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Jan's questions

Martin Hope
Jan
Asked: 2024-04-22 22:22:22 +0800 CST
  • 5

假设我们还有另一个包含财务数据的数据框：

timestamp,close,security_code,volume,bid_volume,ask_volume
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.383985+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.384040+00:00,4968.5,ES,1,1,0
2024-02-05 01:00:01.385840+00:00,4968.5,ES,2,0,2
2024-02-05 01:00:01.385840+00:00,4968.5,ES,1,0,1
2024-02-05 01:00:01.385840+00:00,4968.5,ES,1,0,1
2024-02-05 01:00:01.385840+00:00,4968.5,ES,1,0,1
2024-02-05 01:00:01.385840+00:00,4968.5,ES,2,0,2

人们可以像这样计算 POC（控制点，交易最多合约的地方）

def poc(self, df):
    """ Calculate the POC. """
    return df['close'].expanding().agg({'poc': lambda s:mode(s)[0]})['poc']

然而，这没有考虑体积。如果该特定水平上的交易量较大，则应考虑到这一点。如何mode相应地改变/lambda？

python
Martin Hope
Jan
Asked: 2024-03-16 03:46:28 +0800 CST
  • 4

假设，我有

class State(Enum):
    TAKEPROFIT  = 1
    STOPPEDOUT  = 2
    WINNER      = 3
    LOSER       = 4

如何返回egState.STOPPEDOUT和的组合State.LOSER？
似乎|不支持：

return State.STOPPEDOUT | State.LOSER

投掷

TypeError: unsupported operand type(s) for |: 'State' and 'State'
python
Martin Hope
Jan
Asked: 2023-11-20 22:03:28 +0800 CST
  • 5

我有以下 lisp 代码

(defun sum (vec)
  "Summiert alle Elemente eines Vektors."
  (apply '+ vec))

(defun square (item)
  "Hilfsfunktion zum Quadrieren eines Elements."
  (* item item))

(defun calcVarianz (vec)
  "Berechnet die Varianz eines Vektors."
  (loop with len = (length vec)
        with mean = (/ (sum vec) len)
        with some_func = (lambda (x) (* x x))
        ; causes the error
        for item in vec
        collecting (square (- item mean)) into squared
        collecting (some_func item) into some_vector
        ; some_func cannot be found
        finally (return (/ (sum squared) (- len 1)))))

效果很好（即计算向量的方差）。
现在，我想知道是否可以在构造中将sum和square函数定义为 lambda loop，但一路上陷入困境。这可以用例如

with sum = (lambda (x) ...)

出现错误

The function COMMON-LISP-USER::SOME_FUNC is undefined.
   [Condition of type UNDEFINED-FUNCTION]

我在这里缺少什么？

lisp