How to Add a New Row to an Empty Numpy Array
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...
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 Answers
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
Here is my solution:
arr = []
arr.append([1,2,3])
arr.append([4,5,6])
np_arr = np.array(arr)
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
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'])
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
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))
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)
Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.
