问题
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?
回答1:
(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
orB
is empty and the other one contains a single element, then it returnTrue
. For some reason, the comparisonA==B
returns an empty array, for which theall
operator returnsTrue
. - Another risk is if
A
andB
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
回答2:
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.)
回答3:
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.
回答4:
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
回答5:
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
来源:https://stackoverflow.com/questions/10580676/comparing-two-numpy-arrays-for-equality-element-wise