I have a specific scenario that I need to scan a specific portion of an array for a maximum value of that portion and return the position of that value with regards to the entire array. for example
searchArray = [10,20,30,40,50,60,100,80,90,110]
I want to scan for the max value in portion 3 to 8, (40,50,60,100,80,90)
and then return the location of that value.
so in this case max value is 100 and location is 6
is there a way to get that using python alone or with help oy numpy
First slice your list and then use index on the max function:
searchArray = [10,20,30,40,50,60,100,80,90,110]
slicedArray = searchArray[3:9]
print slicedArray.index(max(slicedArray))+3
This returns the index of the sliced array, plus the added beginSlice
max which iterates through the whole list to find the highest value and then calling index which iterates through half the list again to find where that element was is redundant and should be avoided if performance matters. Check out my answer using an enumerate generator to get the indexes before searching the max value and avoid an unnecessary second iteration over the list.Try this...Assuming you want the index of the max in the whole list -
import numpy as np
searchArray = [10,20,30,40,50,60,100,80,90,110]
start_index = 3
end_index = 8
print (np.argmax(searchArray[start_index:end_index+1]) + start_index)
Use enumerate to get an enumerated list of tuples (actually it's a generator, which means that it always only needs memory for one single entry and not for the whole list) holding the indexes and values, then use max with a custom comparator function to find the greatest value:
searchArray = [10,20,30,40,50,60,100,80,90,110]
lower_bound = 3 # the lower bound is inclusive, i.e. element 3 is the first checked one
upper_bound = 9 # the upper bound is exclusive, i.e. element 8 (9-1) is the last checked one
max_index, max_value = max(enumerate(searchArray[lower_bound:upper_bound], lower_bound),
key=lambda x: x[1])
print max_index, max_value
# output: 6 100
[lower_bound:upper_bound]) - it just avoids searching the list again for the item returned by max just to find out its index. Instead I generate tuples of index and value on the fly using enumerate and let max process them. That way it only performs one iteration and directly returns both the index and the value.
max and greatest value.
enumerate(..., lower_bound) as well as other subtle issues... best answerI'd do it like this:
sliced = searchArray[3:9]
m = max(sliced)
pos = sliced.index(m) + 3
I've added an offset of 3 to the position to give you the true index in the unmodified list.
With itemgetter:
pos = max(enumerate(searcharray[3:9], 3), key=itemgetter(1))[0]
i guess this what you want
maxVal = max(searchArray[3:8]) // to get max element
position = searchArray.index(max(ary[3:8])) //to get the position of the index
