You can use np.where to return a tuple of arrays of x and y indices where a given condition holds in an array.

If a is the name of your array:

>>> np.where(a == 1)
(array([0, 0, 1, 1]), array([0, 1, 2, 3]))

If you want a list of (x, y) pairs, you could zip the two arrays:

>>> list(zip(*np.where(a == 1)))
[(0, 0), (0, 1), (1, 2), (1, 3)]

Or, even better, @jme points out that np.asarray(x).T can be a more efficient way to generate the pairs.

You should return 'the element has not been found' only after checking all the values, so you must remove the else, and push the second return outside of the for loops

This is an implementation without using numpy. It is not that efficient, but works fine.

rows = eval(input("How many rows in the list:"))
m = []
for row in range(rows):
  value = eval(input("Enter a row:"))
  m.append(value)

large = m[0][0]
x = 0
y = 0

for i in range(0, rows):
  for j in range(0, len(m[i])):
    if(large < m[i][j]):
     large = m[i][j]
     y = i
     x = j
print(x, y, large)

You need to iterate over your main list and then you can use list.index() to find the sub-list index, for example:

def index_2d(data, search):
    for i, e in enumerate(data):
        try:
            return i, e.index(search)
        except ValueError:
            pass
    raise ValueError("{!r} is not in list".format(search))

And it will act exactly as list.index() but for a 2D array, so in your case:

position = index_2d(board, "18")  # (4, 3)
print(board[position[0]][position[1]])  # 18

position = index_2d(board, "181")  # ValueError: '181' is not in list

l = [[98, 25, 33,  9, 41],
[67, 32, 67, 27, 85],
[38, 79, 52, 40, 58],
[84, 76, 44,  9,  2]]

def fnd(l,value):
    for i,v in enumerate(l):
        if value in v:
            return {'row':i+1,'col':v.index(value)+1}
    return {'row':-1,'col':-1}
print(fnd(l,40))

{'row': 3, 'col': 4}

The main problem here is that break only exits the innermost loop. So, if an element is found, break will skip checking other elements in the same column, but still the outer loop will advance to next row. What you really want is either this:

found = False
for row in matrix:
    for element in row:
        if element == number:
            found = True
            break
    if found:
        break
if found:
    print("Found")
else:
    print("Not found")

(notice the other break) or, possibly, a more readable solution using a function:

def searchfor(matrix, number):
    for row in matrix:
        for element in row:
            if element == number:
                return True
    return False

if searchfor(matrix, number):
    print("Found")
else:
    print("Not found")

Edit: It just occured to me that it is possible to write it without either a flag variable or a function, but it is not a particularly elegant way. Still, for completeness, here you are:

for row in matrix:
    for element in row:
        if element == number:
            break
    else:
        continue
    break

if element == number:
    print("Found")
else:
    print("Not found")

The continue statement will execute only if the inner loop was not exited by break, and it will advance the outer loop to the next row; otherwise the second break will end the outer loop.

You could use np.unique:

>>> a = np.array([[1,0,0],[1,3,0],[2,7,4]])
>>> np.unique(a)
array([0, 1, 2, 3, 4, 7])

First you need to access to your columns , so you can do that job with zip(*sudokuBoard) then for insert a value , you must check for existence the value in a proper row and column ! Note that you have your rows in sudokuColumn !

columns=map(list,zip(*sudokuBoard))
sudokuBoard=[[0 for sudokuRow in range(0,int(boardSize))] for sudokuColumn in range(0,int(boardSize))]

def insert_value(your_list,value,row,col):
  if value not in columns[col] and value not in your_list[row]:
      your_list[row][col]=value
  else:
      raise ValueError("you can not insert a duplicate value !!")

>>> mylist = [[], ['shotgun', 'weapon'], ['pistol', 'weapon'], ['cheesecake', 'f
ood'], []]
>>> print mylist[2][1]
weapon

Remember a couple of things,

1. don't name your list, list... it's a python reserved word
2. lists start at index 0. so mylist[0] would give []
   similarly, mylist[1][0] would give 'shotgun'
3. consider alternate data structures like dictionaries.

If you have

a=[[1,1],[2,1],[3,1]]
b=[[1,2],[2,2],[3,2]]

Then

a[1][1]

Will work fine. It points to the second column, second row just like you wanted.

I'm not sure what you did wrong.

To multiply the cells in the third column you can just do

c = [a[2][i] * b[2][i] for i in range(len(a[2]))] 

Which will work for any number of rows.

Edit: The first number is the column, the second number is the row, with your current layout. They are both numbered from zero. If you want to switch the order you can do

a = zip(*a)

or you can create it that way:

a=[[1, 2, 3], [1, 1, 1]]

You need to iterate over all the indices of your list to see if an element is a value in one of the nested lists. You can simply iterate over the inner lists and check for the presence of your element, e.g.:

if not any(0 in x for x in board):
    pass  # the board is full

Using any() will serve as a short-stop whenever it encounters an element with a 0 in it so you don't need to iterate over the rest.

You can do a simple for loop, and use the built-in in statement:

def retrieve_sub_array(element):
 for sub_array in array:
  if element in sub_array:
   return sub_array