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How to add a new row to an empty numpy array

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How do I add rows to an empty NumPy array?

If we have an empty array and want to append new rows to it inside a loop, we can use the numpy. empty() function. Since no data type is assigned to a variable before initialization in Python, we have to specify the data type and structure of the array elements while creating the empty array.

How do I add rows to a NumPy array?

Use the numpy. append() Function to Add a Row to a Matrix in NumPy. The append() function from the numpy module can add elements to the end of the array. By specifying the axis as 0, we can use this function to add rows to a matrix.

How do you add to an empty array in Python?

You can add elements to an empty list using the methods append() and insert() : append() adds the element to the end of the list. insert() adds the element at the particular index of the list that you choose.

How do you add elements to an empty array?

You can create empty list by [] . In order to add new item use append . For add other list use extend .


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'])