在熊猫数据帧上使用规范 pdf 的函数中的 Dtype 错误

Question

我在计算函数时遇到问题，而函数本身非常简单。

我有以下数据帧：

import pandas as pd
import numpy as np
import math as m
from scipy.stats import norm

dff = pd.DataFrame({'SKU': ['001', '002', '003','004','005'],
'revenue_contribution_in_percentage': [0.2, 0.2, 0.3,0.1,0.2],
'BuyPrice' : [7.78,9.96,38.87,6.91,14.04],
'SellPrice' : [7.9725,12.25,43,7.1,19.6],
'margin' : [0.9725,2.2908,5.8305,0.2764,5.1948],
'Avg_per_week' : [71.95,75.65,105.7,85.95,66.1],
'StockOnHand' : [260,180,260,205,180],
'StockOnOrder': [0,0,0,0,0],
'Supplier' : ['ABC', 'ABC', 'ABC','ABC','ABC'],
'SupplierLeadTime': [12,12,12,12,12],
'cumul_value':[0.20,0.4,0.6,0.8,1],
'class_mention':['A','A','B','D','C'],
'std_week':[21.585,26.4775,21.14,31.802, 26.44],
'review_time' : [5,5,5,5,5],
'holding_cost': [0.35, 0.35, 0.35,0.35,0.35],
'aggregate_order_placement_cost': [1000, 1000,1000,1000,1000],
'periods' : [7,7,7,7,7]})
dff['holding_cost'] = 0.35
dff1 = dff.sort_values(['Supplier'])
df2 = pd.DataFrame(dff1)
df2['forecast_dts'] = 5
df2['sigma_rtlt'] = 0.5

我需要将其中一些参数传递到函数中：

#
a0 = -5.3925569
a1 = 5.6211054
a2 = -3.883683
a3 = 1.0897299
b0 = 1
b1 = -0.72496485
b2 = 0.507326622
b3 = 0.0669136868
b4 = -0.00329129114

z = np.sqrt(np.log(25
/
(norm.pdf((df2['forecast_dts'])*(1-0.98)/df2['sigma_rtlt']) -
((df2['forecast_dts']*(1-0.98)/df2['sigma_rtlt']))* (1-norm.cdf(df2['forecast_dts']*(1-0.98)/df2['sigma_rtlt']))) ^ 2))

num = (a0 + a1 * z + a2 * z ^ 2 + a3 * z ^ 3)
den = (b0 + b1 * z + b2 * z ^ 2 + b3 * z ^ 3 + b4 * z ^ 4)
k = num / den
return k

但随后计算

calc = calc_invUnitNormalLossApprox()*df2['sigma_rtlt']

返回错误：

File "/usr/local/lib/python3.7/site-packages/pandas/core/ops/__init__.py", line 1280, in na_op
dtype=x.dtype, typ=type(y).__name__
TypeError: cannot compare a dtyped [float64] array with a scalar of type [bool]

在这一点上，我不确定那里发生了什么，特别是因为我知道公式本身是正确的，我假设我对规范pdf和cdf的使用有问题，但我无法弄清楚。

任何帮助将不胜感激。

Answer 1

我认为使用^运算符，您正在尝试执行bitwise XOR我认为您需要使用**运算符。

此代码有效

def calc():
a0 = -5.3925569
a1 = 5.6211054
a2 = -3.883683
a3 = 1.0897299
b0 = 1
b1 = -0.72496485
b2 = 0.507326622
b3 = 0.0669136868
b4 = -0.00329129114

z = np.sqrt(np.log(25
/
(norm.pdf((df2['forecast_dts'])*(1-0.98)/df2['sigma_rtlt']) -
((df2['forecast_dts']*(1-0.98)/df2['sigma_rtlt']))* (1-norm.cdf(df2['forecast_dts']*(1-0.98)/df2['sigma_rtlt']))) ** 2))

num = (a0 + a1 * z + a2 * z ** 2 + a3 * z ** 3)
den = (b0 + b1 * z + b2 * z ** 2 + b3 * z ** 3 + b4 * z ** 4)
k = num / den
return k

不是 ： 我已将^运算符更改为**