python - 用于标准偏差的Python代码以及来自SQLITE3的数据

from math import *
import sqlite3
ages=sqlite3.connect('person.sqlite3')

def main():
    ageslist=ages.execute("SELECT age from person")
#average age
    for row in ageslist:
        row[0]
    average = (sum(row[0]))/len(row[0])
#subtracts average x from x or opposite and square, depending on n
    for n in range(len(ageslist) - 1):
        if numbers[n] > average:
            numbers.append((ageslist[n] - average)**2)
        if numbers[n] < average:
            numbers.append((average - ageslist[n])**2)
#takes square rt of the sum of all these numbers and divides by n-1
    Stdv = math.sqrt(sum(ageslist))/(len(ageslist)-1)
    end=time()
    print(Stdv)

main()

我正在尝试从SQLITE3数据库中找到年龄的标准偏差。但是，我得到当前错误：
 平均=（sum（row [0]））/ len（row [0]）
TypeError：“ int”对象不可迭代

我该如何纠正？

最佳答案

发送到数据库连接的查询返回一个迭代器。在从内存中刷新该迭代器之前，您只能传递一次。这是对代码的一些更正，以执行您要执行的操作。

conn = sqlite3.connect('person.sqlite3')

def main():
    ages_iterator = conn.execute("SELECT age from person")
    # this turns the iterator into an actual list, which you need for stdev
    age_list = [a[0] for a in ages_iterator]

    # average age
    average = (sum(age_list))/len(age_list)

    # subtracts average x from x square
    # because you are squaring the difference, the it does not matter if it is
    # greater or less than the average
    numbers = [(age-average)**2 for age in age_list]

    #takes square rt of the sum of all these numbers and divides by n-1
    Stdv = math.sqrt(sum(numbers))/float(len(numbers)-1)
    end=time()
    print(Stdv)

main()

关于python - 用于标准偏差的Python代码以及来自SQLITE3的数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/40600077/