I am brand new to programming. Is there any way to simplify the conditional operators
a = 50; b = 70; c = 60;
# Classification
if (a == b and b == c and c ==a):
print('Equilateral triangle')
elif (a == b or b == c or c == a):
print('Isosceles triangle')
elif (a!=b and b!=c and c!=a):
print('Scalene triangle')
You can shorten the comparisons that use and by using chained comparisons, and by dropping one of the tests (they are mutually exclusive), using else instead:
if a == b == c:
print('Equilateral triangle')
elif a != b != c != a:
print('Scalene triangle')
else:
print('Isosceles triangle')
Note that Python's if syntax doesn't require any parentheses around the test expressions.
Next, you could look at these values as a set, and test how many elements are in the set:
unique_lengths = len({a, b, c})
if unique_lengths == 1:
print('Equilateral triangle')
elif unique_lengths == 2:
print('Isosceles triangle')
else:
print('Scalene triangle')
This can then be turned into list lookup, mapping 1, 2 and 3 to triangle class names; I slotted None into the 0 position:
classes = [None, 'Equilateral', 'Isosceles', 'Scalene']
print(classes[len({a, b, c})], 'triangle')
a, b, c into a set when there are shared values among those 3 variables means you'll have a set with fewer than 3 values in it, and we can exploit that to count how many sides are equal.Using a list:
lis = [a,b,c]
lis.sort()
if lis[0] == lis[-1]:
print("Equilateral triangle")
elif lis[0] == lis[1] or lis[-1] == lis[1]:
print("Isosceles triangle")
else:
print("Scalene triangle")
if a == b == c