I can use functools.reduce or min / max to get min and max of members in a list. But to get both in one pass I need to write a loop:
from functools import reduce
class foo:
def __init__(self,value): self.value = value
x = []
x.append(foo(1))
x.append(foo(2))
x.append(foo(3))
min_value = reduce(lambda a,b: a if a.value < b.value else b,x).value
max_value = reduce(lambda a,b: a if a.value > b.value else b,x).value
print(min_value)
print(max_value)
min_value2 = min(x,key=lambda a: a.value).value
max_value2 = max(x,key=lambda a: a.value).value
print(min_value2)
print(max_value2)
min_value3 = x[0].value
max_value3 = x[0].value
for f in x:
if f.value < min_value3: min_value3 = f.value
if f.value > max_value3: max_value3 = f.value
print(min_value3)
print(max_value3)
Is it possible to get min and max in one pass without writing a plain loop?
>Solution :
You could use a tuple as your aggregator. Something like this maybe?
min_value, max_value = reduce(lambda a, b:
(a[0] if a[0].value < b.value else b, a[1] if a[1].value > b.value else b),
x,
(x[0], x[1]))
The output should be a tuple where the first is the minimum and the second the maximum.
Example in the REPL, demonstrating that the objects requested are returned, and that the values are correct:
>>> class foo:
... def __init__(self,value): self.value = value
...
>>> ls = [foo(1), foo(2), foo(3)]
>>> min_value, max_value = reduce(lambda a, b: (a[0] if a[0].value < b.value else b, a[1] if a[1].value > b.value else b), ls, (ls[0], ls[1]))
>>> min_value
<__main__.foo object at 0x10bd20940>
>>> min_value.value
1
>>> max_value.value
3