First I create my array
myarray = np.random.random_integers(0,10, size=20)
Then, I want to set 20% of the elements in the array to 0 (or some other number). How should I do this? Apply a mask?
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5 Answers
You can calculate the indices with np.random.choice, limiting the number of chosen indices to the percentage:
indices = np.random.choice(np.arange(myarray.size), replace=False,
size=int(myarray.size * 0.2))
myarray[indices] = 0
4 Comments
Gen Tan
Cool. I should also set
replace=False for np.random.choice right? If I do not, some of the indices generated will be repeated
Mike Müller
Yes. Updated my answer.
Holi
For arbitrarily shaped arrays we can use unravel_index: myarray[np.unravel_index(indices, myarray.shape)] = 0
duckmayr
@Holi 's comment is such a useful addition it should either be incorporated into this answer or made its own separate answer, IMO
For others looking for the answer in case of nd-array, as proposed by user holi:
my_array = np.random.rand(8, 50)
indices = np.random.choice(my_array.shape[1]*my_array.shape[0], replace=False, size=int(my_array.shape[1]*my_array.shape[0]*0.2))
We multiply the dimensions to get an array of length dim1*dim2, then we apply this indices to our array:
my_array[np.unravel_index(indices, my_array.shape)] = 0
The array is now masked.
Comments
Use np.random.permutation as random index generator, and take the first 20% of the index.
myarray = np.random.random_integers(0,10, size=20)
n = len(myarray)
random_idx = np.random.permutation(n)
frac = 20 # [%]
zero_idx = random_idx[:round(n*frac/100)]
myarray[zero_idx] = 0
1 Comment
user3352632
DeprecationWarning: This function is deprecated. Please call randint(0, 10 + 1) instead. Now it is: myarray = np.random.randint(0,10, size=20)
If you want the 20% to be random:
random_list = []
array_len = len(myarray)
while len(random_list) < (array_len/5):
random_int = math.randint(0,array_len)
if random_int not in random_list:
random_list.append(random_int)
for position in random_list:
myarray[position] = 0
return myarray
This would ensure you definitely get 20% of the values, and RNG rolling the same number many times would not result in less than 20% of the values being 0.
Comments
Assume your input numpy array is A and p=0.2. The following are a couple of ways to achieve this.
Exact Masking
ones = np.ones(A.size)
idx = int(min(p*A.size, A.size))
ones[:idx] = 0
A *= np.reshape(np.random.permutation(ones), A.shape)
Approximate Masking
This is commonly done in several denoising objectives, most notably the Masked Language Modeling in Transformers pre-training. Here is a more pythonic way of setting a certain proportion (say 20%) of elements to zero.
A *= np.random.binomial(size=A.shape, n=1, p=0.8)
Another Alternative:
A *= np.random.randint(0, 2, A.shape)
