通过检查熊猫中指定的8种条件从输入字典生成df

Danish 发表于 Dev

丹麦文

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] 
     }

d1：是用户输入的内容，用户可能输入了错误的开始日期，结束日期和n_days。

我们必须更换 end_date = start_date + n_days

可能存在start_date，end_date和n_days不可用的情况。由指定的非可用性start_date = 0, end_date = 0, n_days = 0。

然后我们将有8个条件

1. if (start_date != 0) and (end_date != 0) and (n_days != 0):
          end_date = start_date + n_days

2. if (start_date != 0) and (end_date != 0) and (n_days == 0):
          pass
    
3. if (start_date != 0) and (end_date == 0) and (n_days != 0):
          end_date = start_date + n_days

4. if (start_date != 0) and (end_date == 0) and (n_days == 0):
          print("Please enter required inputs")

5. if (start_date == 0) and (end_date != 0) and (n_days != 0):
          start_date = end_date - n_days

6. if (start_date == 0) and (end_date != 0) and (n_days == 0):
          print("Please enter required inputs")

7. if (start_date == 0) and (end_date == 0) and (n_days != 0):
          print("Please enter required inputs")

8. if (start_date == 0) and (end_date == 0) and (n_days == 0):
          print("Please enter required inputs")

我想通过检查d1并入上述所有条件。之后，我想根据上面处理的d1在df以下进行准备。

从上面，字典作为函数的输入，我想在df以下生成作为输出。

预期产量：

df：

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：

类型可以是线性，常数，多项式和指数。

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

@Shubham Sharma和@Let的尝试很好地回答了类似的问题，请尝试根据特定条件和输入字典生成数据框-熊猫

非常感谢@Shubham Sharma和@让我们尝试

Shubham Sharma

延长我previous answer的question，我们可以创建自定义validation机能的研究，是以输入参数由用户提供的字典，并返回一个两对tuple具有有效start_date和end_date以其他方式筹集的ValueError：

def validate(d):
    n_days, start, end = d['n_days'], d['start_date'], d['end_date']

    if start == 0:
        if end != 0 and n_days != 0:
            end = pd.to_datetime(end)
            start = end - pd.Timedelta(days=n_days)
        else:
            raise ValueError('Invalid user input')
    else:
        start = pd.to_datetime(start)
        if end != 0 and n_days != 0:
            end = pd.to_datetime(end)
        elif end == 0 and n_days == 0:
            raise ValueError('Invalid user input')
        else:
            end = start + pd.Timedelta(days=n_days)

    return start.tz_localize(None).floor('D'),\
                 end.tz_localize(None).floor('D')

然后getDF从中修改函数，previous answer并包含try-except要捕获的语句ValueError：

def getDF(d):
    try:
        start, end = validate(d)
        date = pd.date_range(start, end, freq='D')
        days = (date - date[0]).days + 1
        return pd.DataFrame({'Date': date, 'Day': days, 'function_type': d['type'],
                            'function_value': funcValue(d, days)})
    except ValueError as err:
        print(err)

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

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