## 问题：如何将新行添加到空的numpy数组

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

``````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]``````

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

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?

## 回答 0

“启动”所需阵列的方法是：

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

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

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

``````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``````

numpythonic的实现方法取决于您的应用程序，但它更像是：

``````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``````

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
``````

## 回答 1

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

Here is my solution:

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

## 回答 2

``````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]]``````

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

## 回答 3

``````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'])``````

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'])
``````

## 回答 4

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

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

## 回答 5

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

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 = vstack((x,y))
``````