有没有简单的方法来获得蟒蛇的峰值和最低点？ [等候接听]

5 回答

• -1

  如果这是期望的结果，请 .

  data = [{'date':1, 'products_sold': 2}, {'date':2, 'products_sold': 5},{'date':5, 'products_sold': 2}]
  start_date = 1
  end_date = 2
  max_value_in_period = max(x['products_sold'] for x in data if x['date'] >= start_date and x['date'] <= end_date)
  print(max_value_in_period)
  

  回复于 2024-04-20T17:39:09+08:00
• 1

  我试过@Patrick Artner的评论：

  a = (20180101,10)
  b = (20180102,20)
  c = (20180103,15)
  d = (a,b,c)
  maximum = max( d, key = lambda x:x[1])
  minimum = min(d, key= lambda x:x[1])
  print(minimum)
  

  也许这给了一些灵感 .

  回复于 2024-04-20T17:39:09+08:00
• 0

  Pandas是你想要的lib .

  让我举个例子给你看看：

  import numpy as np
  import pandas as pd
  
  # let's build a dummy dataset
  index = pd.date_range(start="1/1/2015", end="31/12/2018")
  df = pd.DataFrame(np.random.randint(100, size=len(index)),
                    columns=["sales"], index=index)
  
  >>> df.head()
              sales
  2015-01-01     32
  2015-01-02      0
  2015-01-03     12
  2015-01-04     77
  2015-01-05     86
  

  现在假设您希望按月汇总销售额：

  >>> df["sales"].groupby(pd.Grouper(freq="1M")).sum()
  
  2015-01-31    1441
  2015-02-28    1164
  2015-03-31    1624
  2015-04-30    1629
  2015-05-31    1427
  [...]
  

  或一学期的基础

  df["sales"].groupby(pd.Grouper(freq="6M", closed="left", label="right")).sum()    
  2015-06-30    8921
  2015-12-31    9365
  2016-06-30    9820
  2016-12-31    8881
  2017-06-30    8773
  2017-12-31    8709
  2018-06-30    9481
  2018-12-31    9522
  2019-06-30      51
  

  出于某种原因，_s1831_带有六个月频率的频率有一些问题，31/12销售，它在2019年将它们放入一个新的垃圾箱，调查它会让你知道我是否找到任何东西......或者如果有人想评论请做

  或者你想知道哪个是最好的学期：

  >>> df["sales"].groupby(pd.Grouper(freq="6M")).sum().idxmax()              
  Timestamp('2016-06-30 00:00:00', freq='6M')
  

  回复于 2024-04-20T17:39:09+08:00
• 0

  你应该使用 pandas

  假设您的日期列称为“日期”，并且它是日期时间dtypes：

  import pandas as pd
  df = pd.DataFrame(data)
  df = df.set_index('date')
  df.groupby(pd.Grouper(freq='1M')).max()
  

  会给你每个月最大freq可以改变为你喜欢的任何频率 .

  回复于 2024-04-20T17:39:09+08:00
• 1

  这可能是一个有偏见的答案，但大熊猫非常适合处理这样的数据 . 虽然您可以使用元组，列表等完成此类操作，但pandas提供了更多功能 . 例如：

  import pandas as pd
  data = [[20180101,15], [20180102,10], [20180103,12],[20180104,10]]
  df = pd.DataFrame(data=data, columns=['date', 'products'])
  # if your data is in csv, excel, database... whatever... you can easily pull
  # df = pd.read_csv('name') || pd.read_excel() || pd.read_sql()
  df
  Out[2]: 
         date  products
  0  20180101        15
  1  20180102        10
  2  20180103        12
  3  20180104        10
  
  # It helps to use datetime format to perform operations on the data
  # Operations make reference to an "index" in the dataframe
  df.index = pd.to_datetime(df['date'], format="%Y%m%d")  #strftime format
  df
  Out[3]: 
                  date  products
  date                          
  2018-01-01  20180101        15
  2018-01-02  20180102        10
  2018-01-03  20180103        12
  2018-01-04  20180104        10
  
  # Now we can drop that date column...
  df.drop(columns='date', inplace=True)
  df
  Out[4]: 
              products
  date                
  2018-01-01        15
  2018-01-02        10
  2018-01-03        12
  2018-01-04        10
  
  # Yes, there are ways to do the above in shorthand... lots of info on pandas on SO
  # I want you to see the individual steps we are taking to keep simple
  
  # Now is when the fun begins
  df.rolling(2).sum()  # prints a rolling 2-day sum
  Out[5]: 
              products
  date                
  2018-01-01       NaN
  2018-01-02      25.0
  2018-01-03      22.0
  2018-01-04      22.0
  
  df.rolling(3).mean()  # prints a rolling 3-day average
  Out[6]: 
               products
  date                 
  2018-01-01        NaN
  2018-01-02        NaN
  2018-01-03  12.333333
  2018-01-04  10.666667
  
  df.resample('W').sum()  # Resamples the data so you can look on a weekly basis
  Out[7]: 
              products
  date                
  2018-01-07        47
  
  df.rolling(2).max() # max number of products over a rolling two-day period
  Out[9]: 
              products
  date                
  2018-01-01       NaN
  2018-01-02      15.0
  2018-01-03      12.0
  2018-01-04      12.0
  

  回复于 2024-04-20T17:39:09+08:00