CODEWITHSUNDEEP
python
skyline33
skyline33

Reputation: 573

find min values excluding zero by each element index in a list of objects

I have this function below which makes object instances as list,

lst = []

class Ppl:
    def __init__(self, pid):
        # local variables
        self.pid = pid
        self.pos = [3*pid, 10+pid-4*pid, 5*pid] 

for index in range(3):        
    lst.append(Ppl(index))

for index in range(len(lst)):
    print(lst[index].pos)

The above will output.

[0, 10, 0]
[3, 7, 5]
[6, 4, 10]

now I want to make a comprehension list based on the above lists to get the minimum values excluding zero.. so that the expected output is

[3, 4, 5]

I have this function below which works but it includes 0.

lst2 = list(map(min, *[x.pos for x in lst]))

print(lst2)
>> [0, 4, 0]

So is there a way to improve the above code or is there a better solution ?

Upvotes: 1

Views: 1746

Answers (3)

sanyassh
sanyassh

Reputation: 8500

If you are restricted to use one-liner, here is possible solution:

lst2 = list(map(lambda *args: min(arg for arg in args if arg != 0), *[x.pos for x in lst]))

min replaced with a lambda that applies min after filtering zero values.

Upvotes: 1

Shehan Ishanka
Shehan Ishanka

Reputation: 593

Try the below code snippet.

import numpy as np
lst2 = np.array([x.pos for x in lst])
lst2[lst2==0]=np.max(lst2)
print(np.min(lst2,axis=0))

Output: [3 4 5]

Upvotes: 1

Jan
Jan

Reputation: 43169

You could easily define yourself a function for this:

def min2(iterator, threshold = 0):
    minvalue = None
    for x in iterator:
        if (x > threshold) and (x < minvalue or minvalue is None):
            minvalue = x

    return minvalue

Test it with some asserts:

assert min2([0, 10, 0]) == 10
assert min2([3, 7, 5]) == 3
assert min2([6, 4, 10]) == 4
assert min2([10, 10, 101], threshold=100) == 101

Upvotes: 1

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