Numpy.around（）将ndarray len 45转换为ndarray len 1

Question

I have this ndarray

>>> y_pred_test=[1.         3.         1.986      1.         1.79266667 1.048
 1.         3.         2.         2.         3.         3.
 1.         1.976      3.         2.         1.         1.
 2.03       1.         1.976      1.16966667 1.06       2.
 2.         2.         2.         2.         2.02       3.
 1.         1.         2.02       1.02       1.         1.12511111
 3.         2.07       2.         3.         1.24177778 1.
 2.         2.         2.        ]

>>> type(y_pred_test)

numpy.ndarray

>>> len(y_pred_test)

45

and need to round it, so I use np.around

>>> np.around([y_pred_test], decimals=0, out=y_pred_test_round[:,])

>>> print(y_pred_test_round)`

[[1. 3. 2. 1. 2. 1. 1. 3. 2. 2. 3. 3. 1. 2. 3. 2. 1. 1. 2. 1. 2. 1. 1. 2.
  2. 2. 2. 2. 2. 3. 1. 1. 2. 1. 1. 1. 3. 2. 2. 3. 1. 1. 2. 2. 2.]]

But the problem is that now I have an ndarray with len 1

>>> type(y_pred_test_round)

numpy.ndarray

>>> len(y_pred_test_round)

1

I also tried

for i in range(len(y_pred_test)):
    np.around([y_pred_test], decimals=0, out=y_pred_test_round[:,i])

and get this error

ValueError: non-broadcastable output operand with shape (45,) doesn't match the broadcast shape (1,45)

I can't find the way to fix it, anyone can help please?

Answer 1

Change

np.around([y_pred_test], decimals=0, out=y_pred_test_round[:,])

to

np.around(y_pred_test, decimals=0, out=y_pred_test_round[:,])

You don't need to put y_pred_test into [] in your case.

Answer 2

Thanks a lot, finally I solved this way

y_pred_test_round = np.around(y_pred_test, decimals=0) print(y_pred_test_round)

[1. 3. 2. 1. 2. 1. 1. 3. 2. 2. 3. 3. 1. 2. 3. 2. 1. 1. 2. 1. 2. 1. 1. 2. 2. 2. 2. 2. 2. 3. 1. 1. 2. 1. 1. 1. 3. 2. 2. 3. 1. 1. 2. 2. 2.]

len(y_pred_test_round)

45