sudoku()
validate_entries(puzzle)
def validate_entries(puzzle):
check = [1, 2, 3, 4, 5, 6, 7, 8, 9]
b1, b2, b3, b4, b5, b6, b7, b8, b9 = [], [], [], [], [], [], [], [], []
for i in range(9):
z = []
for x in range(9):
z.append(puzzle[i][x])
puzzle.append(z)
for i in range(3):
b1 += (puzzle[i][:3])
b4 += (puzzle[i][3:6])
b7 += (puzzle[i][6:])
for i in range(3,6):
b2 += (puzzle[i][:3])
b5 += (puzzle[i][3:6])
b8 += (puzzle[i][6:])
for i in range(6,9):
b3 += (puzzle[i][:3])
b6 += (puzzle[i][3:6])
b9 += (puzzle[i][6:])
puzzle.append(b1)
puzzle.append(b2)
puzzle.append(b3)
puzzle.append(b4)
puzzle.append(b5)
puzzle.append(b6)
puzzle.append(b7)
puzzle.append(b8)
puzzle.append(b9)
for iter in puzzle:
if sorted(iter) != check:
print("No")
return
print("Yes")
insert_sudoku()
Inspired by this article
EDIT: Indentation might be off from copy + pasting the code.

What my answer adds is the use of a list comprehension to extract the tiles from the board.
"""
# good
board=[
[2,9,5,7,4,3,8,6,1],
[4,3,1,8,6,5,9,2,7],
[8,7,6,1,9,2,5,4,3],
[3,8,7,4,5,9,2,1,6],
[6,1,2,3,8,7,4,9,5],
[5,4,9,2,1,6,7,3,8],
[7,6,3,5,2,4,1,8,9],
[9,2,8,6,7,1,3,5,4],
[1,5,4,9,3,8,6,7,2]
]
"""
# bad
board = [
[1,9,5,7,4,3,8,6,2],
[4,3,1,8,6,5,9,2,7],
[8,7,6,1,9,2,5,4,3],
[3,8,7,4,5,9,2,1,6],
[6,1,2,3,8,7,4,9,5],
[5,4,9,2,1,6,7,3,8],
[7,6,3,5,2,4,1,8,9],
[9,2,8,6,7,1,3,5,4],
[2,5,4,9,3,8,6,7,1]
]
def check(l):
# each digit 1-9 should occur once
for n in range(1,10):
try:
l.index(n)
except ValueError:
return False
return True
# check the rows
for row in board:
print(check(row))
# check the columns
for column in [ [ board[r][c] for r in range(9) ] for c in range(9) ]:
print(check(column))
# check the tiles
for tile in [[board[r][c] for r in range(row, row + 3) for c in range(col, col + 3)] for row in range(0, 9, 3) for col in range(0, 9, 3)]:
print(check(tile))

This is my solution. I also want to confirm the time and space complexity of this code:
"""
Sample input 1:
295743861
431865927
876192543
387459216
612387495
549216738
763524189
928671354
154938672
Output: YES!
Sample input 2
195743862
431865927
876192543
387459216
612387495
549216738
763524189
928671354
254938671
Output: NOPE!!
"""
##################Solution############################################
def get_input():
#Get the input in form of strings and convert into list
print("Enter the board here: ")
lines = []
while True:
line = input()
if line:
lines.append(line)
else:
break
final = [(list(i)) for i in lines]
return final
def row_check(board):
# row check function which will be used in other two functions as well
text = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
x = True
for row in board:
if sorted(row) == text:
x = True
else:
x = False
return x
def col_check(board):
# Getting the list of 9 lists containing column elements
i = 0
j = 0
cols = [[], [], [], [], [], [], [], [], []]
for j in range(0, 9):
for i in range(0, 9):
cols[j].append(board[i][j])
return (row_check(cols))
def tile_check(board):
#Getting the list of 9 lists converting each tile of 3x3 into 9 element list
lst =[[],[],[],[],[],[],[],[],[]]
i = 0
j = 0
k = 0
while k<9:
for r in range(i,i+3):
for c in range(j, j+3):
lst[k].append(board[r][c])
j = j +3
k = k +1
if j == 9:
j = 0
i = i+3
return (row_check(lst))
#main program
c = get_input()
if row_check(c) and col_check(c) and tile_check(c):
print("YES!")
else:
print("NOPE!!")