For a 2D numpy array A, the loop for a in A will loop through all the rows in A. This functionality is what I want for my code, but I'm having difficulty with the edge case where A only has one row (i.e., is essentially a 1-dimensional array). In this case, the for loop treats A as a 1D array and iterates through its elements. What I want to instead happen in this case is a natural extension of the 2D case, where the loop retrieves the (single) row in A. Is there a way to format the array A such that the for loop functions like this?
3 Answers
I think you can use np.expand_dims to achieve your goal
X = np.expand_dims(X, axis=0)
Comments
Depending on if you declare the array yourself you can do this:
A = np.array([[1, 2, 3]])
Else you can check the dim of your array before iterating over it
B = np.array([1, 2, 3])
if B.ndim == 1:
B = B[None, :]
Or you can use the function np.at_least2d
C = np.array([1, 2, 3])
C = np.atleast_2d(C)
Comments
If your array trully is a 2D array, even with one row, there is no edge case:
import numpy
a = numpy.array([[1, 2, 3]])
for line in a:
print(line)
>>> [1 2 3]
You seem to be confusing numpy.array([[1, 2, 3]]) which is a 2D array of one line and numpy.array([1, 2, 3]) which would be a 1D array.
np.atleast_2donAbefore iterating.shape(A) = (1, n), i.e., a matrix with a single row, your loop will work just fine. The issue arises if you have a vector, i.e.,shape(A) = (n,). You can checklen(shape(A))beforehand and apply a suitablereshapeif you only got a vector.