## The Question :

*489 people think this question is useful*

A numpy matrix can be reshaped into a vector using reshape function with parameter -1. But I don’t know what -1 means here.

For example:

a = numpy.matrix([[1, 2, 3, 4], [5, 6, 7, 8]])
b = numpy.reshape(a, -1)

The result of `b`

is: `matrix([[1, 2, 3, 4, 5, 6, 7, 8]])`

Does anyone know what -1 means here?
And it seems python assign -1 several meanings, such as: `array[-1]`

means the last element. Can you give an explanation?

*The Question Comments :*

## The Answer 1

*653 people think this answer is useful*

The criterion to satisfy for providing the new shape is that *‘The new shape should be compatible with the original shape’*

numpy allow us to give one of new shape parameter as -1 (eg: (2,-1) or (-1,3) but not (-1, -1)). It simply means that it is an unknown dimension and we want numpy to figure it out. And numpy will figure this by looking at the *‘length of the array and remaining dimensions’* and making sure it satisfies the above mentioned criteria

Now see the example.

z = np.array([[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]])
z.shape
(3, 4)

Now trying to reshape with (-1) . Result new shape is (12,) and is compatible with original shape (3,4)

z.reshape(-1)
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])

Now trying to reshape with (-1, 1) . We have provided column as 1 but rows as unknown . So we get result new shape as (12, 1).again compatible with original shape(3,4)

z.reshape(-1,1)
array([[ 1],
[ 2],
[ 3],
[ 4],
[ 5],
[ 6],
[ 7],
[ 8],
[ 9],
[10],
[11],
[12]])

The above is consistent with `numpy`

advice/error message, to use `reshape(-1,1)`

for a single feature; i.e. single column

Reshape your data using `array.reshape(-1, 1)`

if your data has a **single feature**

New shape as (-1, 2). row unknown, column 2. we get result new shape as (6, 2)

z.reshape(-1, 2)
array([[ 1, 2],
[ 3, 4],
[ 5, 6],
[ 7, 8],
[ 9, 10],
[11, 12]])

Now trying to keep column as unknown. New shape as (1,-1). i.e, row is 1, column unknown. we get result new shape as (1, 12)

z.reshape(1,-1)
array([[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]])

The above is consistent with `numpy`

advice/error message, to use `reshape(1,-1)`

for a single sample; i.e. single row

Reshape your data using `array.reshape(1, -1)`

if it contains a **single sample**

New shape (2, -1). Row 2, column unknown. we get result new shape as (2,6)

z.reshape(2, -1)
array([[ 1, 2, 3, 4, 5, 6],
[ 7, 8, 9, 10, 11, 12]])

New shape as (3, -1). Row 3, column unknown. we get result new shape as (3,4)

z.reshape(3, -1)
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])

And finally, if we try to provide both dimension as unknown i.e new shape as (-1,-1). It will throw an error

z.reshape(-1, -1)
ValueError: can only specify one unknown dimension

## The Answer 2

*88 people think this answer is useful*

Used to reshape an array.

Say we have a 3 dimensional array of dimensions 2 x 10 x 10:

r = numpy.random.rand(2, 10, 10)

Now we want to reshape to 5 X 5 x 8:

numpy.reshape(r, shape=(5, 5, 8))

will do the job.

Note that, once you fix first dim = 5 and second dim = 5, you don’t need to determine third dimension. To assist your laziness, python gives the option of -1:

numpy.reshape(r, shape=(5, 5, -1))

will give you an array of shape = (5, 5, 8).

Likewise,

numpy.reshape(r, shape=(50, -1))

will give you an array of shape = (50, 4)

You can read more at http://anie.me/numpy-reshape-transpose-theano-dimshuffle/

## The Answer 3

*62 people think this answer is useful*

According to `the documentation`

:

newshape : int or tuple of ints

The new shape should be compatible with the original shape. If an
integer, then the result will be a 1-D array of that length. One shape
dimension can be **-1. In this case, the value is inferred from the
length of the array and remaining dimensions.**

## The Answer 4

*20 people think this answer is useful*

numpy.reshape(a,newshape,order{})

check the below link for more info.
https://docs.scipy.org/doc/numpy/reference/generated/numpy.reshape.html

for the below example you mentioned the output explains the resultant vector to be a single row.(-1) indicates the number of rows to be 1.
if the

a = numpy.matrix([[1, 2, 3, 4], [5, 6, 7, 8]])
b = numpy.reshape(a, -1)

output:

matrix([[1, 2, 3, 4, 5, 6, 7, 8]])

this can be explained more precisely with another example:

b = np.arange(10).reshape((-1,1))

output:(is a 1 dimensional columnar array)

array([[0],
[1],
[2],
[3],
[4],
[5],
[6],
[7],
[8],
[9]])

or

b = np.arange(10).reshape((1,-1))

output:(is a 1 dimensional row array)

array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]])

## The Answer 5

*13 people think this answer is useful*

It is fairly easy to understand. The “-1” stands for “unknown dimension” which can should be infered from another dimension.
In this case, if you set your matrix like this:

a = numpy.matrix([[1, 2, 3, 4], [5, 6, 7, 8]])

Modify your matrix like this:

b = numpy.reshape(a, -1)

It will call some deafult operations to the matrix a, which will return a 1-d numpy array/martrix.

However, I don’t think it is a good idea to use code like this. Why not try:

b = a.reshape(1,-1)

It will give you the same result and it’s more clear for readers to understand: Set b as another shape of a. For a, we don’t how much columns it should have(set it to -1!), but we want a 1-dimension array(set the first parameter to 1!).

## The Answer 6

*10 people think this answer is useful*

**Long story short**: you set some dimensions and let NumPy set the remaining(s).

(userDim1, userDim2, ..., -1) -->>
(userDim1, userDim1, ..., TOTAL_DIMENSION - (userDim1 + userDim2 + ...))

## The Answer 7

*8 people think this answer is useful*

*It simply means that you are not sure about what number of rows or columns you can give and you are asking numpy to suggest number of column or rows to get reshaped in.*

numpy provides last example for -1
https://docs.scipy.org/doc/numpy/reference/generated/numpy.reshape.html

check below code and its output to better understand about (-1):

CODE:-

import numpy
a = numpy.matrix([[1, 2, 3, 4], [5, 6, 7, 8]])
print("Without reshaping -> ")
print(a)
b = numpy.reshape(a, -1)
print("HERE We don't know about what number we should give to row/col")
print("Reshaping as (a,-1)")
print(b)
c = numpy.reshape(a, (-1,2))
print("HERE We just know about number of columns")
print("Reshaping as (a,(-1,2))")
print(c)
d = numpy.reshape(a, (2,-1))
print("HERE We just know about number of rows")
print("Reshaping as (a,(2,-1))")
print(d)

OUTPUT :-

Without reshaping ->
[[1 2 3 4]
[5 6 7 8]]
HERE We don't know about what number we should give to row/col
Reshaping as (a,-1)
[[1 2 3 4 5 6 7 8]]
HERE We just know about number of columns
Reshaping as (a,(-1,2))
[[1 2]
[3 4]
[5 6]
[7 8]]
HERE We just know about number of rows
Reshaping as (a,(2,-1))
[[1 2 3 4]
[5 6 7 8]]

## The Answer 8

*8 people think this answer is useful*

import numpy as np
x = np.array([[2,3,4], [5,6,7]])
# Convert any shape to 1D shape
x = np.reshape(x, (-1)) # Making it 1 row -> (6,)
# When you don't care about rows and just want to fix number of columns
x = np.reshape(x, (-1, 1)) # Making it 1 column -> (6, 1)
x = np.reshape(x, (-1, 2)) # Making it 2 column -> (3, 2)
x = np.reshape(x, (-1, 3)) # Making it 3 column -> (2, 3)
# When you don't care about columns and just want to fix number of rows
x = np.reshape(x, (1, -1)) # Making it 1 row -> (1, 6)
x = np.reshape(x, (2, -1)) # Making it 2 row -> (2, 3)
x = np.reshape(x, (3, -1)) # Making it 3 row -> (3, 2)

## The Answer 9

*7 people think this answer is useful*

The final outcome of the conversion is that the number of elements in the final array is same as that of the initial array or data frame.

-1 corresponds to the unknown count of the row or column.
We can think of it as `x`

(unknown). `x`

is obtained by dividing the number of elements in the original array by the other value of the ordered pair with -1.

Examples:

12 elements with `reshape(-1,1)`

corresponds to an array with `x`

=12/1=12 rows and 1 column.

12 elements with `reshape(1,-1)`

corresponds to an array with 1 row and `x`

=12/1=12 columns.