x = [5 ,8 , 3 ,29, 445, 54]
def high():
for num in x:
if num > z:
z = num
return z
high()
print(z)
I want a function that returns the highest number from a list without using the max() inbuilt python function. Whenever I try and run this, i get this error:
line 6, in high
if num > z:
UnboundLocalError: local variable 'z' referenced before assignment
You can't use a variable without defining it first so you need to define z = x[0]
The second problem is with print(z) z is a local variable you can't call it like that you need to save whatever high() returns first in a variable then print it
x = [5 ,8 , 3 ,29, 445, 54]
def high():
z = x[0]
for num in x:
if num > z:
z = num
return z
z = high()
print(z)
z = x[0] it makes more sense to start processing the list at x[1] i.e. for num in x[1:]:
x[1:] would cause the entire list to be copied (minus the first element), which in general would be way slower than checking the first element. That optimization would only make sense if it used an index to access the list elements, avoiding the copy, but that would probably be slower as well.num in x[1:] the interpreter would have been smart enough to realise it didn't need a copy. Thanks for letting me know, I'll try to avoid that construct.You need to declare always any variable before using that variable.
x = [5 ,8 , 3 ,29, 445, 54]
def high(x):
z = 0 # add this line and..
for num in x:
if num > z:
z = num
return z
z = high(x) # ..this line
print(z)
Thanks for reaching out. First of all, z was not defined in your function. We define z and set it to an integer of 0. The code
x = [5 ,8 , 3 ,29, 445, 54]
def high(list_of_numbers):
z = 0
for num in x:
if num > z:
z = num
return z
print(high(x))
first you need to declare a varriable before using that varriable globally using global keyword.like z=x[0]
x=[5,8,3,29,445,54]
def high():
global z
z=x[0]
for num in x:
if(num>z):
z=num
return z
high()
print(z)
numis larger thanz, but what isz?