Fastest way to add a column in numpy array with a specific value?

Question

This is most likely a duplicate; apologies as I couldn't find the answer. I have a (3000, 3) array and I want to make a (3000, 4) array with the last column being a specific value. I can do a very simple np.insert(x, 1, val, axis=1) to insert the value before column 1, but I can't add it to the last column with np.insert(). I can do np.repeat(val, np.shape(x)[0]) or using np.hstack() but they require creating an array first of the same length as the original array (which varies) and a bit clunky (maybe this is the only way), and I'm guessing there is a better way if I am only concerned with one value I want to append.

Is there anything better than np.concatenate((x, np.repeat(val, np.shape(x)[0])[:, np.newaxis]), axis=1)?

I would normally go for the np.hstack method. It wouldn't really require any extra memory. But may I ask, if the value is the same for every row, why do you need this column in the array at all? — Seb
– Seb, Commented Nov 10, 2019 at 23:18
I just have other arrays with different values for the last column (computed in different places) to then combine afterwards so it's just easier to to have this procedure at this particular step. — ru111
– ru111, Commented Nov 10, 2019 at 23:21
I see. If you do want an array with shape (3000, 4), there is no way around allocating the memory for that extra column. So I'd suggest to just go with np.hstack. — Seb
– Seb, Commented Nov 10, 2019 at 23:25
What about np.c_[z,np.ones_like(z[:,0])*val] ? I don't know if it's "better" but it's slightly more concise IMO — Arun
– Arun, Commented Nov 10, 2019 at 23:38
Copy the array to a (3000,4), and the fill value to the last column . — hpaulj
– hpaulj, Commented Nov 10, 2019 at 23:42

tstanisl · Accepted Answer · 2019-11-10 23:46:08Z

3

You can allocate a larger array, copy old values from x and fill the last column with a scalar with a little help of broadcasting

y=np.empty_like(x, shape=(3000,4))
y[:,:-1]=x
y[:,-1]=your_value

answered Nov 10, 2019 at 23:46

tstanisl

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Arun · Accepted Answer · 2019-11-11 00:06:27Z

0

I would suggest using np.c_ and np.ones_like to cut out some verbosity, but the solution is similar to your suggestion

x=np.c_[x,val*np.ones_like(x[:,0])]

It doesn't require allocating memory for a new matrix.

answered Nov 11, 2019 at 0:06

Arun

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