squares = [i*i for i in range(1, 100)] print(squares)

``` list=[] for i in range(1,101): list.append(i**2) print(list) ``` ``` list=[] for i in range(1,101): list.append(i**2) sum=0 for j in list: sum+=j print(sum) ``` 冒泡算法倒序排列 ``` [9, 8, 7, 6, 5, 4, 3, 2, 1] ```

2019-01-01 00:00:00 - INFO - [1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 5041, 5184, 5329, 5476, 5625, 5776, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 8281, 8464, 8649, 8836, 9025, 9216, 9409, 9604, 9801, 10000] 2019-01-01 00:00:00 - INFO - 338350

计算平方序列以及序列之和： ``` def func1(start, end): list = [] for i in range(start, end): k = i**2 list.append(k) return(list) def func2(start, end): sum = 0 for value in func1(start, end): sum += value print(sum) return(sum) if __name__ == '__main__': func2(1, 101) ``` 输出为338350

@chevishere01 最后一个答案是987654321，一道排序题

def initialize(context): run_daily(period,time='every_bar') # 把两个股票代码作为list存入g.security中 g.security = ['000001.XSHE','000002.XSHE'] def period(context): for stk in g.security: order(stk,100) # 获得股票持仓成本 cost=context.portfolio.positions[stk].avg_cost # 获得股票现价 price=context.portfolio.positions[stk].price # 计算收益率 ret=price/cost-1 # 如果收益率小于-0.01，即亏损达到1%则卖出股票，幅度可以自己调，一般10% if ret< -0.01: order_target(stk,0) print('触发止损') #用for语句生成一个100项的平方数数列的list，即[1,4,9,16,...,10000] list = [] for i in range(1,101,1): list.append(i**2) print(list) #用for语句计算100项的平方数数列的和，即1+4+9+16+...+10000的结果 sum = 0 for i in range(1,101,1): sum += i**2 print(sum) sum2=0 for i in list: sum2 += i print(sum2) 2019-01-01 00:00:00 - INFO - [1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 5041, 5184, 5329, 5476, 5625, 5776, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 8281, 8464, 8649, 8836, 9025, 9216, 9409, 9604, 9801, 10000] 2019-01-01 00:00:00 - INFO - 338350 2019-01-01 00:00:00 - INFO - 338350 2019-01-01 00:00:00 - INFO - [9, 8, 7, 6, 5, 4, 3, 2, 1]

打卡学习

if ret< -0.01: order_target(stk,0) print('触发止损')

``` code ```if ret< -0.01: order_target(stk,0) print('触发止损')

我有一个问题，买入多只股票后，右侧的窗口显示的还是第一只股票的每日股价，如何预览多只股票呢

生成平方数 def get_data(L): original = range(1,101,1) for i in original: L.append(i*i)

想问大家，买入多股票之后，止损的股票是多个还是一个？

squares = [j**2 for j in range(1,101)] print(squares)

squares = [j**2 for j in range(1,101)] print(squares) sum = 0 for i in squares: sum += i print(sum)

相关内容已经和现在环境不一致了，当前环境下示例代码输出如下： for j in range(1,7,2): print(j) print(range(1,7,2)) print(type(range(1,7,2))) 输出： 1 3 5 range(1, 7, 2) < class 'range'>

按照例题中的写法，止损只针对击穿止损线的单只股票，如果两只都击穿了那两只各止各的。 因为在for循环的调用中，“cost = context.portfolio.positions[stk].avg_cost”等等一系列的数据读取和操作，只针对当前循环到的个股。

2025-11-9

打卡

用for语句生成一个100项的平方数数列的list，即[1,4,9,16,...,10000] a=[] for i in range(1,101): a.append(i ** 2) print(a)