CODEWITHSUNDEEP
python
Greg J
Greg J

Reputation: 15

Return the closest value in a list

A list that is being output is below. What I would like to do is find the closest value to a number in the second column plus a step size. How would I go about doing this? I have tried using the min function but it was not iterable.

stepsize = .5
return closest(column[1] + stepsize)

(72817, 10.009872745252624, 40.999890710175876, 9.73)
(103394, 10.044319950550072, 11.450070211613395, 8.1)
(52251, 10.047512210212679, 73.31363177934391, 9.75)
(98118, 10.558521350586966, 24.665802379879878, 9.13)
(101401, 10.595011636219427, 17.691450116195412, 8.34)
(90376, 10.718434008267023, 32.37843662097162, 9.24)
(53624, 10.78156459297739, 65.73007957579946, 8.43)
(99855, 10.960898039784297, 20.83812851138556, 9.05)
(91937, 10.999664485957146, 31.048708072900475, 9.04)
(97166, 11.049670008406684, 25.46563972962055, 9.92)

so it should return 10.558521... This list goes on for a very long time so step sizes will vary.

Upvotes: 0

Views: 122

Answers (2)

Padraic Cunningham
Padraic Cunningham

Reputation: 180540

Get just the first column using zip then sort by abs difference of each element x + .5 - the number you want it closest to:

col = (zip(*l))[1] # l is your list of tuples

print sorted(col,key=lambda x: abs(x + 0.5 - 10.6))`
[10.047512210212679, 10.044319950550072, 10.009872745252624,10.558521350586966,10.595011636219427, 10.718434008267023, 10.78156459297739, 10.960898039784297, 10.999664485957146, 11.049670008406684]

sorted(col,key=lambda x: abs(x + 0.5 - 10.6))[0] will be the closest

To put it in a function:

def closest(l,step,val):
    col = (zip(*l))[1]
    return  sorted(col,key=lambda x: abs(x + step - val))[0]

Upvotes: 4

Adam Smith
Adam Smith

Reputation: 54243

stepsize = 0.5
grid = [(72817, 10.009872745252624, 40.999890710175876, 9.73),
        (103394, 10.044319950550072, 11.450070211613395, 8.1),
        (52251, 10.047512210212679, 73.31363177934391, 9.75),
        (98118, 10.558521350586966, 24.665802379879878, 9.13),
        (101401, 10.595011636219427, 17.691450116195412, 8.34),
        (90376, 10.718434008267023, 32.37843662097162, 9.24),
        (53624, 10.78156459297739, 65.73007957579946, 8.43),
        (99855, 10.960898039784297, 20.83812851138556, 9.05),
        (91937, 10.999664485957146, 31.048708072900475, 9.04),
        (97166, 11.049670008406684, 25.46563972962055, 9.92)]

def closest(value, data):
    """Returns the closest element in data to the value passed"""
    return min(data, key=lambda x: abs(x-value))

results = [(row[1], closest(row[1]+stepsize, map(lambda x: x[1], grid))) for row in grid]
# or simplified using Padraic's zip method to grab the relevant column:
## data = (zip(*grid))[1]
## results = [(datum, closest(datum+stepsize, data)) for datum in data]

Results in:

results = [(10.009872745252624, 10.558521350586966),
           (10.044319950550072, 10.558521350586966),
           (10.047512210212679, 10.558521350586966),
           (10.558521350586966, 11.049670008406684),
           (10.595011636219427, 11.049670008406684),
           (10.718434008267023, 11.049670008406684),
           (10.78156459297739, 11.049670008406684),
           (10.960898039784297, 11.049670008406684),
           (10.999664485957146, 11.049670008406684),
           (11.049670008406684, 11.049670008406684)]

Upvotes: 0

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