将多个列表附加在一起的最Python方式和最快方式(可能是相同的)是什么?例如,给定以下列表:
a = [1, 2]
b = [3, 4]
c = [5, 6]
d = [7, 8]
我们得到一个列表:
combined = [1, 2, 3, 4, 5, 6, 7, 8]
答案 0 :(得分:2)
在Python 3.5+中,您可以使用常规解压缩:
y
或者在Python 3.5+之前,您可以使用SELECT DISTINCTROW Query1.Period, Query1.Portfolio_1
FROM Query1
WHERE (((Query1.Period)="May 2018"))
ORDER BY Query1.Portfolio_2, Query1.Department, Query1.Position_NO;
:
Private Sub newChangePeriodCriteria_Click()
Dim qdf As DAO.QueryDef
Dim qdfOLD As String
Set qdf = CurrentDb.QueryDefs("Query1_Mth")
With qdf
qdf.SQL = sqlString
.SQL = Replace(.SQL, "Period='May 2018'", "Period='Jun 2018'")
' Code to do stuff with SQL-string/query
' .SQL = qdfOLD ' Reset SQL to old setting
qdfOLD = .SQL
' DoCmd.RunSQL (Replace("Period='May 2018'", "Period='Jun 2018'"))
End With
Set qdf = Nothing
End Sub
答案 1 :(得分:0)
我听不懂你 您是说如何合并它们? 例如
a = [1, 2]
b = [3, 4]
c = [5, 6]
d = [7, 8]
combined = a + b + c + d
因此将合并为
[1, 2, 3, 4, 5, 6, 7, 8]
答案 2 :(得分:0)
OR:
LateUpdate
现在:
Update
返回:
a = [1, 2]
b = [3, 4]
c = [5, 6]
d = [7, 8]
a.extend(b)
a.extend(c)
a.extend(d)