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Using standard Python arrays, I can do the following:

arr = []
arr.append([1,2,3])
arr.append([4,5,6])
# arr is now [[1,2,3],[4,5,6]]

However, I cannot do the same thing in numpy. For example:

arr = np.array([])
arr = np.append(arr, np.array([1,2,3]))
arr = np.append(arr, np.array([4,5,6]))
# arr is now [1,2,3,4,5,6]

I also looked into vstack, but when I use vstack on an empty array, I get:

ValueError: all the input array dimensions except for the concatenation axis must match exactly

So how do I do append a new row to an empty array in numpy?

Fred Foo
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Tony Stark
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    If it's empty, why bother? Just start from an array holding only the first row. – jonrsharpe Mar 13 '14 at 22:50
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    I just want to know whether it is possible to append to an empty numpy array. Sometimes it's cleaner to write code like this since the append operations are in a loop. – Tony Stark Mar 13 '14 at 22:55
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    Given the way numpy arrays work, you are much better building an empty array then putting the data in, e.g. See http://stackoverflow.com/questions/568962/how-do-i-create-an-empty-array-matrix-in-numpy – jonrsharpe Mar 13 '14 at 22:56
  • how one can create array of single dimension? `np.empty(3,float)` gives `array([0.00e+000, 0.00e+000, 2.77e-322])`, instead of `array([])`. – Mahesha999 Oct 17 '20 at 21:49

7 Answers7

333

The way to "start" the array that you want is:

arr = np.empty((0,3), int)

Which is an empty array but it has the proper dimensionality.

>>> arr
array([], shape=(0, 3), dtype=int64)

Then be sure to append along axis 0:

arr = np.append(arr, np.array([[1,2,3]]), axis=0)
arr = np.append(arr, np.array([[4,5,6]]), axis=0)

But, @jonrsharpe is right. In fact, if you're going to be appending in a loop, it would be much faster to append to a list as in your first example, then convert to a numpy array at the end, since you're really not using numpy as intended during the loop:

In [210]: %%timeit
   .....: l = []
   .....: for i in xrange(1000):
   .....:     l.append([3*i+1,3*i+2,3*i+3])
   .....: l = np.asarray(l)
   .....: 
1000 loops, best of 3: 1.18 ms per loop

In [211]: %%timeit
   .....: a = np.empty((0,3), int)
   .....: for i in xrange(1000):
   .....:     a = np.append(a, 3*i+np.array([[1,2,3]]), 0)
   .....: 
100 loops, best of 3: 18.5 ms per loop

In [214]: np.allclose(a, l)
Out[214]: True

The numpythonic way to do it depends on your application, but it would be more like:

In [220]: timeit n = np.arange(1,3001).reshape(1000,3)
100000 loops, best of 3: 5.93 µs per loop

In [221]: np.allclose(a, n)
Out[221]: True
askewchan
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  • what if I have to do this 10^5 or 10^6 times? it seems that neither of these methods will hold. any suggestion? – Rho Phi Dec 05 '16 at 09:00
  • @Roberto, usually there is some way to determine the size or shape (at the very least, values would be preferable) of the array in advance. Do you think you can do that? Appending should really be a one or two time operation. – askewchan Dec 07 '16 at 16:40
  • sometimes you cannot guess the dimensions, it's life. However you can allocate a big enough array and give values to its views. I do not like it though, because there are unwanted values that one has to find a way to "mask". This idea of masking really does not fit my taste. – Rho Phi Dec 08 '16 at 17:17
  • No need to mask, just slice! `a = a[:N]` Though I strongly believe you should find a way to vectorize it (post a new question with your specifics if you need help) or just use lists until the loop is over. – askewchan Dec 08 '16 at 20:15
  • how one can create array of single dimension? `np.empty(3,float)` gives `array([0.00e+000, 0.00e+000, 2.77e-322])`, instead of `array([])`. – Mahesha999 Oct 17 '20 at 21:49
  • More I use python, more Matlab makes sense. ``arr = []; arr = [arr; x];`` – malintha Nov 03 '21 at 04:37
  • You show how append a 2D array of one row (`np.array([[1,2,3]])`) to `arr`. Is it possible to append a 1D array (`np.array([1,2,3])`)? – mins Dec 19 '22 at 18:26
  • When I try to do this, the n.empty create an array with random values (probably whatever was in those memory locations). The np.append will add the rows to the end of those random values. Not sure how to append replacing the values. – Alex Mar 10 '23 at 22:23
46

Here is my solution:

arr = []
arr.append([1,2,3])
arr.append([4,5,6])
np_arr = np.array(arr)
just4fun
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31

In this case you might want to use the functions np.hstack and np.vstack

arr = np.array([])
arr = np.hstack((arr, np.array([1,2,3])))
# arr is now [1,2,3]

arr = np.vstack((arr, np.array([4,5,6])))
# arr is now [[1,2,3],[4,5,6]]

You also can use the np.concatenate function.

Cheers

mrcl
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    Won't work if the second array has dimension >=2 like ones((2, 2)). It appears to me there is no way to avoid boundary cases if you are building up arrays from empty by concatenation. – Taozi Oct 20 '15 at 21:25
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    Not a good solution as one needs to check the dimension every time. – SKR Nov 22 '18 at 19:32
2

I want to do a for loop, yet with askewchan's method it does not work well, so I have modified it.

x = np.empty((0,3))
y = np.array([1,2,3])
for i in ...
    x = np.vstack((x,y))
WDC
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Qingdong Wang
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1

using an custom dtype definition, what worked for me was:

import numpy

# define custom dtype
type1 = numpy.dtype([('freq', numpy.float64, 1), ('amplitude', numpy.float64, 1)])
# declare empty array, zero rows but one column
arr = numpy.empty([0,1],dtype=type1)
# store row data, maybe inside a loop
row = numpy.array([(0.0001, 0.002)], dtype=type1)
# append row to the main array
arr = numpy.row_stack((arr, row))
# print values stored in the row 0
print float(arr[0]['freq'])
print float(arr[0]['amplitude'])
boclodoa
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1

In case of adding new rows for array in loop, Assign the array directly for firsttime in loop instead of initialising an empty array.

for i in range(0,len(0,100)):
    SOMECALCULATEDARRAY = .......
    if(i==0):
        finalArrayCollection = SOMECALCULATEDARRAY
    else:
        finalArrayCollection = np.vstack(finalArrayCollection,SOMECALCULATEDARRAY)

This is mainly useful when the shape of the array is unknown

Rajesh_Saladi
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-1

This is more efficient way when taking into account memory:

shape = (n, inp_len)
arr= np.empty(shape)

for i in range(n):
    arr[i] = np.expand_dims(arr, axis=0)
İbrahim KARABAYIR
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