We can find the nearest value in the list by using the min() function. Define a function that calculates the difference between a value in the list and the given value and returns the absolute value of the result. Then call the min() function which returns the closest value to the given value.
To find the closest number, you would have the smallest absolute value of the difference between two numbers, so you need to take the difference of all the numbers and keep in mind which difference (and the number it corresponds to) has the smallest absolute value.
The math. isclose() method checks whether two values are close to each other, or not. Returns True if the values are close, otherwise False. This method uses a relative or absolute tolerance, to see if the values are close.
So what you have to do is you take the average and you subtract the 5 numbers from the average. Now the result may come in negative as some numbers can be greater than average. For example if 12 is average and you have 11 and 15, you can see 12-11=1 and 12-15=-3 .
If we are not sure that the list is sorted, we could use the built-in min()
function, to find the element which has the minimum distance from the specified number.
>>> min(myList, key=lambda x:abs(x-myNumber))
4
Note that it also works with dicts with int keys, like {1: "a", 2: "b"}
. This method takes O(n) time.
If the list is already sorted, or you could pay the price of sorting the array once only, use the bisection method illustrated in @Lauritz's answer which only takes O(log n) time (note however checking if a list is already sorted is O(n) and sorting is O(n log n).)
I'll rename the function take_closest
to conform with PEP8 naming conventions.
If you mean quick-to-execute as opposed to quick-to-write, min
should not be your weapon of choice, except in one very narrow use case. The min
solution needs to examine every number in the list and do a calculation for each number. Using bisect.bisect_left
instead is almost always faster.
The "almost" comes from the fact that bisect_left
requires the list to be sorted to work. Hopefully, your use case is such that you can sort the list once and then leave it alone. Even if not, as long as you don't need to sort before every time you call take_closest
, the bisect
module will likely come out on top. If you're in doubt, try both and look at the real-world difference.
from bisect import bisect_left
def take_closest(myList, myNumber):
"""
Assumes myList is sorted. Returns closest value to myNumber.
If two numbers are equally close, return the smallest number.
"""
pos = bisect_left(myList, myNumber)
if pos == 0:
return myList[0]
if pos == len(myList):
return myList[-1]
before = myList[pos - 1]
after = myList[pos]
if after - myNumber < myNumber - before:
return after
else:
return before
Bisect works by repeatedly halving a list and finding out which half myNumber
has to be in by looking at the middle value. This means it has a running time of O(log n) as opposed to the O(n) running time of the highest voted answer. If we compare the two methods and supply both with a sorted myList
, these are the results:
$ python -m timeit -s " from closest import take_closest from random import randint a = range(-1000, 1000, 10)" "take_closest(a, randint(-1100, 1100))" 100000 loops, best of 3: 2.22 usec per loop $ python -m timeit -s " from closest import with_min from random import randint a = range(-1000, 1000, 10)" "with_min(a, randint(-1100, 1100))" 10000 loops, best of 3: 43.9 usec per loop
So in this particular test, bisect
is almost 20 times faster. For longer lists, the difference will be greater.
What if we level the playing field by removing the precondition that myList
must be sorted? Let's say we sort a copy of the list every time take_closest
is called, while leaving the min
solution unaltered. Using the 200-item list in the above test, the bisect
solution is still the fastest, though only by about 30%.
This is a strange result, considering that the sorting step is O(n log(n))! The only reason min
is still losing is that the sorting is done in highly optimalized c code, while min
has to plod along calling a lambda function for every item. As myList
grows in size, the min
solution will eventually be faster. Note that we had to stack everything in its favour for the min
solution to win.
>>> takeClosest = lambda num,collection:min(collection,key=lambda x:abs(x-num))
>>> takeClosest(5,[4,1,88,44,3])
4
A lambda is a special way of writing an "anonymous" function (a function that doesn't have a name). You can assign it any name you want because a lambda is an expression.
The "long" way of writing the the above would be:
def takeClosest(num,collection):
return min(collection,key=lambda x:abs(x-num))
def closest(list, Number):
aux = []
for valor in list:
aux.append(abs(Number-valor))
return aux.index(min(aux))
This code will give you the index of the closest number of Number in the list.
The solution given by KennyTM is the best overall, but in the cases you cannot use it (like brython), this function will do the work
Iterate over the list and compare the current closest number with abs(currentNumber - myNumber)
:
def takeClosest(myList, myNumber):
closest = myList[0]
for i in range(1, len(myList)):
if abs(i - myNumber) < closest:
closest = i
return closest
It's important to note that Lauritz's suggestion idea of using bisect does not actually find the closest value in MyList to MyNumber. Instead, bisect finds the next value in order after MyNumber in MyList. So in OP's case you'd actually get the position of 44 returned instead of the position of 4.
>>> myList = [1, 3, 4, 44, 88]
>>> myNumber = 5
>>> pos = (bisect_left(myList, myNumber))
>>> myList[pos]
...
44
To get the value that's closest to 5 you could try converting the list to an array and using argmin from numpy like so.
>>> import numpy as np
>>> myNumber = 5
>>> myList = [1, 3, 4, 44, 88]
>>> myArray = np.array(myList)
>>> pos = (np.abs(myArray-myNumber)).argmin()
>>> myArray[pos]
...
4
I don't know how fast this would be though, my guess would be "not very".
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