Answer 1

您可以将图像的数组类型更改为元组：

normalized = resized/255,0

更改为：

normalized = resized/255

Answer 2

删除我白名单上没有的所有电子邮件，

function deleteAllEmailsNotWhiteListed() {
  const acct = getGlobal('abrevEmail');
  var idA = [];
  GmailApp.getTrashThreads().forEach(t => { Gmail.Users.Threads.remove("me", t.getId()) });
  for (let k = 0; k < 2; k++) {
    switch (k) {
      case 0:
        var threads = GmailApp.getInboxThreads();
        break;
      case 1:
        var threads = GmailApp.getSpamThreads();
        break;
    }
    if (threads) {
      for (let i = 0; i < threads.length; i++) {
        const messages = threads[i].getMessages();
        if (messages) {
          for (let j = 0; j < messages.length; j++) {
            let msg = messages[j];
            let id = msg.getId();
            let sub = msg.getSubject().replace(/[^\s\w"!,?']/g, '');
            let from = msg.getFrom();
            let to = msg.getTo();
            let body = msg.getBody();
            let m = { id: id, from: from, to: to, subject: sub, account: acct }
            if (!isWhiteListed(m)) {
              idA.push(id);
            }
          }
        }
      }
    }
  }
  if (idA.length > 0) {
    var request = { "ids": idA };
    Gmail.Users.Messages.batchDelete(request, "me");
  }
}

function isWhiteListed(msg) {
  const ss = SpreadsheetApp.openById(getGlobal('ssid'));
  const sh = ss.getSheetByName(getGlobal('wlid'));
  const wlog = ss.getSheetByName(getGlobal('wlog'));
  const sr = 2;
  const rg = sh.getRange(sr, 1, sh.getLastRow() - sr + 1, 1);
  const wl = rg.getDisplayValues().flat().map(e => e.toLowerCase());
  if (wl.length == 0) throw ('Error: WhiteList has no content');
  let obj = decomposeEmailsWithTitle({ raw: msg.from, email: '', title: '' });
  let status = wl.indexOf(obj.email.toLowerCase()) > -1;
  wlog.appendRow([Utilities.formatDate(new Date(), Session.getScriptTimeZone(), "E MMM dd, yyyy HH:mm:ss"), obj.email, msg.to, msg.subject, status, msg.account]);
  return status;
}

这是一个仅关注删除电子邮件的版本

function deleteAllEmailsNotWhiteListed() {
  const dt = new Date();
  const ydtv = new Date(dt.getFullYear(),dt.getMonth(), dt.getDate()-1).valueOf();
  const acct = getGlobal('abrevEmail');
  var idA = [];
  GmailApp.getTrashThreads().forEach(t => { Gmail.Users.Threads.remove("me", t.getId()) });
  for (let k = 0; k < 2; k++) {
    switch (k) {
      case 0:
        var threads = GmailApp.getInboxThreads();
        break;
      case 1:
        var threads = GmailApp.getSpamThreads();
        break;
    }
    if (threads) {
      for (let i = 0; i < threads.length; i++) {
        const messages = threads[i].getMessages();
        if (messages) {
          for (let j = 0; j < messages.length; j++) {
            let msg = messages[j];
            let id = msg.getId();
            let d = msg.getDate();
            let dtv = new Date(d.getFullYear(),d.getMonth(),d.getDate()).valueOf()
            if (dtv <= ydtv) {//modifed
              idA.push(id);
            }
          }
        }
      }
    }
  }
  if (idA.length > 0) {
    var request = { "ids": idA };
    Gmail.Users.Messages.batchDelete(request, "me");
  }
}

Answer 3

使用：

<div class="collapse in " [ngClass]="(active_tab=='assignservice' || active_tab=='manage')?'show':''" id="collapseExampleOrganization" aria-expanded="true" style="">
    <ul>
        <li class="nav-item" [ngClass]="{'active': active_tab=='manage'}">
            <a routerLink="/main/organization/manage" (click)="activemenu('manage')"> <i class="la la-building-o"></i>
               <p>Manage</p></a></li>
        <li class="nav-item" [ngClass]="{'active': active_tab=='assignservice'}"><a routerLink="/main/organization/assignservice" (click)="activemenu('assignservice')"><i class="la la-user"></i><p>Add organization</p></a></li>
    </ul>
</div>

该代码是ngclass 的一个很好的例子，如果 - else 条件。

[ngClass]="(active_tab=='assignservice' || active_tab=='manage')?'show':''"

[ngClass]="{'active': active_tab=='assignservice'}"

Answer 4

手写圆的某些部分（在线间距区域）可以提取，假设“许多字母在同一行上对齐”。我认为，在您的图像中，将提取圆的上部和下部。

然后，如果从提取的部分开始跟踪黑线（假设曲率平滑），则可以检测到连接的手写圆圈。

但是……实际上，我认为这样的过程会遇到许多困难：尤其是关于通过删除曲线切断角色的事实。

Answer 5

from pyspark.sql import SparkSession
from pyspark.sql import functions as F

from pyspark.sql.types import *
from datetime import datetime
from pyspark.sql import *
from collections import *
from pyspark.sql.functions import udf,explode
from pyspark.sql.types import StringType
from pyspark.context import SparkContext
from pyspark.sql.session import SparkSession

spark = SparkSession(sc)

df= spark.createDataFrame(
    [
        ('0018aad4', [{"val1":"blue", "val2":"red"}],[300,500]), 
         ('0018aad', [{"val1":"blue", "val2":"null"}],[300])
       
        ],("ID","List","Tlist")
    )

df2 = df.select(df.ID,explode(df.List).alias("Dict"),df.Tlist )
df2.withColumn("Val1", F.col("Dict").getItem("val1")).withColumn("Val2", F.col("Dict").getItem("val2")).show(truncate=False)


+--------+----------------------------+----------+----+----+
|ID      |Dict                        |Tlist     |Val1|Val2|
+--------+----------------------------+----------+----+----+
|0018aad4|{val2 -> red, val1 -> blue} |[300, 500]|blue|red |
|0018aad |{val2 -> null, val1 -> blue}|[300]     |blue|null|

+--------+--------------------------------------+-----------+ -----+ ----+

这是您要寻找的。

Answer 6

不确定我会说这是可扩展的，但是对于字母，我将

const letters = ["A","B".."Z"]

function generateRandomLetterArray(){
let randomLetterArr = []

for(i=0, i<{how every many random letters you want}, i++){

let randomNumber = {math random number between 0 and 25 (because there are 26 letters)}

randomLetter = letters[randomNumber]
randomLetterArr.push(randomLetter)
}
}

在此时添加到一个数组

Answer 7

您已经有一个用户数据，并且想在输入中绑定它以创建用户配置文件吗？如果是，您可以在模板中使用 vor 指令在数组上迭代。

演示：

new Vue({
  el: '#app',
  data: {
    users: [{
      name: "Fred",
      phone: "00997878",
      email: "[email protected]",
      age: "20"
    },{
      name: "Tom",
      phone: "0998899",
      email: "[email protected]",
      age: "23"
    }]
  },
  methods: {
    createprofile() {
        console.log(this.users);
    }
  }
})

<script src="https://cdnjs.cloudflare.com/ajax/libs/vue/2.5.17/vue.js"></script>
<div id="app">
  <div v-for="(user, index) in users" :key="index">
    <input id="email" v-model="user.email" type="text" placeholder="Email">
    <input id="name" v-model="user.name" type="text" placeholder="Name">
    <input id="phone" v-model="user.phone" type="text" placeholder="Phone no.">
    <input id="age" v-model="user.age" type="text" placeholder="Age">
  </div><br>
  <button @click="createprofile()">Submit</button>
</div>

Answer 8

您可以用计数器对象来计算外观，然后进行最频繁的外观：

from collections import Counter

column = ("S", "A", "A", "A", "D", "S-A", "A")

counter = Counter(column)
# returns a sorted list of elements and number of appearances
frequencies = counter.most_common()

# grab the first (most common) element and its frequency.
most_frequent_element, frequency = fequencies[0]

print(most_frequent_element, frequency)

输出：

A 4

Answer 9

多亏了T3RR0R，它起作用了。

@echo off
chcp 65001

for /f "tokens=*" %%a IN ('dir /b /o:-d *.log') do (
    call :ChercheSucc %%a
    )
)

:RienTrouve
echo Pas de succès
pause
exit 1 /b

:TrouveFichier
set FichierTrouve=%~1
echo %FichierTrouve%
echo Trouvé !
pause
exit 0 /b

:ChercheSucc
findstr /i /m /c:"SUCC" %~1 >nul
if %errorlevel%==0 goto TrouveFichier

Answer 10

这就是熊猫数据框架中的预期行为。

在您的情况下， procinject1 存储数据框的字符串表示形式，即使相应的数据框为空，这也是非空框的。

例如，请选中以下代码段，其中我创建一个空的数据框 df 并检查其字符串表示：

df = pd.DataFrame()
print(df.to_string(index = False)) 
print(df.to_string(index = True))

既适用 index = false and code> index = index = true 案例，输出将相同，如下所示（这是预期的行为）。因此，您对应的 len（）将始终返回非零。

Empty DataFrame
Columns: []
Index: []

但是，如果您使用非空置框架，则为 index = false and index = true 案例的输出将有所不同：

data = [{'A': 10, 'B': 20, 'C':30}, {'A':5, 'B': 10, 'C': 15}]
df = pd.DataFrame(data)
print(df.to_string(index = False)) 
print(df.to_string(index = True))

然后是的输出index = false 和 index = true 案例将是 -

A  B  C
10 20 30
5 10 15

    A   B   C
0  10  20  30
1   5  10  15

由于Pandas对空数据框的处理方式有所不同，要解决您的问题，您应该首先检查您的数据帧是否为空，请使用 pandas.dataframe.empty.empty 。

然后，如果数据帧实际上是非空的，则可以打印该数据框的字符串表示形式，同时保留 index = false 以隐藏索引列。

Answer 11

在通过互联网搜索时，我发现它可能与保险库策略设置有关。因此，我终于能够使这件事变得有效，我不得不将 data 附加在库中成功阅读的路径。因为当我研究组织的策略文档时，数据是在事实上附加的。

const vault = require("node-vault")({
    apiVersion: "v1",
    endpoint: "vaultURL",
  });

const GITHUB_TOKEN = '';

const run = async () => {
 try {
  const result = await vault.githubLogin({ token: GITHUB_TOKEN });
  
  vault.token = result.auth.client_token;
  console.log('Client Token', vault.token);

  const { data : { data } }  = await vault.read("root/data/path");  // <---- important
    console.log('data', data);
 } catch (error) {
   console.log(error.message);
 }
};

run();

我最初的道路是
秘密/apikey

必须使用
秘密/data/apikey

用于参考“>这个在github上的答案。

Answer 12

这是根据设计工作的，状态将随机生成的唯一值通常用于预防。该状态还用于编码有关应用程序中用户状态的信息，请参见文档以获取更多详细信息 - https://learn.microsoft.com/en-us/graph/auth-v2-user

Answer 13

可以通过创建扩展 AbstractCelleditor 并实施 TableCelleDitor 的私人类来解决。然后创建并自定义滚动条。这是示例。

public class MainForm {
    private JTable table1;
    private JPanel panel1;
    private JTextArea text = new JTextArea();

    public MainForm() {
        text.setEditable(false);
        DefaultTableModel tableModel = new DefaultTableModel(new Object[][]{}, new String[]{"Meno", "Priezvisko", "Datum", "Popis"});
        table1.setModel(tableModel);
        table1.setRowHeight(40);
        tableModel.addRow(new Object[]{"Jaimi", "Parker", "2022", "fdjfadufaouifasoifjasifhasiofa \nasdasdasdasdasdasdas \nasdasdasdasdasdasd\nasdasdasdasd "});
        tableModel.addRow(new Object[]{"William", "Mcdonald", "2021", "fdjfadufaouasfadfdsfafifasoifjasifhasiofa"});
        tableModel.addRow(new Object[]{"Matt", "Ashley", "2020", "asfasfafdsfgdafgdfgasgsdg"});
        tableModel.addRow(new Object[]{"Ariana", "Burns", "2019", "fdjfadufaouifasfdsgdgagsgsdgsdgsdagsdgsdgsdgsagsdgoifjasifhasiofa"});

        TableColumn column = table1.getColumn("Popis");
        column.setCellEditor(new HScrollableTCE());
    }

    public static void main(String[] args) {
        JFrame frame = new JFrame("MainForm");
        frame.setContentPane(new MainForm().panel1);
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        frame.pack();
        frame.setVisible(true);
    }

    private class HScrollableTCE extends AbstractCellEditor implements TableCellEditor
    {
        JScrollPane scroll = new JScrollPane(text, JScrollPane.VERTICAL_SCROLLBAR_AS_NEEDED, JScrollPane.HORIZONTAL_SCROLLBAR_AS_NEEDED);

        public Component getTableCellEditorComponent(JTable table, Object value, boolean isSelected, int rowIndex, int vColIndex)
        {
            if (value != null) {
                scroll.setHorizontalScrollBar(scroll.createHorizontalScrollBar());
                scroll.getHorizontalScrollBar().setPreferredSize(new Dimension(0, 5));

                scroll.setBorder(new EmptyBorder(0,0,0,0));
                scroll.setToolTipText(value.toString());
                text.setText(value.toString());
                return scroll;
            }
            return null;
        }
        public Object getCellEditorValue()
        {
            return text.getText();
        }
    }

}

Answer 14

def load_csv(filename):
    category = pd.read_csv(filename, nrows=0).columns.str.replace('(\.\d+)
我想到了。但这太丑了。
,'')
    category_subset = pd.read_csv(filename, skiprows=1, nrows=0).columns.str.replace('\:.*','').str.replace('(\.\d+)
我想到了。但这太丑了。
,'')
    df = pd.read_csv(filename, header=2)
    df.columns = category + ": " + category_subset + ": " + df.columns
    return df

我想到了。但这太丑了。

Answer 15

您可以尝试下面的代码以显示在随机方法返回价值后要购买餐点的人的名称。

# get the input names of friends separated by comma
names=input("Enter the names of every person separated by a comma")
# get the list of names by splitting the string above using the comma delimiter
name_collection=names.split(",")
# get the random index of the person who is going to pay for the bills
choice=random.randint(0,len(name_collection))
#print the name of the person who is going to buy the meal
print(name_colection[choice])