### Question :

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]
```

##
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])
```

##
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:

`np.ravel`

: returns view, if possible`np.reshape(-1)`

: returns view, if possible`np.flatten`

: returns copy`np.flat`

: returns`numpy.flatiter`

. similar to`iterable`

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)
```

##
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 '(?,)'
```

##
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]`

##
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
```