[python] Add single element to array in numpy

I have a numpy array containing:

[1, 2, 3]

I want to create an array containing:

[1, 2, 3, 1]

That is, I want to add the first element on to the end of the array.

I have tried the obvious:

np.concatenate((a, a[0]))

But I get an error saying ValueError: arrays must have same number of dimensions

I don't understand this - the arrays are both just 1d arrays.

a[0] isn't an array, it's the first element of a and therefore has no dimensions.

Try using a[0:1] instead, which will return the first element of a inside a single item array.

When appending only once or once every now and again, using np.append on your array should be fine. The drawback of this approach is that memory is allocated for a completely new array every time it is called. When growing an array for a significant amount of samples it would be better to either pre-allocate the array (if the total size is known) or to append to a list and convert to an array afterward.

Using np.append:

b = np.array([0])
for k in range(int(10e4)):
    b = np.append(b, k)
1.2 s ± 16.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Using python list converting to array afterward:

d = [0]
for k in range(int(10e4)):
    d.append(k)
f = np.array(d)
13.5 ms ± 277 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Pre-allocating numpy array:

e = np.zeros((n,))
for k in range(n):
    e[k] = k
9.92 ms ± 752 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

When the final size is unkown pre-allocating is difficult, I tried pre-allocating in chunks of 50 but it did not come close to using a list.

85.1 ms ± 561 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Let's say a=[1,2,3] and you want it to be [1,2,3,1].

You may use the built-in append function

np.append(a,1)

Here 1 is an int, it may be a string and it may or may not belong to the elements in the array. Prints: [1,2,3,1]

If you want to add an element use append()

a = numpy.append(a, 1) in this case add the 1 at the end of the array

If you want to insert an element use insert()

a = numpy.insert(a, index, 1) in this case you can put the 1 where you desire, using index to set the position in the array.

t = np.array([2, 3])
t = np.append(t, [4])

This command,

numpy.append(a, a[0])

does not alter a array. However, it returns a new modified array. So, if a modification is required, then the following must be used.

a = numpy.append(a, a[0])

Try this:

np.concatenate((a, np.array([a[0]])))

http://docs.scipy.org/doc/numpy/reference/generated/numpy.concatenate.html

concatenate needs both elements to be numpy arrays; however, a[0] is not an array. That is why it does not work.

This might be a bit overkill, but I always use the the np.take function for any wrap-around indexing:

>>> a = np.array([1, 2, 3])
>>> np.take(a, range(0, len(a)+1), mode='wrap')
array([1, 2, 3, 1])

>>> np.take(a, range(-1, len(a)+1), mode='wrap')
array([3, 1, 2, 3, 1])