列表中的数值大于平均值

Question

Python: 我无法弄清楚如何显示用户给定列表中出现的高于计算平均值的数字列表以及该列表中出现的超过 90 的任何数字。这就是我到目前为止一直在做的事情。单独算平均还行，加上这两个函数就不行了

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

您可以使用列表理解来对照您的平均值检查每个值

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]

例如

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

“90 岁以上”的想法相同

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

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

列表中的数值大于平均值

Number values in a list greater than the average

python

parameters

function