numpy arrays 切片使用重塑

Question

I was learning how to reshape an array. I found several tutorials on the topic. I have confusion after reading all those stuff.

Suppose I declared an array using numpy

arr = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])

I reshaped it

newarr = arr.reshape(4, 3)

The resultant array newarr which has 4 rows and 3 columns

I found another command for reshaping

arr.reshape(data.shape[d1],d2)

Here data is the array name d1 = dimension1 and d2= dimension2.

As per my knowledge, shape[0] returns numbers of rows of a 2-d array, shape[1] returns numbers of columns of a 2d array

if I write down

newarr = arr.reshape(arr.shape[4],3)

I got an error tuple index out of range . Doesn't the previous command mean that create a 2d array with 4 rows and 3 columns?

I also tried the same command for

 newarr = arr.reshape(arr.shape[0],1)  

It works fine with 12 rows and one column, So does this mean shape[0] returns numbers of columns?

I read various tutorials on slicing an array but this is not clear to me.

What happen with newarr = arr.reshape(arr.shape[4],3) command?

Answer 1

arr is unidimensional, then only arr.shape[0] is defined

Try printing arr.shape

Answer 2

I found another command for reshaping

arr.reshape(data.shape[d1],d2)

This is not correct, not in the general case anyway. This command makes the first dimension of newarr match the d1´th dimension of the array data .

You should stick to

newarr = arr.reshape(d1, d2)

Answer 3

Your arr = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]) is an 1D array, there arent any rows/columns. Its just array of elements with indexing. Your arr.shape has only one dimension with 12 values (12,) . so, when you call arr.shape[0] it returns number of elements in the dimension and it is 12.

So, finally you are asking an arr = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]) to reshape to newarr = arr.reshape(arr.shape[4],3) where arr.shape[4] is out of bound.