Question

我正在尝试使用for循环来获取每周和python中的月份的平均值：

sum = 0
sumA = 0
sumB = 0
sumC = 0
sumD = 0

week1 = (35,38,30,34,27,40,39)
week2 = (35,38,30,34,27,40,39)
week3 = (35,38,30,34,27,40,39)
week4 = (35,38,30,34,27,40,39)

for x in (week1):
    sum = sum + week1[x]
    avg1 = (sum + week1[x]) / 7
for y in (week2):
    sumA = sumA + week2[y]
    avg2 = (sumA + week2[y]) / 7
for z in (week3):
    sumB = sumB + week3[z]
    avg3 = (sumB + week3[z]) / 7
for k in (week4):
    sumC = sumC + week4[k]
    avg4 = (sumC + week4[k]) / 7

sumD = sum + sumA + sumB + sumC
avg = (sum + sumA + sumB + sumC) / 28

就是这样但不正确。我能得到一些帮助吗

Answer 1

假设您使用的是Python 2.x，两个整数的/运算符使用整数divsion，i。即除法的结果向下舍入到下一个整数。在交互式解释器中尝试：

>>> 5/3
1

要获得正确的浮点除法，请使用

from __future__ import division

或首先将其中一个操作数转换为float

avg = float(sum + sumA + sumB + sumC) / 28

Answer 2

您不需要这些循环。这是一个简单的例子：

>>> week1 = (35,38,30,34,27,40,39)
>>> average1 = sum(week1) / len(week1)
>>> average1
34

如评论中所述：

上面的例子（在Python 2.x中）如果你想要“真正的”除法（例如34.71），则需要一个部分转换为浮动。

在Python 3.x中，单/除法默认为'true'除法，因此上面的代码段是正确的（尽管average1的结果值不同）

Answer 3

这里有几个问题。首先，for x in lst产生lst的元素，而不是索引。其次，您在元素中添加两次，一次更新sum，然后再次更新avg。只需在列表外计算avg即可。第三，除以float而不是int以防止截断：

for x in (week1):
    sum = sum + x
avg1 = sum / 7.

Answer 4

list = [....]
avg = sum(list)/float(len(list))

Answer 5


>>> def ave(numbers):
...  return sum(numbers) / len(numbers)
... 
>>> week1 = (35,38,30,34,27,40,39)
>>> week2 = (35,38,30,34,27,40,39)
>>> week3 = (35,38,30,34,27,40,39)
>>> week4 = (35,38,30,34,27,40,39)
>>> ave(week1)
34
>>> ave(week1+week2+week3+week4)
34
>>>

<小时/> 正如其他人所指出的，如果你想要一个非截断的结果

，请使用from __future__ import division

查找python列表中的最小数字

5 个答案: