如何执行Python循环中迭代的元素之和

Question

我对Python非常陌生，而且在理解它的逻辑工作方式方面也遇到了麻烦。我试图构建一个简单的代码，给出两个列表，找出每个列表之间的差异，并返回最小差异的平均值。

假设我有两份清单：

list1=[1, 4, 10]
list2=[2, 3, 6]

使用下面的代码，我可以用第二个列表的编号来迭代第一个列表的编号：

for x in list1:

   diff=[abs(y-x) for y in list2]
   print (diff)

[1, 2, 5] # that is, 1-2, 1-3 and 1-6
[2, 1, 2] # that is, 4-2, 4-3 and 4-6
[8, 7, 4] # that is, 10-2, 10-3, 10-6

通过下面的代码，我可以在每次迭代中找到最小的差异：

for x in list1:

   diff=[abs(y-x) for y in list2]
   mindiff=min(int(s) for s in diff)

   print (mindiff)

1 
1 
4 

我也没意见。现在，我想把所有的最小差加起来，除以我计算出来的差的数目。这是我不明白的部分。换句话说，我如何构建一个函数来总结for循环所做的所有迭代？在此之后，(在本例中为6 (1+1+4))，我可以轻松地将list2的最大范围的差异之和除以。

我该怎么做呢？

提前谢谢你的回答

Answer 1

下面的代码可能会帮助您解决问题。

list1=[1, 4, 10]
list2=[2, 3, 6]

n = len(list2)
sum_min = 0

for x in list1:

    diff=[abs(y-x) for y in list2]
    mindiff=min(int(s) for s in diff)

    sum_min += mindiff  # iterating will keep on adding min of differences 1 + 1 + 4

    print (mindiff)

print(sum_min/n)  # mean minimum difference

我认为，为了计算列表中最小差异的平均值，您需要将最小差异的总和除以每组中的差异数(即list2长度)。如果你想修改代码，我想你已经知道了怎么做。

Answer 2

list1=[1, 4, 10]
list2=[2, 3, 6]

sumOfMinDiffs = 0 # Initialize the sum value to start out at zero

for x in list1:

   diff=[abs(y-x) for y in list2]
   mindiff=min(int(s) for s in diff)
   print (mindiff)

   sumOfMinDiffs += mindiff # Keep track of the sum


maximumRangeOfList2 = max(list2) - min(list2) # I think this is what you mean by "maximum range of list 2"

print(sumOfMinDiffs/maximumRangeOfList2) # Tah dah, this is your answer
# print(sumOfMinDiffs/len(list2)) # This is your answer if you just want to divide by the length of list2

上述代码用于执行以下操作：

• 在循环之外保留一个和变量，并在每个循环上增加它。
• 在for循环的末尾，计算“列表2的最大范围”，并将其存储在一个变量中。
• 对于您的最终答案，只需将和变量除以范围变量即可。

Answer 3

我认为您可以在for循环之外使用一个变量来实现您想要的结果，将每个迭代的值和一个计算迭代次数的变量相加。

list1=[1, 4, 10]
list2=[2, 3, 6]


min_sum = 0
n = 0
for x in list1:

    diff=[abs(y-x) for y in list2]
    mindiff=min(int(s) for s in diff)
    min_sum += mindiff
    n+=1

    print (mindiff)

print(min_sum/n)

希望它能帮上忙