根据特定条件和输入字典生成数据框 - pandas

Question

我有一本字典，如下所示。

d1 = { 'start_date' : '2020-10-01T20:00:00.000Z',
       'end_date'  : '2020-10-05T20:00:00.000Z',
       'n_days'    : 6,
       'type'      : 'linear',
       "coef": [0.1,0.1,0.1,0.1,0.1,0.1]    
     }

从上面的字典作为函数的输入，我想生成下面的df作为输出。

预期输出：

Date                Day           function_type         function_value
2020-10-01          1             linear                (0.1*1)+0.1 = 0.2
2020-10-02          2             linear                (0.1*2)+0.1 = 0.3
2020-10-03          3             linear                (0.1*3)+0.1 = 0.4
2020-10-04          4             linear                (0.1*4)+0.1 = 0.5
2020-10-05          5             linear                (0.1*5)+0.1 = 0.6

注意：

type 可以是线性的、常数的、多项式的和指数的。

a0, a1, a2, a3, a4, a5 = d1['coef']

If constant:
funtion_value = a0

If exponential: 
funtion_value = e**(a0+a1T)

if polynomial:
funtion_value = a0+a1T+a2(T**2)+a3(T**3)+a4(T**4)+a5(T**5)

T: value of Day column

Answer 1

定义一个函数funcValue，它根据字典中的type 从给定的输入字典d 和天数列T 计算函数值 列：

def funcValue(d, T):
    a0, a1, a2, a3, a4, a5 = d['coef']
    func = {
        'constant': a0,
        'linear': a0 + a1*T,
        'polynomial': a0 + a1*T + a2*(T**2)+ a3 * (T**3) + a4*(T**4) + a5*(T**5),
        'exponential':  np.power(np.e, a0 + a1*T)
    }

    return func[d['type']]

然后定义一个函数getDF，它根据用户定义的字典d中提供的信息生成所需的数据帧：

def getDF(d):
    date = pd.date_range(d['start_date'], d['end_date'], freq='D').tz_localize(None).floor('D')
    days = (date - date[0]).days + 1
    return pd.DataFrame({'Date': date, 'Day': days, 'function_type': d['type'],
                         'function_value': funcValue(d, days)})

结果：

print(getDF(d1))

        Date  Day function_type  function_value
0 2020-10-01    1        linear             0.2
1 2020-10-02    2        linear             0.3
2 2020-10-03    3        linear             0.4
3 2020-10-04    4        linear             0.5
4 2020-10-05    5        linear             0.6

Answer 2

使用 timedelta 并生成列表列表会有所帮助：

from datetime import timedelta

d1 = { 'start_date' : '2020-10-01T20:00:00.000Z',
       'end_date'  : '2020-10-05T20:00:00.000Z',
       'n_days'    : 6,
       'type'      : 'linear',
       "coef":[0.1,0.1,0.1,0.1,0.1,0.1] 
     }

def value(tp, d, cf):
    if tp == "linear":
        val = (cf*d)+cf 
    elif tp == "exp":
        val = d**cf
    elif tp == "constant":
        val = d
    elif tp == "polynomial":
        val = cf*d**2+cf*d+cf
    return val

start = datetime.strptime(d1["start_date"], '%Y-%m-%dT%H:%M:%S.%f%z')
# end = datetime.strptime(d1["end_date"], '%Y-%m-%dT%H:%M:%S.%f%z')
end = start + timedelta(days=d1["n_days"])

df = [[start + timedelta(days=i),i,d1["type"],value(d1["type"],i,d1["coef"][i])] for i in range((end-start).days+1)]
df = pd.DataFrame(df,columns = ["Date","Day","function_type","function_value"])

输出：

    Date                        Day function_type   function_value
0   2020-10-01 20:00:00+00:00   0   linear          0.1
1   2020-10-02 20:00:00+00:00   1   linear          0.2
2   2020-10-03 20:00:00+00:00   2   linear          0.3
3   2020-10-04 20:00:00+00:00   3   linear          0.4
4   2020-10-05 20:00:00+00:00   4   linear          0.5