Python: how to find value in list smaller than target

Question

For example I have a non-ordered list of values [10, 20, 50, 200, 100, 300, 250, 150]

I have this code which returns the next greater value:

def GetN         


        

Answer 1

nextHighest = lambda seq,x: min([(i-x,i) for i in seq if x<=i] or [(0,None)])[1]
nextLowest  = lambda seq,x: min([(x-i,i) for i in seq if x>=i] or [(0,None)])[1]

Here's how this works: Looking at nextHighest, the argument to min is a list comprehension, that calculates the differences between each value in the list and the input x, but only for those values >= x. Since you want the actual value, then we need the list elements to include both the difference to the value, and the actual value. Tuples are compared value by value, left-to-right, so the tuple for each value i in the sequence becomes (i-x,i) - the min tuple will have the actual value in the [1]'th element.

If the input x value is outside the range of values in seq (or if seq is just empty), then the list comprehension will give us an empty list, which will raise a ValueError in min. In case this happens, we add the or [(0,None)] term inside the argument to min. If the list comprehension is empty, it will evaluate to False, in which case min will instead look at the sequence containing the single tuple (0,None). In the case, the [1]'th element is None, indicating that there were no elements in seq higher than x.

Here are some test cases:

>>> t = [10, 20, 50, 200, 100, 300, 250, 150]
>>> print nextHighest(t,55)
100
>>> print nextLowest(t,55)
50
>>> print nextHighest([],55)
None
>>> print nextLowest([],55)
None
>>> print nextHighest(t,550)
None

Answer 2

Let the unordered list be myList:

answer = max(x for x in myList if x < temp)

Answer 3

a=[4,3,8,2,5]
temp=4
def getSmaller(temp,alist):
    alist.sort()
    for i in range(len(alist)):
        if(i>0 and alist[i]==temp):
            print alist[i-1]
        elif(i==0 and alist[i]==temp):
            print alist[i]
getSmaller(temp,a)

Answer 4

Edit: Ah, I used templist instead of list -- hence the confusion. I didn't mean it to be a one-line function; you still have to do the conversions. (Of course, as Mike DeSimone rightly points out, using list as a variable name is a terrible idea!! So I had a good reason for being confusing. :)

To be more explicit about it, here's a slightly streamlined version of the function (fixed to test properly for an empty list):

def GetNextHighTemp(self, temp, templist):
    templist = (int(t) for t in templist if t != '')
    templist = [t for t in templist if t < int(temp)]
    if templist: return max(templist)
    else: return None                   # or raise an error

Thanks to Mike for the suggestion to return None in case of an empty list -- I like that.

You could shorten this even more like so:

def GetNextHighTemp(self, temp, templist):
    try: return str(max(int(t) for t in templist if t != '' and int(t) < int(temp)))
    except ValueError: return None      # or raise a different error

Answer 5

A better and much faster (code and cpu wise) way is to use bisect module which does binary search but for that you will need to sort the list first, here is the sample usage:

import bisect

mylist = [10, 20, 50, 200, 100, 300, 250, 150]
mylist.sort()

index = bisect.bisect(mylist, 55)
print "Greater than target", mylist[index]
print "Smaller than or equal to target", mylist[index-1]

output:

Greater than target 100
Smaller than or equal to target 50

Also you will need to check the returned index, if it is 0 it means you have passed target lower than the lowest

Answer 6

If I understand you correctly, you want the greatest value that is less than your target; e.g. in your example, if your target is 55, you want 50, but if your target is 35, you want 20. The following function should do that:

def get_closest_less(lst, target):
    lst.sort()
    ret_val = None
    previous = lst[0]
    if (previous <= target):
        for ndx in xrange(1, len(lst) - 1):
            if lst[ndx] > target:
                ret_val = previous
                break
            else:
                previous = lst[ndx]
    return str(ret_val)

If you need to step through these values, you could use a generator to get the values in succession:

def next_lesser(l, target):
    for n in l:
        if n < target:
            yield str(n)

Both these worked properly from within a simple program.