如何在時間范圍內取平均值python？

Question

我是 Python 的初學者，所以我懇請您的幫助。 我想要一個文檔，其中第一列是 2011.01，第二列是該月的 ARD“事件”數，第三列是該月顯示的所有 ARD 的平均值。 如果沒有，例如 2012.07 0 0

我已經嘗試了 3 個小時，現在我開始緊張了。

我非常感謝你的幫助

import pandas as pd
from numpy import mean
from numpy import std
from numpy import cov
from matplotlib import pyplot
from scipy.stats import pearsonr
from scipy.stats import spearmanr
data = pd.read_csv('ARD.txt',delimiter= "\t")

month = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12']
day = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23', '24', '25', '26', '27', '28', '29', '30', '31']
year = ['2011', '2012', '2013', '2014', '2015', '2016', '2017', '2018', '2019', '2020', '2021']

ertek = data[:1].iloc[0].values
print(ertek)
print(data.head)

def list_to_string ( y, m, d):
    str = ""
    s = [y, m, d]
    str.join(s)
    return str

for x in year:
for y in month:
    for i in day:
        x = 1
        ertek = data[:x].iloc[0].values
        list_to_string(x, y, i)
        if ertek[0] == list_to_string[x, y, i]:
            print("")
            x += 1
        else:
            print("")

結果：

['2011.01.05.' 0.583333333]
<bound method NDFrame.head of              Date       ARB
0     2011.01.05.  0.583333
1     2011.01.06.  0.583333
2     2011.01.07.  0.590909
3     2011.01.09.  0.625000
4     2011.01.10.  0.142857
...           ...       ...
1284  2020.12.31.  0.900000
1285  2020.12.31.  0.900000
1286  2020.12.31.  0.900000
1287  2020.12.31.  0.900000
1288  2020.12.31.  0.900000

[1289 rows x 2 columns]>
 Traceback (most recent call last):
  File "C:\Users\Kókai Dávid\Desktop\python,java\python\stock-trading-ml-master\venv\Scripts\orosz\oroszpred.py", line 29, in <module>
   list_to_string(x, y, i)
   File "C:\Users\Kókai Dávid\Desktop\python,java\python\stock-trading-ml-master\venv\Scripts\orosz\oroszpred.py", line 21, in list_to_string
    str.join(s)
  TypeError: sequence item 0: expected str instance, int found

 Process finished with exit code 1

Answer 1

我不太確定我正在使用list_to_string函數跟蹤您的意圖； 如果是用於字符串日期比較，讓我們完全避開它

df.iloc[:,0] = df.iloc[:,0].apply(pd.to_datetime)