列表中大于平均值的数值

Question

Python: I am having trouble figuring out how to display a list of numbers that occur above the calculated average from a user-given list and any from that list that occurs over 90. this is what I have been working on so far. The average calculation works ok alone, its the addition of these two functions I can't work out.

from math import sqrt

def getNumbers():
    nums = []    

    xStr = input("Enter a number (<Enter> to quit) >> ")
    while xStr != "":
        x = eval(xStr)
        nums.append(x)   
        xStr = input("Enter a number (<Enter> to quit) >> ")
    return nums
    
def mean(nums):
    sum = 0.0
    for num in nums:
        sum = sum + num
    return sum / len(nums)

def grt_mean(mean):
    grt_mean_ls = nums []
    for num in nums:
        if num > mean:
            grt_mean_ls = nums
            nums.append(grt_mean_ls)
            
def above90(nums):
    above_list = nums[]
    for num in nums:
        if num >90:
            above_list = nums
            nums.append(above_list)  

def display(nums):
    print(nums)
      
def main():
    abmean = grt_mean(mean)
    ab90 = above(nums)
    data = getNumbers()
    xbar = mean(data)
    display(data)
    
    print("\nThe mean is", xbar)
    print("The numbers above 90 are: ab90")
    print("the numbers above the average are:")

main()

Answer 1

You could use a list comprehension to check each value against your mean

def mean(nums):
    if len(nums) == 0:
        return 0.0
    return sum(nums) / len(nums)

def grt_mean(nums):
    m = mean(nums)
    return [i for i in nums if i > m]

For example

>>> nums = [1,2,3,4,5,6,7,8,9]
>>> mean(nums)
5.0
>>> grt_mean(nums)
[6, 7, 8, 9]

Same idea for "above 90"

def above90(nums):
    return [i for i in nums if i > 90]

>>> above90([88, 89, 90, 91, 92, 93])
[91, 92, 93]