# Comparing two NumPy arrays for equality, element-wise

## Question:

What is the simplest way to compare two NumPy arrays for equality (where equality is defined as: A = B iff for all indices i: `A[i] == B[i]`

)?

Simply using `==`

gives me a boolean array:

```
>>> numpy.array([1,1,1]) == numpy.array([1,1,1])
array([ True, True, True], dtype=bool)
```

Do I have to `and`

the elements of this array to determine if the arrays are equal, or is there a simpler way to compare?

## Answers:

```
(A==B).all()
```

test if all values of array (A==B) are True.

Note: maybe you also want to test A and B shape, such as `A.shape == B.shape`

**Special cases and alternatives** (from dbaupp’s answer and yoavram’s comment)

It should be noted that:

- this solution can have a strange behavior in a particular case: if either
`A`

or`B`

is empty and the other one contains a single element, then it return`True`

. For some reason, the comparison`A==B`

returns an empty array, for which the`all`

operator returns`True`

. - Another risk is if
`A`

and`B`

don’t have the same shape and aren’t broadcastable, then this approach will raise an error.

In conclusion, if you have a doubt about `A`

and `B`

shape or simply want to be safe: use one of the specialized functions:

```
np.array_equal(A,B) # test if same shape, same elements values
np.array_equiv(A,B) # test if broadcastable shape, same elements values
np.allclose(A,B,...) # test if same shape, elements have close enough values
```

The `(A==B).all()`

solution is very neat, but there are some built-in functions for this task. Namely `array_equal`

, `allclose`

and `array_equiv`

.

(Although, some quick testing with `timeit`

seems to indicate that the `(A==B).all()`

method is the fastest, which is a little peculiar, given it has to allocate a whole new array.)

Let’s measure the performance by using the following piece of code.

```
import numpy as np
import time
exec_time0 = []
exec_time1 = []
exec_time2 = []
sizeOfArray = 5000
numOfIterations = 200
for i in xrange(numOfIterations):
A = np.random.randint(0,255,(sizeOfArray,sizeOfArray))
B = np.random.randint(0,255,(sizeOfArray,sizeOfArray))
a = time.clock()
res = (A==B).all()
b = time.clock()
exec_time0.append( b - a )
a = time.clock()
res = np.array_equal(A,B)
b = time.clock()
exec_time1.append( b - a )
a = time.clock()
res = np.array_equiv(A,B)
b = time.clock()
exec_time2.append( b - a )
print 'Method: (A==B).all(), ', np.mean(exec_time0)
print 'Method: np.array_equal(A,B),', np.mean(exec_time1)
print 'Method: np.array_equiv(A,B),', np.mean(exec_time2)
```

**Output**

```
Method: (A==B).all(), 0.03031857
Method: np.array_equal(A,B), 0.030025185
Method: np.array_equiv(A,B), 0.030141515
```

According to the results above, the numpy methods seem to be faster than the combination of the **==** operator and the **all()** method and by comparing the numpy methods **the fastest** one seems to be the **numpy.array_equal** method.

If you want to check if two arrays have the same `shape`

AND `elements`

you should use `np.array_equal`

as it is the method recommended in the documentation.

Performance-wise don’t expect that any equality check will beat another, as there is not much room to optimize

`comparing two elements`

. Just for the sake, i still did some tests.

```
import numpy as np
import timeit
A = np.zeros((300, 300, 3))
B = np.zeros((300, 300, 3))
C = np.ones((300, 300, 3))
timeit.timeit(stmt='(A==B).all()', setup='from __main__ import A, B', number=10**5)
timeit.timeit(stmt='np.array_equal(A, B)', setup='from __main__ import A, B, np', number=10**5)
timeit.timeit(stmt='np.array_equiv(A, B)', setup='from __main__ import A, B, np', number=10**5)
> 51.5094
> 52.555
> 52.761
```

So pretty much equal, no need to talk about the speed.

The `(A==B).all()`

behaves pretty much as the following code snippet:

```
x = [1,2,3]
y = [1,2,3]
print all([x[i]==y[i] for i in range(len(x))])
> True
```

Usually two arrays will have some small numeric errors,

You can use `numpy.allclose(A,B)`

, instead of `(A==B).all()`

. This returns a bool True/False

Now use `np.array_equal`

. From documentation:

```
np.array_equal([1, 2], [1, 2])
True
np.array_equal(np.array([1, 2]), np.array([1, 2]))
True
np.array_equal([1, 2], [1, 2, 3])
False
np.array_equal([1, 2], [1, 4])
False
```

On top of the other answers, you can now use an assertion:

```
numpy.testing.assert_array_equal(x, y)
```

You also have similar function such as `numpy.testing.assert_almost_equal()`

https://numpy.org/doc/stable/reference/generated/numpy.testing.assert_array_equal.html

Just for the sake of completeness. I will add the

pandas approach for comparing two arrays:

```
import numpy as np
a = np.arange(0.0, 10.2, 0.12)
b = np.arange(0.0, 10.2, 0.12)
ap = pd.DataFrame(a)
bp = pd.DataFrame(b)
ap.equals(bp)
True
```

FYI: In case you are looking of How to

compare Vectors, Arrays or Dataframes in R.

You just you can use:

```
identical(iris1, iris2)
#[1] TRUE
all.equal(array1, array2)
#> [1] TRUE
```