In A Python List Which Is Sorted, Find The Closest Value To Target Value And Its Index In The List
I am trying to get both the closest value and its index in a sorted list in python. In MATLAB this is possible with: [closest_val,index] = min(abs(array - target)) I was wondering
Solution 1:
this is an option:
lst = [
13.09409,
12.18347,
11.33447,
10.32184,
9.544922,
8.813385,
]
target = 11.5
res = min(enumerate(lst), key=lambda x: abs(target - x[1]))
# (2, 11.33447)
enumerate
iterates over your list in index, value
pairs. the key
of the min
method tells it to only consider the value
.
note that python starts indexing at 0
; matlab at 1
as far as i remember. if you want that same behavior:
res = min(enumerate(lst, start=1), key=lambda x: abs(target - x[1]))
# (3, 11.33447)
Solution 2:
bisect
wasn't used in the linked question because the list was not sorted. Here, we don't have the same problem, and we can use bisect
for the speed it provides:
import bisect
def find_closest_index(a, x):
i = bisect.bisect_left(a, x)
if i >= len(a):
i = len(a) - 1
elif i and a[i] - x > x - a[i - 1]:
i = i - 1
return (i, a[i])
find_closest_index([1, 2, 3, 7, 10, 11], 0) # => 0, 1
find_closest_index([1, 2, 3, 7, 10, 11], 7) # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 8) # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 9) # => 4, 10
find_closest_index([1, 2, 3, 7, 10, 11], 12) # => 5, 11
EDIT: In case of descending array:
def bisect_left_rev(a, x, lo=0, hi=None):
if lo < 0:
raise ValueError('lo must be non-negative')
if hi is None:
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] > x: lo = mid+1
else: hi = mid
return lo
def find_closest_index_rev(a, x):
i = bisect_left_rev(a, x)
if i >= len(a):
i = len(a) - 1
elif i and a[i] - x < x - a[i - 1]:
i = i - 1
return (i, a[i])
find_closest_index_rev([11, 10, 7, 3, 2, 1], 0) # => 5, 1
find_closest_index_rev([11, 10, 7, 3, 2, 1], 7) # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 8) # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 9) # => 1, 10
find_closest_index_rev([11, 10, 7, 3, 2, 1], 12) # => 0, 11
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