CODEWITHSUNDEEP
pythonmatlabnumpy
LostInTheFrequencyDomain
LostInTheFrequencyDomain

Reputation: 1147

Replace a loop in python with the equivalent of a matlab find

Assume I have a sorted array of tuples which is sorted by the first value. I want to find the first index where a condition on the first element of the tuple holds. i.e. How do I replace the following code 

test_array = [(1,2),(3,4),(5,6),(7,8),)(9,10)]
min_value = 5
index = 0
for c in test_array:
        if c[0] > min_value:
           break
        else:
            index = index + 1

With the equivalent of a matlab find ?

i.e. At the end of this loop I expect to get 3 but I'd like to make this more efficient. I an fine with using numpy for this. I tried using argmax but to no avail.

Thanks

Upvotes: 1

Views: 265

Answers (3)

Yuval Atzmon
Yuval Atzmon

Reputation: 5945

You can use numpy to indicate the elements that obey the conditions and then use argmax(), to get the index of the first one

import numpy
test_array = numpy.array([(1,2),(3,4),(5,6),(7,8),(9,10)])
min_value = 5

print (test_array[:,0]>min_value).argmax()

if you would like to find all of the elements that obey the condition, use can replace argmax() by nonzero()[0]

Upvotes: 0

Jean-François Fabre
Jean-François Fabre

Reputation: 140168

Since the list is sorted and if you know the max possible value for the second element (or if there can only be 1 element with the same first value), you could apply bisect on the list of tuples (returns the sorted insertion position in the list)

import bisect
test_array = [(1,2),(3,4),(5,6),(7,8),(9,10)]
min_value = 5

print(bisect.bisect_left(test_array,(min_value,10000)))

Hardcoding to 10000 is bad, so if you only have integers you can do that instead:

print(bisect.bisect_left(test_array,(min_value+1,)))

result: 3

if you had floats (also works with integers) you could use sys.float_info.epsilon like this:

print(bisect.bisect_left(test_array,(min_value*(1+sys.float_info.epsilon),)))

It has O(log(n)) complexity so it's much better than a simple for loop when there are a lot of elements.

Upvotes: 4

heyiamt
heyiamt

Reputation: 211

In general, numpy's where is used in a fashion similar to MATLAB's find. However, from an efficiency standpoint, I where cannot be controlled to return only the first element found. So, from a computational perspective, what you're doing here is not arguably less inefficient.

The where equivalent would be

index = numpy.where(numpy.array([t[0] for t in test_array]) >= min_value)
index = index[0] - 1

Upvotes: 0

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