I am looking for an efficient way to do the following:
If my input is:
np.array([9,0,1,0,3,0])
I want my output to be:
np.array([0,3,2,3,1,3]) # 9 is the highest, so it gets rank 0
# 3 is the second highest, so it gets rank 1
# 1 is third highest, so it gets rank 2
# 0's are forth highest so they get rank 3
I am trying to apply the following to 2D matrix:
Input:
a = np.array([[9,0,1,0,3,0],
[0,1,2,3,4,5],
[0.01,0.3,2,100,1,1],
[0,0,0,0,1,1],
[4,4,4,4,4,4]])
Output:
>>> get_order_array(a)
array([[0, 3, 2, 3, 1, 3],
[5, 4, 3, 2, 1, 0],
[4, 3, 1, 0, 2, 2],
[1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 0]])
I do can achieve the above with the following solution; however, I feel that its is very inefficient, so I was hoping that someone can suggest a better way to achieve my goal.
def get_order(x):
unique_x = np.unique(x)
step_1 = np.argsort(unique_x)[::-1]
temp_dict = dict(zip(unique_x, step_1))
return np.vectorize(temp_dict.get)(x)
def get_order_array(x):
new_array = np.empty(x.shape, dtype=np.int)
for i in xrange(x.shape[0]):
new_array[i] = get_order(x[i])
return new_array
@Jaime's answer is great (as usual!). Here's an alternative, using scipy.stats.rankdata.
In rankdata's terminology, you want a "dense" ranking. You also want to rank the values in the opposite order than usual. To accomplish the reverse order, we'll pass -a to rankdata. We'll also subtract 1 from the ranking so the ranks begin at 0 instead of 1. Finally, you want to rank the rows of a two-dimensional array. rankdata works on one-dimensional data, so we'll have to loop over the rows.
Here's the code:
import numpy as np
from scipy.stats import rankdata
def get_order_array(a):
b = np.empty(a.shape, dtype=int)
for k, row in enumerate(a):
b[k] = rankdata(-row, method='dense') - 1
return b
if __name__ == "__main__":
a = np.array([[9,0,1,0,3,0],
[0,1,2,3,4,5],
[0.01,0.3,2,100,1,1],
[0,0,0,0,1,1],
[4,4,4,4,4,4]])
print get_order_array(a)
Output:
[[0 3 2 3 1 3]
[5 4 3 2 1 0]
[4 3 1 0 2 2]
[1 1 1 1 0 0]
[0 0 0 0 0 0]]
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