Question

我有一个元组列表，例如。

export class AuthGuard implements CanActivate {
    constructor(private loggedInService: LoggedinService) {
    }

    canActivate(next: ActivatedRouteSnapshot,
                state: RouterStateSnapshot):  boolean{
        let url: string = state.url;
        return this.checkLogin(url);
    }

    checkLogin(url: string): boolean {
        if (this.loggedInService.isLoggedIn) {
            return true;
        } else {
            this.loggedInService.redirectUrl = url;
            return false;
        }
    }
}

我还有另一个列表

l1 = [(1, 'a', 'x'), (2, 'b', 'y'), (3, 'a', 'z'), (4, 'c', 'xyz')]

我想根据l2 = ['a', 'b', 'c']中出现的顺序对每个元组的第二个值排序l1。最终结果应该是

l2

如您所见，第二个值为l1 = [(1, 'a', 'x'), (3, 'a', 'z'), (2, 'b', 'y'), (4, 'c', 'xyz')]的元组首先出现，然后是第二个值为a的元组，最后是具有b的元组。

我写了以下内容：

这很简单。但是我想知道是否有任何pythonic方式可以完成此任务？

Answer 1

您可以将sorted与相应的t函数一起使用：

key

Answer 2

您可以创建一个使用值作为键的<%@ Control Language="C#" AutoEventWireup="true" CodeFile="CMSArticleGrid.ascx.cs" Inherits="UserControls_CMS_CMSArticleGrid" %> <div> <div class="page-header"> <h1> Vis nyheter</h1> </div> <br /> <div class="grid_top_border"> <asp:DataGrid ID="grdCMS" runat="server" Width="940px" AutoGenerateColumns="false" OnItemCommand="grdCMS_ItemCommand" AllowPaging="true" AllowSorting="true" OnSortCommand="pageSort" OnPageIndexChanged="pageChange" PageSize="20" PagerStyle-Mode="NumericPages" OnItemDataBound="item_binding" PagerStyle-Font-Size="Large" EnableViewState="true"> <HeaderStyle Font-Bold="true" Font-Size="Medium" BackColor="#D0CACD" /> <PagerStyle Mode="NumericPages" Font-Size="Large"></PagerStyle> <ItemStyle Height="20px" /> <Columns> <asp:TemplateColumn HeaderText="Artikkeldato" SortExpression="CreatedDate"> <ItemTemplate> <%#Eval("CreatedDate")%> </ItemTemplate> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Forfatter" SortExpression="Authour"> <ItemTemplate> <%#Eval("Authour")%> </ItemTemplate> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Brukernavn" SortExpression="UserName"> <ItemTemplate> <%#Eval("UserName")%> </ItemTemplate> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Godkjent" SortExpression="Verified"> <ItemTemplate> <%#Eval("Verified")%> </ItemTemplate> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Tittel" SortExpression="Title"> <ItemTemplate> <asp:Label ID="id" runat="server" Text='<%# Eval("ID") %>' Visible="false"></asp:Label> <a class="lnkNobb" id="linkAticle" href="CMSArticleList.aspx?ElementId=<%# Eval("ID") %>"> <%# Eval("Title") %></a> </ItemTemplate> <%--<ItemTemplate> <%#Eval("Title")%> </ItemTemplate>--%> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Kat.Sjef" SortExpression="CatManager"> <ItemTemplate> <%--<asp:DropDownList ID="ddlprofile" AutoPostBack="true" Width="80px" runat="server"> </asp:DropDownList>--%> <asp:ListBox ID="lstprofile" runat="server" SelectionMode="Multiple" ToolTip="Press ctrl to select multiple users"></asp:ListBox> </ItemTemplate> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Godkjent"> <ItemTemplate> <asp:CheckBox ID="cbxVerify" runat="server" /> </ItemTemplate> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Rediger"> <ItemTemplate> <asp:LinkButton ID="lnkEdit" CommandName="Edit" CommandArgument='<%#Eval("Id") + "@" +Eval("Verified")%>' Text="Rediger artikkelen" runat="server"></asp:LinkButton> </ItemTemplate> </asp:TemplateColumn> <asp:TemplateColumn HeaderText="Slett"> <ItemTemplate> <asp:LinkButton ID="lnkDelete" CommandName="Delete" CommandArgument='<%#Eval("Id") + "@" +Eval("Verified")%>' Text="Slett artikkelen" runat="server" OnClientClick="javascript:ConfirmDelete();"></asp:LinkButton> </ItemTemplate> </asp:TemplateColumn> </Columns> </asp:DataGrid> </div> <div class="container_24 grid_24"> <asp:Button ID="btnVerify" text="Godkjenne" runat="server" OnClick="btnVerify_click"/> </div> <asp:HiddenField ID="hfConfDel" runat="server" /> </div> and below is my code from code behind:- protected void Page_Load(object sender, EventArgs e) { if (Session["User"] == null) { Response.Redirect("~/Webpages/Info.aspx"); } if (!IsPostBack) { BindGrid(); } } protected void btnVerify_click(object sender, EventArgs e) { string userid = ((Session["User"]) as WebUser).UserId.ToString(); // if (userid == "999922" || userid == "999944" || userid == "999945") { foreach (DataGridItem item in grdCMS.Items) { CheckBox cbx = (CheckBox)item.FindControl("cbxVerify"); Label id = (Label)item.FindControl("id"); ListBox lstprofile = (ListBox)item.FindControl("lstprofile"); int countSelected = 0; StringBuilder sb = new StringBuilder(); sb.Append("#"); foreach (ListItem lstitem in lstprofile.Items) { if (lstitem.Selected) { countSelected += 1; sb.Append("" + lstitem.Value + ""); sb.Append("#"); } } if (cbx.Checked) { WebCMS.updateVerified("Y", id.Text, sb.ToString()); } else { WebCMS.updateVerified("N", id.Text, sb.ToString()); } Page.ClientScript.RegisterStartupScript(GetType(), "key", "<script>showAlert('Artikkel oppdatert')</script>"); } BindGrid(); getVerified(); //}else{ // Page.ClientScript.RegisterStartupScript(GetType(), "key", "<script>showAlert('du er ikke autorisert til å godkjenne')</script>"); //} } public void BindGrid() { try { if (base.Session["User"] is WebUser) { long userId = (base.Session["User"] as WebUser).UserId; if ((base.Session["User"] as WebUser).Type == WebUser.UserType.SUPP) { this.listItem = WebCMS.GetArticlesList((long)Convert.ToInt32((base.Session["User"] as WebUser).UserId.ToString())); this.grdCMS.Columns[5].Visible = false; this.grdCMS.Columns[6].Visible = false; this.btnVerify.Visible = false; } else { this.grdCMS.Columns[3].Visible = false; if ((base.Session["User"] as WebUser).Type == WebUser.UserType.ADMIN) { this.listItem = WebCMS.GetArticles(); } else { this.listItem = WebCMS.GetArticlesOfCatChief(userId); } } if ((this.listItem != null) && (this.listItem.Count > 0)) { this.grdCMS.DataSource = this.listItem; this.grdCMS.DataBind(); } this.getVerified(); } } catch { } }，并使用dict给出的索引对其进行排序：

enumerate(l2)

以这种方式使用dict的优势在于，不会在每次对l1 = [(1, 'a', 'x'), (2, 'b', 'y'), (3, 'a', 'z'), (4, 'c', 'xyz')] l2 = ['a', 'b', 'c'] sorting ={key:idx for idx,key in enumerate(l2)} l3 = sorted(l1, key = lambda x: sorting.get(x[1],9999999999)) print(l3)的查找中都使用index(...)。

您将使用l2来避免丢失键而造成麻烦，并提供一个默认值（确保该值不会出现在列表中），以便将它们放在末尾。

输出：

dict.get()

Answer 3

您还可以只使用列表l1的 sort 方法对每个元组的第二个值进行排序：

l1 = [(1, 'a', 'x'), (2, 'b', 'y'), (3, 'a', 'z'), (4, 'c', 'xyz')]

l1.sort(key=lambda x: x[1])

print(l1)

如果这还不够，并且您需要第二个列表l2来定义订单，则可以使用列表的 index 方法：

l1 = [(1, 'a', 'x'), (2, 'b', 'y'), (3, 'a', 'z'), (4, 'c', 'xyz')]
l2 = ['a', 'b', 'c']

l1.sort(key=lambda x: l2.index(x[1]))

print(l1)

根据另一个列表对元组列表进行排序

3 个答案: