有没有简单的方法来获取python中的峰值和最低值?[闭门]

这可能是一个有偏见的答案,但熊猫真的很适合处理这样的数据。而您可以使用元组、列表等来完成这种操作。 熊猫提供了更多的功能。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

Pandas 这就是你想要的自由。

让我举一个例子:

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,假设您希望每月汇总销售额:

>>> 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
[...]

还是以学期为基础

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 Grouper binning with six months freq与31/12销售有一些问题,它在2019年将其放入一个新的bin,如果我发现任何问题,查看它会让你知道。。。或者如果其他人想发表评论,请发表评论

或者你想知道哪个学期最好:

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

你应该使用 pandas

假设您的日期列名为“date”,并且是datetime数据类型:

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

会给你每个月的最大频率可以改变到任何你喜欢的频率。

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)

也许这会给我们一些启发。

如果这是期望的结果,请回答。

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)