python - How to add a new row to an empty numpy array

187
votes

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?

7
pythonnumpyscipy
If it's empty, why bother? Just start from an array holding only the first row. – jonrsharpe
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
Given the way numpy arrays work, you are much better building an empty array then putting the data in, e.g. See stackoverflow.com/questions/568962/… – jonrsharpe
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

7 Answers

275
votes

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
35
votes

Here is my solution:

arr = []
arr.append([1,2,3])
arr.append([4,5,6])
np_arr = np.array(arr)
27
votes

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

1
votes

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'])
1
votes

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

1
votes

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))
0
votes

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)