Say:
p = array([4, 0, 8, 2, 7])
Want to find the index of max value, except few indexes, say:
excptIndx = [2, 3]
Ans: 4, as 7 will be max.
if excptIndx = [1, 3], Ans: 2, as 8 will be max.
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Don't want to mark as duplicate but this is relevant: How to select inverse of indexes of a numpy array?sshashank124– sshashank1242020-01-15 05:43:50 +00:00Commented Jan 15, 2020 at 5:43
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4 Answers
In numpy, you can mask all values at excptIndx and run argmax to obtain index of max element:
import numpy as np
p = np.array([4, 0, 8, 2, 7])
excptIndx = [2, 3]
m = np.zeros(p.size, dtype=bool)
m[excptIndx] = True
a = np.ma.array(p, mask=m)
print(np.argmax(a))
# 4
2 Comments
Dr.PB
I need the index, not max value.
Trisoloriansunscreen
This gives the index.
The setup:
In [153]: p = np.array([4,0,8,2,7])
In [154]: exceptions = [2,3]
Original indexes in p:
In [155]: idx = np.arange(p.shape[0])
delete exceptions from both:
In [156]: np.delete(p,exceptions)
Out[156]: array([4, 0, 7])
In [157]: np.delete(idx,exceptions)
Out[157]: array([0, 1, 4])
Find the argmax in the deleted array:
In [158]: np.argmax(np.delete(p,exceptions))
Out[158]: 2
Use that to find the max value (could just as well use np.max(_156)
In [159]: _156[_158]
Out[159]: 7
Use the same index to find the index in the original p
In [160]: _157[_158]
Out[160]: 4
In [161]: p[_160] # another way to get the max value
Out[161]: 7
For this small example, the pure Python alternatives might well be faster. They often are in small cases. We need test cases with a 1000 or more values to really see the advantages of numpy.
Another method
Set the exceptions to a small enough value, and take the argmax:
In [162]: p1 = p.copy(); p1[exceptions] = -1000
In [163]: np.argmax(p1)
Out[163]: 4
Here the small enough is easy to pick; more generally it may require some thought.
Or taking advantage of the np.nan... functions:
In [164]: p1 = p.astype(float); p1[exceptions]=np.nan
In [165]: np.nanargmax(p1)
Out[165]: 4
Comments
A solution is
mask = np.isin(np.arange(len(p)), excptIndx)
subset_idx = np.argmax(p[mask])
parent_idx = np.arange(len(p))[mask][subset_idx]
See http://seanlaw.github.io/2015/09/10/numpy-argmin-with-a-condition/
1 Comment
bluewhale
Well he can just declare
excptIndx = {2, 3}p = np.array([4,0,8,2,7]) # given
exceptions = [2,3] # given
idx = list( range(0,len(p)) ) # simple array of index
a1 = np.delete(idx, exceptions) # remove exceptions from idx (i.e., index)
a2 = np.argmax(np.delete(p, exceptions)) # get index of the max value after removing exceptions from actual p array
a1[a2] # as a1 and a2 are in sync, this will give the original index (as asked) of the max value
1 Comment
byaruhaf
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value