How to find the *smallest* number in the list that is close to a given number in Python?-CodePudding

I wanted to know if there is a way to find the smallest number in the list which is closest to a given number.

Foe e.x.

my_list=[1,10,15,20,45,56]
my_no=9

Output should be equal to 1 Explanation: since 9 comes between 1 and 10 and I want the smaller number, i.e. 1.

Similarly if my_no=3, output is equal to 1

I am new to python, so any help will be much appreciated. Thanks!

CodePudding user response：

I've used bisect to do this:

from bisect import bisect_right
my_list=[1,10,15,20,45,56]
my_no=9
targ=bisect_right(my_list, my_number)
#edge case where you give a number lesser than the smallest number in the list. e.g. my_no=0 in your case
if(targ): 
    print(my_list[targ-1])

Output=1

CodePudding user response：

You can reduce from functools:

from functools import reduce

i = reduce(lambda l, r: l if l <= my_no <= r else r, my_list)
print(i)

# Output
1

Test

# my_no = 18
>>> reduce(lambda l, r: l if l <= my_no <= r else r, my_list)
15

# my_no = 59
>>> reduce(lambda l, r: l if l <= my_no <= r else r, my_list)
56

Note: this code doesn't work if my_no < my_list[0]. You have to check it before.

CodePudding user response：

If you want to stick to basics and want to approach it using for loop and if statement then it can be achieved as follows:

my_list=[1,10,15,20,45,56]
given_number = 9

output=my_list[0]
for x in my_list:
  if(x < given_number and x > output):
    output= x

print(output)

# Output
1

CodePudding user response：

List comprehension is another elegant way to do this.

my_list=[1,10,15,20,45,56]
my_no = 9 
output = max([x for x in my_list if x<=my_no])
print(output) #1