Numpy: how I can determine if all elements of numpy array are equal to a number
Question:
I need to know if all the elements of an array of numpy are equal to a number
It would be like:
numbers = np.zeros(5) # array[0,0,0,0,0]
print numbers.allEqual(0) # return True because all elements are 0
I can make an algorithm but, there is some method implemented in numpy library?
Answers:
You can break that down into np.all(), which takes a boolean array and checks it’s all True, and an equality comparison:
np.all(numbers == 0)
# or equivalently
(numbers == 0).all()
If you want to compare floats, use np.isclose instead:
np.all(np.isclose(numbers, numbers[0]))
Are numpy methods necessary? If all the elements are equal to a number, then all the elements are the same. You could do the following, which takes advantage of short circuiting.
numbers[0] == 0 and len(set(numbers)) == 1
This way is faster than using np.all()
np.array_equal() also works (for Python 3).
tmp0 = np.array([0]*5)
tmp1 = np.array([0]*5)
np.array_equal(tmp0, tmp1)
returns True
The following version checks the equality of all entries of the array without requiring the repeating number.
numbers_0 = np.zeros(5) # array[0,0,0,0,0]
numbers.ptp() == 0.0 # True
# checking an array having some random repeating entry
numbers_r = np.ones((10, 10))*np.random.rand()
numbers_r.ptp() == 0.0 # True
I need to know if all the elements of an array of numpy are equal to a number
It would be like:
numbers = np.zeros(5) # array[0,0,0,0,0]
print numbers.allEqual(0) # return True because all elements are 0
I can make an algorithm but, there is some method implemented in numpy library?
You can break that down into np.all(), which takes a boolean array and checks it’s all True, and an equality comparison:
np.all(numbers == 0)
# or equivalently
(numbers == 0).all()
If you want to compare floats, use np.isclose instead:
np.all(np.isclose(numbers, numbers[0]))
Are numpy methods necessary? If all the elements are equal to a number, then all the elements are the same. You could do the following, which takes advantage of short circuiting.
numbers[0] == 0 and len(set(numbers)) == 1
This way is faster than using np.all()
np.array_equal() also works (for Python 3).
tmp0 = np.array([0]*5)
tmp1 = np.array([0]*5)
np.array_equal(tmp0, tmp1)
returns True
The following version checks the equality of all entries of the array without requiring the repeating number.
numbers_0 = np.zeros(5) # array[0,0,0,0,0]
numbers.ptp() == 0.0 # True
# checking an array having some random repeating entry
numbers_r = np.ones((10, 10))*np.random.rand()
numbers_r.ptp() == 0.0 # True