Question

我有一个样本DF：

df = pd.DataFrame(np.random.randint(0,10,size = (6,2)),columns=["A","B"])
df["col"] = ["Apple","Apple","Mango","Apple","Mango","Apple"]
df

OP：

    A   B   col
0   3   9   Apple
1   8   8   Apple
2   7   9   Mango
3   2   4   Apple
4   4   5   Mango
5   1   6   Apple

我试图获取列"A" and "B"中有重复值的地方col的值。例如，列col在索引-Apple中的值为0,1,3,5，而我试图在列A and B中获取相应的值，即

 {"Apple":[[3,9],[8,8],[2,4],[1,6]], "Mango": [[7,9],[4,5]]}

我有一个迭代方法，在大型Dfs上花费很长时间。

当前方法：

->在col列中查找唯一值

unique_values = list(df["col"].unique())

->遍历此列表，并遍历DF的每一行以获取所需的OP：

op = {}
for i in range(len(unique_values)):
    for index, rows in df.iterrows():
        if rows ["col"] == unique_values[i]:
            inner_op = []
            inner_op.append(rows["A"]);inner_op.append(rows["B"])
            if rows["col"] in op:
                op[rows["col"]].append(inner_op)
            else:
                op[rows["col"]] = [inner_op]

最终操作：

{'Apple': [[3, 9], [8, 8], [2, 4], [1, 6]], 'Mango': [[7, 9], [4, 5]]}

有没有建议采用更有效的熊猫方法？

Answer 1

创建由列表SendRiepilogo填充的列，将值转换为numpy数组，然后将其转换为列表，然后由GroupBy.agg与Series.to_dict聚合public static async void SendRiepilogo(string piva, int idOrdine) { var order = GetOrdine(piva, idOrdine); if (order == null) { return; } try { var negozio = getNegozio(order.idNegozio); var from = new MailAddress("ordini@visualorder.it", "VisualOrder"); var to = new MailAddress(order.cliente.FirstOrDefault().email); using MemoryStream ms = new MemoryStream(); QRCodeGenerator qrGenerator = new QRCodeGenerator(); QRCodeData qrCodeData = qrGenerator.CreateQrCode("vo/" + idOrdine, QRCodeGenerator.ECCLevel.Q); Base64QRCode qrCode = new Base64QRCode(qrCodeData); byte[] byteQr = Convert.FromBase64String(qrCode.GetGraphic(20)); MemoryStream streamQr = new MemoryStream(byteQr); var qrImage = new LinkedResource(streamQr, MediaTypeNames.Image.Jpeg) { ContentId = "qrImage" }; string nome = order.cliente.FirstOrDefault().nome; var orderEmail = new { idOrdine, order, nome, negozio }; byte[] byteLogo = Convert.FromBase64String(System.Text.Encoding.UTF8.GetString(negozio.logo)); MemoryStream streamLogo = new MemoryStream(byteLogo); var logoImage = new LinkedResource(streamLogo, MediaTypeNames.Image.Jpeg) { ContentId = "logoImage" }; string template = File.ReadAllText("Views/Emails/EmailRiepilogo.cshtml"); var htmlBody = Engine.Razor.RunCompile(template, "riepilogo", null, orderEmail); AlternateView alternateView = AlternateView.CreateAlternateViewFromString(htmlBody, null, MediaTypeNames.Text.Html); alternateView.LinkedResources.Add(qrImage); alternateView.LinkedResources.Add(logoImage); var message = new MailMessage(from, to) { Subject = "Riepilogo ordine", Body = htmlBody }; message.IsBodyHtml = true; message.AlternateViews.Add(alternateView); using var smtp = new SmtpClient("smtps.aruba.it", 587) { EnableSsl = true, Credentials = new NetworkCredential("XXX", "XXX") }; await smtp.SendMailAsync(message); // sending email to user await smtp.SendMailAsync(MessageNegozio(order, idOrdine, negozio)); // sending email to shop } catch (Exception e) { return; } ConfirmEmail(piva, idOrdine); // setting "EMAIL SENT" flag in DB to true return; }：

熊猫根据重复的行获取列值

1 个答案: