Creating an array by adding to shape of existing array

Question

I have a numpy array:

X = np.array([[1,0,1],
              [1,1,1],
              [0,1,0],
              [1,0,1]])

which has a shape of (4,3)

I would like to change this shape into (4,4) by adding 1 to the second dimension of the array, via:

X_b = np.ones((X.shape+(0,1)))

but what I get is:

ValueError: could not broadcast input array from shape (4,3) into shape (4,2,0,1)

What is the right way to do it?

Basically I want X_b to have a shape of (4,4) if X.shape = (4,3)

tel · Accepted Answer · 2018-11-21 21:55:08Z

1

To fix your code, do this instead:

X_b = np.ones(X.shape + np.array((0,1)))

The catch here is that X.shape returns a plain Python tuple. By adding (0,1) you were actually performing tuple concatenation, instead of pairwise addition like you intended.

Of course, you could also just stick an extra column on to your existing array with append:

X_b = np.append(X, [[1]]*X.shape[0], axis=1)

edited Nov 21, 2018 at 21:55

answered Nov 21, 2018 at 20:13

tel

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8 Comments

oakca Over a year ago

X_b[:,:-1] = X and then use X_b as input... probably doing the same thing. your answer looks good. Just waiting the minutes

tel Over a year ago

@oakca assuming X_b was created with np.ones, then yep. The one "gotcha" is that X_b will be of dtype float64, which is the default for np.ones. Thus the dtypes of X and X_b may not match (eg if X.dtype==int)

xnx Over a year ago

Or np.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.

tel Over a year ago

@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, both append and hstack are thin (5-10 lines) wrappers around concatenate.

oakca Over a year ago

Here you can find more about the thema stackoverflow.com/questions/8486294/…

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