how does searchsort in python work?
Question:
To make my question clear
say if I have an array a as
Out[123]: [1, 3, 4, 6, 9, 10, 54]
When I try to search the numbers in the list, searchsort returns correct value but when I try something not in the list, it returns an absurd value
here is some of the results
In [131]: a
Out[131]: [1, 3, 4, 6, 9, 10, 54]
In [132]: searchsorted(a,1)
Out[132]: 0
In [133]: searchsorted(a,6)
Out[133]: 3
In [134]: searchsorted(a,[9,54,1])
Out[134]: array([4, 6, 0])
In [135]: searchsorted(a,[9,54,1,0])
Out[135]: array([4, 6, 0, 0])
***> # here 0 is not in the list, but turns up @ position 0***
In [136]: searchsorted(a,740)
Out[136]: 7
***> # here 0 is not in the list, but turns up @ position 7***
why is this happening?
Answers:
from the docs it states that it uses binary search to spot insertion point of an item in a sorted list.
the word ‘insertion point’ means, if item I is inserted to the insertion point index N in sorted array A, the array A will remain sorted with new item I.
your examples like [9, 54, 1] is meaningless since the array is not sorted.
you can use bisect module in python to do the same thing, without numpy.
searchsorted tells you where the element belongs to guarantee ordering:
Find the indices into a sorted array a such that, if the corresponding elements in v were inserted before the indices, the order of a would be preserved.
inserting 740 at position 7 would preserve ordering, as would inserting 0 at position 0.
searchsorted doesn’t tell you where things are, it tells you where things should go to keep the list sorted.
So 0 would have to be inserted at position 0, before the 1. Similarly, 740 needs to be inserted at position 7, beyond the current end of the list.
You can see this by reading the docs here:
numpy.searchsorted(a, v, side=’left’, sorter=None)
Find indices where elements should be inserted to maintain order.
Find the indices into a sorted array a such that, if the corresponding elements in v were inserted before the indices, the order of a would be preserved.
searchsorted(initial_list,insert_list,side)
default: side = ‘left’
For example: searchsorted(x,v)
x = [1,2,3,4,5]
v = [-10,10,2,3]
This is my result of my example
🙂
To make my question clear
say if I have an array a as
Out[123]: [1, 3, 4, 6, 9, 10, 54]
When I try to search the numbers in the list, searchsort returns correct value but when I try something not in the list, it returns an absurd value
here is some of the results
In [131]: a
Out[131]: [1, 3, 4, 6, 9, 10, 54]
In [132]: searchsorted(a,1)
Out[132]: 0
In [133]: searchsorted(a,6)
Out[133]: 3
In [134]: searchsorted(a,[9,54,1])
Out[134]: array([4, 6, 0])
In [135]: searchsorted(a,[9,54,1,0])
Out[135]: array([4, 6, 0, 0])
***> # here 0 is not in the list, but turns up @ position 0***
In [136]: searchsorted(a,740)
Out[136]: 7
***> # here 0 is not in the list, but turns up @ position 7***
why is this happening?
from the docs it states that it uses binary search to spot insertion point of an item in a sorted list.
the word ‘insertion point’ means, if item I is inserted to the insertion point index N in sorted array A, the array A will remain sorted with new item I.
your examples like [9, 54, 1] is meaningless since the array is not sorted.
you can use bisect module in python to do the same thing, without numpy.
searchsorted tells you where the element belongs to guarantee ordering:
Find the indices into a sorted array a such that, if the corresponding elements in v were inserted before the indices, the order of a would be preserved.
inserting 740 at position 7 would preserve ordering, as would inserting 0 at position 0.
searchsorted doesn’t tell you where things are, it tells you where things should go to keep the list sorted.
So 0 would have to be inserted at position 0, before the 1. Similarly, 740 needs to be inserted at position 7, beyond the current end of the list.
You can see this by reading the docs here:
numpy.searchsorted(a, v, side=’left’, sorter=None)
Find indices where elements should be inserted to maintain order.
Find the indices into a sorted array a such that, if the corresponding elements in v were inserted before the indices, the order of a would be preserved.
searchsorted(initial_list,insert_list,side)
default: side = ‘left’
For example: searchsorted(x,v)
x = [1,2,3,4,5]
v = [-10,10,2,3]
This is my result of my example
🙂