I can find max value, I can find average but I just can't seem to find the min. I know there is a way to find max and min in a loop but right now I can only find the max.
def large(s)
sum=0
n=0
for number in s:
if number>n:
n=number
return n
Is there a way to find the min value using this function?
You can use Python's built-in sum(), min(), and max() functions for this kind of analysis.
However, if you're wanting to do it all in one pass or just want to learn how to write it yourself, then the process is 1) iterate over the input and 2) keep track of the cumulative sum, the minimum value seen so far, and the maximum value seen so far:
def stats(iterable):
'''Return a tuple of the minimum, average, and maximum values
>>> stats([20, 50, 30, 40])
(20, 35.0, 50)
'''
it = iter(iterable)
first = next(it) # Raises an exception if the input is empty
minimum = maximum = cumsum = first
n = 1
for x in it:
n += 1
cumsum += x
if x < minimum:
minimum = x
if x > maximum:
maximum = x
average = cumsum / float(n)
return minimum, average, maximum
if __name__ == '__main__':
import doctest
print doctest.testmod()
The code has one other nuance. It uses the first value from the input iterable as the starting value for the minimum, maximum, and cumulative sum. This is preferred over creating a positive or negative infinity value as initial values for the maximum and minimum. FWIW, Python's own builtin functions are written this way.
stats throws an exception if the list is empty. Although that doesn't seem to be a problem for Python programmers (that's the "pythonic" way), it still bothers me a bit. Personally, I prefer to handle special cases and return a meaningful result, since passing an empty list is not an exceptional condition per se. A matter of style.
first = next(it, default_value).Finding the minimum takes the same algorithm as finding the maximum, but with the comparison reversed. < becomes > and vice versa. Initialize the minimum to the largest value possible, which is float("inf"), or to the first element of the list.
FYI, Python has a builtin min function for this purpose.
You must set n to a very high number (higher than any of the expected) or to take one from the list to start comparison:
def large(s)
n = s.pop()
for number in s:
if number < n:
n = number
return n
Obviously you have already max and min for this purpose.
A straightforward solution:
def minimum(lst):
n = float('+inf')
for num in lst:
if num < n:
n = num
return n
Explanation: first, you initialize n (the minimum number) to a very large value, in such a way that any other number will be smaller than it - for example, the infinite value. It's an initialization trick, in case the list is empty, it will return infinite, meaning with that that the list was empty and it didn't contain a minimum value.
After that, we iterate over all the values in the list, checking each one to see if it is smaller than the value we assumed to be the minimum. If a new minimum is found, we update the value of n.
At the end, we return the minimum value found.
n in a special value (infinity in my case) is to take into account the special case when the list is empty. Your code throws an exception for an empty list, mine returns a special value - it's a matter of styleWhy not just replace large with small and > with <? Also, you might not want to initialize n to 0 if you're looking for the smallest value. Your large function only works for lists of positive numbers. Also, you're missing a ":" after your def line.
def small(s):
if len(s)==0: return None
n = s[0]
for number in s[1:]:
if n < number:
n=number
return n
This handles empty lists by returning None.
Using this function to find minimum is
min=-large(-s)
The logic is just to find maximum of the negative list , which gives the minimum value
You can use same function with iteration just instead of n=0 use n=L[0]
def min(L): n=L[0] for x in L: if x
def min(s):
n=s[0]
for number in s:
if number < n:
n=number
return n
