From ND to 1D arrays

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Question :

From ND to 1D arrays

Say I have an array a:

a = np.array([[1,2,3], [4,5,6]])

array([[1, 2, 3],
       [4, 5, 6]])

I would like to convert it to a 1D array (i.e. a column vector):

b = np.reshape(a, (1,np.product(a.shape)))

but this returns

array([[1, 2, 3, 4, 5, 6]])

which is not the same as:

array([1, 2, 3, 4, 5, 6])

I can take the first element of this array to manually convert it to a 1D array:

b = np.reshape(a, (1,np.product(a.shape)))[0]

but this requires me to know how many dimensions the original array has (and concatenate [0]’s when working with higher dimensions)

Is there a dimensions-independent way of getting a column/row vector from an arbitrary ndarray?

Answer #1:

Use np.ravel (for a 1D view) or np.ndarray.flatten (for a 1D copy) or np.ndarray.flat (for an 1D iterator):

In [12]: a = np.array([[1,2,3], [4,5,6]])

In [13]: b = a.ravel()

In [14]: b
Out[14]: array([1, 2, 3, 4, 5, 6])

Note that ravel() returns a view of a when possible. So modifying b also modifies a. ravel() returns a view when the 1D elements are contiguous in memory, but would return a copy if, for example, a were made from slicing another array using a non-unit step size (e.g. a = x[::2]).

If you want a copy rather than a view, use

In [15]: c = a.flatten()

If you just want an iterator, use np.ndarray.flat:

In [20]: d = a.flat

In [21]: d
Out[21]: <numpy.flatiter object at 0x8ec2068>

In [22]: list(d)
Out[22]: [1, 2, 3, 4, 5, 6]
Answered By: unutbu

Answer #2:

In [14]: b = np.reshape(a, (np.product(a.shape),))

In [15]: b
Out[15]: array([1, 2, 3, 4, 5, 6])

or, simply:

In [16]: a.flatten()
Out[16]: array([1, 2, 3, 4, 5, 6])
Answered By: NPE

Answer #3:

I wanted to see a benchmark result of functions mentioned in answers including unutbu’s.

Also want to point out that numpy doc recommend to use arr.reshape(-1) in case view is preferable. (even though ravel is tad faster in the following result)


TL;DR: np.ravel is the most performant (by very small amount).

Benchmark

Functions:

numpy version: ‘1.18.0’

Execution times on different ndarray sizes

+-------------+----------+-----------+-----------+-------------+
|  function   |   10x10  |  100x100  | 1000x1000 | 10000x10000 |
+-------------+----------+-----------+-----------+-------------+
| ravel       | 0.002073 |  0.002123 |  0.002153 |    0.002077 |
| reshape(-1) | 0.002612 |  0.002635 |  0.002674 |    0.002701 |
| flatten     | 0.000810 |  0.007467 |  0.587538 |  107.321913 |
| flat        | 0.000337 |  0.000255 |  0.000227 |    0.000216 |
+-------------+----------+-----------+-----------+-------------+

Conclusion

ravel and reshape(-1)‘s execution time was consistent and independent from ndarray size.
However, ravel is tad faster, but reshape provides flexibility in reshaping size. (maybe that’s why numpy doc recommend to use it instead. Or there could be some cases where reshape returns view and ravel doesn’t).
If you are dealing with large size ndarray, using flatten can cause a performance issue. Recommend not to use it. Unless you need a copy of the data to do something else.

Used code

import timeit
setup = '''
import numpy as np
nd = np.random.randint(10, size=(10, 10))
'''

timeit.timeit('nd = np.reshape(nd, -1)', setup=setup, number=1000)
timeit.timeit('nd = np.ravel(nd)', setup=setup, number=1000)
timeit.timeit('nd = nd.flatten()', setup=setup, number=1000)
timeit.timeit('nd.flat', setup=setup, number=1000)
Answered By: haku

Answer #4:

One of the simplest way is to use flatten(), like this example :

 import numpy as np

 batch_y =train_output.iloc[sample, :]
 batch_y = np.array(batch_y).flatten()

My array it was like this :

    0
0   6
1   6
2   5
3   4
4   3
.
.
.

After using flatten():

array([6, 6, 5, ..., 5, 3, 6])

It’s also the solution of errors of this type :

Cannot feed value of shape (100, 1) for Tensor 'input/Y:0', which has shape '(?,)' 
Answered By: DINA TAKLIT

Answer #5:

For list of array with different size use following:

import numpy as np

# ND array list with different size
a = [[1],[2,3,4,5],[6,7,8]]

# stack them
b = np.hstack(a)

print(b)

Output:

[1 2 3 4 5 6 7 8]

Answered By: bikram

Answer #6:

Although this isn’t using the np array format, (to lazy to modify my code) this should do what you want… If, you truly want a column vector you will want to transpose the vector result. It all depends on how you are planning to use this.

def getVector(data_array,col):
    vector = []
    imax = len(data_array)
    for i in range(imax):
        vector.append(data_array[i][col])
    return ( vector )
a = ([1,2,3], [4,5,6])
b = getVector(a,1)
print(b)

Out>[2,5]

So if you need to transpose, you can do something like this:

def transposeArray(data_array):
    # need to test if this is a 1D array 
    # can't do a len(data_array[0]) if it's 1D
    two_d = True
    if isinstance(data_array[0], list):
        dimx = len(data_array[0])
    else:
        dimx = 1
        two_d = False
    dimy = len(data_array)
    # init output transposed array
    data_array_t = [[0 for row in range(dimx)] for col in range(dimy)]
    # fill output transposed array
    for i in range(dimx):
        for j in range(dimy):
            if two_d:
                data_array_t[j][i] = data_array[i][j]
            else:
                data_array_t[j][i] = data_array[j]
    return data_array_t
Answered By: xcellsior

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