Getting index of numpy.ndarray

Question

I have a one-dimensional array of the type numpy.ndarray and I want to know the index of it's max entry. After finding the max, I used

peakIndex = numpy.where(myArray==max)

to find the peak's index. But instead of the index, my script spits out

peakIndex = (array([1293]),)

I want my code to spit out just the integer 1293. How can I clean up the output?

score 8 · Accepted Answer · answered Jul 07 '12 at 18:06

8

Rather than using numpy.where, you can use numpy.argmax.

peakIndex = numpy.argmax(myArray)

numpy.argmax returns a single number, the flattened index of the first occurrence of the maximum value. If myArray is multidimensional you might want to convert the flattened index to an index tuple:

peakIndexTuple = numpy.unravel_index(numpy.argmax(myArray), myArray.shape)

answered Jul 07 '12 at 18:06

ecatmur

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That would work, but it will only return the index of the first position with a value equal to the max. Depending on the nature of the dataset and application, this could be problematic. – abought Jul 07 '12 at 18:09
1

He said he wants an integer as solution. – Charles Brunet Jul 07 '12 at 19:46

score 3 · Answer 2 · edited May 23 '17 at 12:33

To find the max value of an array, you can use the array.max() method. This will probably be more efficient than the for loop described in another answer, which- in addition to not being pythonic- isn't actually written in python. (if you wanted to take items out of the array one by one to compare, you could use ndenumerate, but you'd be sacrificing some of the performance benefits of arrays)

The reason that numpy.where() yields results as tuples is that more than one position could be equal to the max... and it's that edge case that would make something simple (like taking array[0]) prone to bugs. Per Is there a Numpy function to return the first index of something in an array?,

"The result is a tuple with first all the row indices, then all the column indices".

Your example uses a 1-D array, so you'd get the results you want directly from the array provided. It's a tuple with one element (one array of indices), and although you can iterate over ind_1d[0] directly, I converted it to a list solely for readability.

>>> peakIndex_1d
array([ 1. ,  1.1,  1.6,  1. ,  1.6,  0.8])
>>> ind_1d = numpy.where( peakIndex_1d == peakIndex_1d.max() )
(array([2, 4]),)
>>> list( ind_1d[0] )
[2, 4]

For a 2-D array with 3 values equal to the max, you could use:

>>> peakIndex
array([[ 0. ,  1.1,  1.5],
       [ 1.1,  1.5,  0.7],
       [ 0.2,  1.2,  1.5]])
>>> indices = numpy.where( peakIndex == peakIndex.max() )
>>> ind2d = zip(indices[0], indices[1])
[(0, 2), (1, 1), (2, 2)]

Getting index of numpy.ndarray

2 Answers2