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

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

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 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])

u may use this answer

c="yam"
index=[(i, fruits.index(c)) for i, fruits in enumerate(array) if c in fruits]

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 !!")

You don't need to define no_classes yourself. Use enumerate():

def in_list(c, classes):
    for i, sublist in enumerate(classes):
        if c in sublist:
            return i
    return -1

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)

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}

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