如何访问二维数组的元素？

Question

I would like to understand how one goes about manipulating the elements of a 2D array.

If I have for example:

a= ( a11 a12 a13 )  and b = (b11 b12 b13) 
     a21 a22 a23             b21 b22 b23

I have defined them in python as for example:

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

I saw that I cannot refer to a[1][1] but to a[1] which gives me a result of [2,1] . So, I don't understand how do I access the second row of these arrays? That would be a21, a22, a23, b21, b22, b23 ? And how would I do in order to multiply them as c1 = a21*b21, c2 = a22*b22 , etc?

Answer 1

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]]

Answer 2

If you want do many calculation with 2d array, you should use NumPy array instead of nest list.

for your question, you can use:zip(*a) to transpose it:

In [55]: a=[[1,1],[2,1],[3,1]]
In [56]: zip(*a)
Out[56]: [(1, 2, 3), (1, 1, 1)]
In [57]: zip(*a)[0]
Out[57]: (1, 2, 3)

Answer 3

Look carefully how many brackets does your array have. I met an example when function returned answer with extra bracket, like that:

>>>approx
array([[[1192,  391]],
       [[1191,  409]],
       [[1209,  438]],
       [[1191,  409]]])

And this didn't work

>>> approx[1,1]
IndexError: index 1 is out of bounds for axis 1 with size 1

This could open the brackets:

>>> approx[:,0]
array([[1192,  391],
       [1191,  409],
       [1209,  438],
       [1191,  409]])

Now it is possible to use an ordinary element access notation:

>>> approx[:,0][1,1]
409

Answer 4

If you have this:

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

You can easily access this by using:

print(a[0][2])
a[0][1] = 7
print(a)

Answer 5

Seems to work here:

>>> a=[[1,1],[2,1],[3,1]]
>>> a
[[1, 1], [2, 1], [3, 1]]
>>> a[1]
[2, 1]
>>> a[1][0]
2
>>> a[1][1]
1

Answer 6

a[1][1] does work as expected. Do you mean a11 as the first element of the first row? Cause that would be a[0][0].