Question

我需要在python中执行以下循环（以伪代码，我正在学习python）

listOfNumbers = [1,2,3]
averagesOfNumbers = [0,0,0]

for i = 1 to 2
   averagesOfNumbers [i] = (listOfNumbers [i] + listOfNumbers[i + 1]) / 2
end i

averageOfNumbers看起来像[1.5，2.5，0]。

如何在Python中做到这一点？

Answer 1

如果这正是您想要的，请查看代码：

List=[1,2,3]
avg=[0]*len(List)
for i in range(len(List)-1):
    avg[i]=(List[i]+List[i+1])/2
print(avg)

希望您得到了代码。输出： [1.5,2.5,0]

Answer 2

这是在Python中执行的操作：

listOfNumbers = [1,2,3]
averagesOfNumbers = [0,0,0]

# In Python, we start counting from 0.
# So, 2 means 0,1 and in total, that 
# is 2 numbers.
for i in range(2):
    averagesOfNumbers [i] = (listOfNumbers [i] + listOfNumbers[i + 1])/2

print(averagesOfNumbers)

Answer 3

哇，太酷了。我以前从未使用过stackoverflow。我一直想知道如何让聪明的人来帮助我！

这是给我正确作业的答案。

计算平均变化，最大增加，最大减少。

i = 0
while (i < rowCount - 1):
   diffPL.append(PL[i+1] - PL[i])
    i = i + 1
averageChange = sum(diffPL)/len(diffPL)
greatestIncrease = max(diffPL)
greatestDecrease = min(diffPL)

感谢所有提出建议的人。

Answer 4

我想您想执行以下操作：

listOfNumbers = [1,2,3]
averagesOfNumbers = [0,0,0]

for i in range(len(listOfNumbers)-1):
    averagesOfNumbers[i] = (listOfNumbers[i] + listOfNumbers[i+1]) / 2

Answer 5

list_of_numbers = [1, 2, 3] 
averages_of_numbers = [0, 0, 0]

for i in range(0, 2): # this will take indexes 0 and 1    
    averages_of_numbers[i] = (list_of_numbers[i] + list_of_numbers[i+1]) / 2

print(averages_of_numbers)

Answer 6

这里最Python化的方法是利用zip。这样可以避免使用不真正使用的索引：

>>> numbers = [1,2,3]
>>> [sum(pair)/2 for pair in zip(numbers, numbers[1:])]
[1.5, 2.5]

如果您确实需要最后加一个零，则可以手动将其添加：

numbers = [1,2,3]
avg = [sum(pair)/2 for pair in zip(numbers, numbers[1:])]
avg.append(0)

在某些情况下（例如，您不需要列表，但需要一个迭代器），甚至可以更清楚地使用map：

>>> map(lambda pair: sum(pair)/2, zip(numbers, numbers[1:]))
<map object ...>

添加列表的相邻元素

6 个答案: