麻烦迭代数组以求3个数组的平均，并与其他3个数组进行比较

Question

import numpy as np
import math
array = [1,2,3,4,5,6,7,8,9,10,11,12]
for k in range(len(array)):
    target1 = array[k]
    target2 = array [k-1]
    if k<11:
        target3 = array[k+1]
    else:
        target3 = array[0]

    targets = [target1, target2, target3]
    target = np.mean(targets)

    sigma = 1
    if ((target-avg)/(math.sqrt(target))>= sigma:
        print(array[k])

    if k < 8:
            range1_target1 = array[k+2]
            range1_target2 = array[k+3]
            range1_target3 = array[k+4]

Answer 1

首先，定义平均函数，一个计算可迭代对象中值的平均值的函数：

def average(iterable):
    sum = 0
    for number in iterable:
        sum += number
    return sum / len(iterable)

这个函数非常简单。它定义了一个名为“sum”的变量，并将其初始值设置为0。然后循环遍历参数“iterable”，将iterable参数中每个元素的值添加到sum。然后，它将所有值的和（sum变量）除以Iterable的长度。它将结果作为浮点返回。

一个例子：函数被调用。iterable参数被分配给列表[25,75,-100,20]。sum变量被初始化为0。通过循环迭代，它将25、75、-100和20加到sum。加起来是20。按照Python的len（）方法，iterable的长度是4。20/4=5.0（一个浮点，因为Python中的一个斜杠总是表示真除法，并且返回一个浮点，即使在转换为int后结果保持不变。使用两个斜杠进行整数除法），因此方法返回5.0。

averages = []

很简单。将全局变量“Average”设置为空列表。

因为您需要每三个数字的平均值，所以应该将迭代范围设置为步骤3，而不是步骤1。这是这样做的。

for counter in range(0, len(array), 3):

范围构造函数最多接受三个参数(start、stop、step)。当它接受三个参数时，它从“start”开始计数，从“step”开始计数，从“stop”开始计数，从“stop”停止。它记录在一个不可变的可迭代对象中计数的每个数字。当它接受两个参数时，“step”被设置为1。当它接受一个参数时，“start”被设置为0，“stop”被设置为该参数，“step”被设置为1。

    averages.append(average(array[counter: counter+3: 1]))

你现在得到了所有的平均数。不需要担心IndexError,因为切片忽略任何超出范围的索引,而不会引发异常。

接下来，我们想把平均值相互比较一下。代码如下：

for i in range(0, len(averages)):
    for j in range(0, len(averages)):
        if i == j:
            #Avoid duplicate averages.
            continue
        if averages[i] < averages[j]:
            #averages[i] < averages[j]. 
            #Do something
            continue
        elif averages[i] == averages[j]:
            #averages[i] == averages[j]. 
            #Do something
            continue
        else:
            #averages[i] > averages[j]. Only possibility left. 
            #Do something
            continue

就是这么做的。我们在平均数上迭代索引，而不是平均数本身，以测试是否重复。如果索引彼此相等，那么我们继续循环，因为我们不想处理它们。否则，我们将Averages[i]与Averages[j]进行比较，并根据条件执行python语句。你没有说明你想做什么，所以我只是处理他们像重复，尽管所有你不应该。代码的其余部分是不言自明的。

现在，完整的代码。无注释：

def average(iterable):
    sum = 0
    for number in iterable:
        sum += number
    return sum / len(iterable)

#Same old array that you mentioned in your question
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

averages = []

for counter in range(0, len(array), 3):
    averages.append(average(array[counter: counter+3: 1]))

for i in range(0, len(averages)):
    for j in range(0, len(averages)):
        if i == j:
            #Avoid duplicate averages.
            continue
        if averages[i] < averages[j]:
            #averages[i] < averages[j]. 
            #Do something
            continue
        elif averages[i] == averages[j]:
            #averages[i] == averages[j]. 
            #Do something
            continue
        else:
            #averages[i] > averages[j]. Only possibility left. 
            #Do something
            continue

def average(iterable):
    sum = 0
    for number in iterable:
        sum += number
    return sum / len(iterable)

#Same old array that you mentioned in your question
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

averages = []

for counter in range(0, len(array), 3):
    averages.append(average(array[counter: counter+3: 1]))

for i in range(0, len(averages)):
    for j in range(0, len(averages)):
        if i == j:
            #Avoid duplicate averages.
            continue
        if averages[i] < averages[j]:
            #averages[i] < averages[j]. 
            #Do something
            print("averages[{}] < averages[{}]".format(i, j))
        elif averages[i] == averages[j]:
            #averages[i] == averages[j]. 
            #Do something
            print("averages[{}] = averages[{}]".format(i, j))
        else:
            #averages[i] > averages[j]. Only possibility left. 
            #Do something
            print("averages[{}] > averages[{}]".format(i, j))

print("Averages: ", averages)

averages[0] < averages[1]
averages[0] < averages[2]
averages[0] < averages[3]
averages[1] > averages[0]
averages[1] < averages[2]
averages[1] < averages[3]
averages[2] > averages[0]
averages[2] > averages[1]
averages[2] < averages[3]
averages[3] > averages[0]
averages[3] > averages[1]
averages[3] > averages[2]
Averages:  [2.0, 5.0, 8.0, 11.0]