在Python列表中找到最小的數字

Question

我正在嘗試使用for循環來獲取python中每周和每月的平均值：

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

就是這樣，但這是不正確的。 我可以幫忙嗎

Answer 1

假設您使用的是Python 2.x，則兩個整數的/運算符使用整數除法，即除法的結果四舍五入為下一個整數。 在交互式解釋器中嘗試：

>>> 5/3
1

要獲得正確的浮點除法，可以使用

from __future__ import division

或將其中一個操作數轉換為float

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

Answer 2

您不需要這些循環。 這是一個簡單的示例：

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

如評論中所示：

上面的示例（在Python 2.x中）如果需要“真”除法（例如34.71），則需要將一部分強制轉換為浮點數。

在Python 3.x中單/除法默認為“真”分裂所以上面的片段將是正確的（盡管具有用於不同結果值average1 ）。

Answer 3

這里有幾個問題。 首先， for x in lst產生元素lst ，而不是索引。 其次，將元素添加兩次，一次是在更新sum ，一次是在更新avg 。 只需計算列表之外的avg 。 第三，用float代替int來防止截斷：

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
>>>

---

正如其他人指出的那樣，如果您想要不被截斷的結果，請使用from __future__ import division