我在 numpy 中有一個名為 np_charge 的數(shù)據(jù)集，其大小為 (1430, 258)。我的想法是創(chuàng)建一個新列，其中第一行包含方程的輸出第一行：np_charge[-1][1]-np_charge[0][1]第二行： np_charge[-1][1]-np_charge[1][1] 。. .最后一行：np_charge[-1][1]-np_charge[1429][1]我試圖在互聯(lián)網(wǎng)上找到解決方案，但找不到任何解決方案。我的解決方案如下所示：e=[]n_rows=np_charge.shape[0]-1for i in range(n_rows): e[i]=(np_charge[-1][1]-np_charge[i][1])但我收到此錯誤代碼：---------------------------------------------------------------------------IndexError Traceback (most recent call last)<ipython-input-988-9463a5cb0618> in <module>() 15 16 for i in range(n_rows):---> 17 e[i]=np_charge[-1][1]-np_charge[i][1] 18 19 IndexError: list assignment index out of range謝謝，R
2 回答
慕森王
TA貢獻1777條經(jīng)驗 獲得超3個贊
正如您在您的評論之一中所闡明的那樣。創(chuàng)建 MVE
from datetime import datetime
from dateutil.relativedelta import relativedelta
import random
a=np.array([datetime.now()-relativedelta(months=random.randint(0,100)) for i in range(10)])
print(a)
輸出：
array([datetime.datetime(2014, 10, 12, 9, 0, 31, 696000),
datetime.datetime(2018, 9, 12, 9, 0, 31, 696000),
datetime.datetime(2014, 4, 12, 9, 0, 31, 696000),
datetime.datetime(2011, 4, 12, 9, 0, 31, 696000),
datetime.datetime(2015, 10, 12, 9, 0, 31, 696000),
datetime.datetime(2011, 11, 12, 9, 0, 31, 696000),
datetime.datetime(2016, 3, 12, 9, 0, 31, 696000),
datetime.datetime(2017, 10, 12, 9, 0, 31, 696000),
datetime.datetime(2010, 6, 12, 9, 0, 31, 696000),
datetime.datetime(2016, 2, 12, 9, 0, 31, 696000)], dtype=object)
tn 我假設數(shù)組的最后一個索引。所以
result=a[len(a)-1]-a
[datetime.timedelta(488),
datetime.timedelta(-943),
datetime.timedelta(671),
datetime.timedelta(1767),
datetime.timedelta(123),
datetime.timedelta(1553),
datetime.timedelta(-29),
datetime.timedelta(-608),
datetime.timedelta(2071),
datetime.timedelta(0)]
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