由于您没有发布任何代码,因此在这种情况下,很难知道如何使用flatmap()
。您可以简单地map()
在数组上并过滤每个嵌套数组属性。
const array = [{ name: 'Parent Brand 1', childBrands: [{ name: 'Child Brand 1', status: 'active' }, { name: 'Child Brand 2', status: 'discontinued' },] }, { name: 'Parent Brand 2', childBrands: [{ name: 'Child Brand 1', status: 'discontinued' }, { name: 'Child Brand 2', status: 'active' },] }];
const filtered = array.map(parent => ({
...parent,
childBrands: parent.childBrands.filter(child => child.status === 'active')
}));
console.log(filtered)
void LoadFile(vector<vector<string>> courseInfo) { ... }
这会创建向量的副本,因为您按值将其传递。然后,您将所有数据存储在副本中,并在功能末尾将副本销毁。您绝对不会修改原始矢量。将其更改为
void LoadFile(vector<vector<string>> &courseInfo) { ... }
不客气:)
默认情况下,Docker将内部网络中的服务连接。要重新使用其他服务,您需要使用服务名称(或Aliases ...),Localhost将无法使用。
第二期发布的内容与RabbitMQ连接无关,您的MYTB_1容器无法访问您的数据卷映射。该服务需要有足够的权利来写入映射的目录
version: '2.2'
services:
mytb:
privileged: true
restart: always
depends_on:
- rabbitmq
image: "thingsboard/tb-postgres"
ports:
- "9091:9090"
- "1883:1883"
- "7070:7070"
- "5683-5688:5683-5688/udp"
environment:
TB_QUEUE_TYPE: rabbitmq
TB_QUEUE_RABBIT_MQ_USERNAME: test
TB_QUEUE_RABBIT_MQ_PASSWORD: test123369
# Use the rabbitmq service name as a host instead of localhost
TB_QUEUE_RABBIT_MQ_HOST: rabbitmq
TB_QUEUE_RABBIT_MQ_PORT: 5672
volumes:
- ./.mytb-data:/data
- ./.mytb-logs:/var/log/thingsboard
# make sure that the service has sufficient rights to
# write to the mapped directories
rabbitmq:
... (rest of file left out)
如果 - 属性 statement utate 这样(出于多种原因);相反,您可以使用case_when
:
data <-
data %>%
mutate(second.indicator = case_when((il.count.description + un.count.description) > (elle.count.description + une.count.description) ~ -1,
(il.count.description + un.count.description) == (elle.count.description + une.count.description) ~ 0,
T ~ 1
)
似乎DOT NET核心,实体框架(EF)和C#在构建复杂查询(旧数据)的情况下有多个LINQ的问题。最终,我通过简单连接的多个列表创建了数据。
所有先前的陈述都出现为&lt;&gt; h__ transparentIdentifier,在与EF打交道时似乎行为的行为。但是,鉴于本地列表,系统丢弃了上述错误,这是可以理解的,因为SQL编译器对列表不了解,我希望LINQ能够作为一组次级征服而不是尝试形成单个查询。
我当前的应用具有appconfig类,该类别扩展 norfollow noreferrer“> sentaritedwidgetwidget 可以使实例可用的实例在使用中。
您可以尝试使用
class AppConfig extends InheritedWidget {
static AppConfig? of(BuildContext context) {
initializeFirebase();
return context.dependOnInheritedWidgetOfExactType<AppConfig>();
}
static Future<void> initializeFirebase() async {
try {
await Firebase.initializeApp();
} catch (e) {}
}
}
Main()中的AppConfig()初始化
这是由于在字符串开始时插入了Bom。我必须指定正确的编码字符集来解决此问题。
请检查 u'\ ufeff'
print(repr(finalSql))
有关详细信息,
insert into OTBI_MANUFACTURERS (\ufeffMANUFACTURER,REGISTRY_ID,MFR_DESCRIPTION,STATUS,PARTY_TYPE,CITY,STATE,POSTAL_CODE,PRIMARY_PHONE_NUMBER,PRIMARY_PHONE_EXTENSION,WEBSITE,CONTACT_NAME,GENERAL_INQUIRY_EMAIL,MANUFACTURER_NAME_LONG,MANUFACTURER_ADDRESS,COUNTRY,ADDRESS1,SUPPLIER_NAME,SUPPLIER_EMAIL,SUPPLIER_PHONE_NUMBER,ADDRESS2) values (:1,:2,:3,:4,:5,:6,:7,:8,:9,:10,:11,:12,:13,:14,:15,:16,:17,:18,:19,:20,:21)'
有三个部分:1)找到零,2)找到平均值,3)填充发现的零。因此:
/****
* in-place fill zeros with the mean of the surrounding neighborhoods
***/
void fillHoles(Mat gray){
// find the zeros
Mat mask = (gray == 0);
// find the mean with filter2d
Mat kernel = (Mat_<double>(3,3) <<
1/8, 1/8, 1/8
1/8, 0 , 1/8
1/8, 1/8, 1/8
);
Mat avg;
cv::filter2d(gray, avg, CV_8U, kernel)
// then fill the zeros, only where indicated by `mask`
cv::bitwise_or(gray, avg, gray, mask);
}
注释我只是意识到这显然是在获得平均值,而不是非零的平均值。对于该操作,您可能需要执行两个过滤器,一个用于总和,一个用于非零计数,然后将两个过滤器划分为:
// find the neighborhood sum with filter2d
Mat kernel = (Mat_<double>(3,3) <<
1, 1, 1
1, 0, 1
1, 1, 1
);
Mat sums;
cv::filter2d(gray, sums, CV_64F, kernel);
// find the neighborhood count with filter2d
Mat counts;
cv::filter2d(gray!=0, counts, CV_64F, kernel);
counts /= 255; // because gray!=0 returns 255 where true
// force counts to 1 if 0, so we can divide later
cv::max(counts, 1, counts);
Mat out;
cv::divide(sums, counts, out);
out.convertTo(gray, CV_8U);
搜索列表
您可以使用以下代码:
idx = np.searchsorted(df.index, num)
df.index[max(0, idx)]
示例
数据集为:
df = pd.DataFrame({'Time': range(10), 'Alt': range(10)}, index=range(0,20, 2))
Output:
Time Alt
0 0 0
2 1 1
4 2 2
6 3 3
8 4 4
10 5 5
12 6 6
14 7 7
16 8 8
18 9 9
如果
num = 7
idx = np.searchsorted(df.index, num)
df.index[max(0, idx)]
Output:
8
要在输入的索引号下方找到索引,则更改:
df.index[max(0, idx)]
8
df.index[max(0, idx-1)]
从上面的示例输出为6,而不是
和为了良好的衡量,这是一个可以处理这一切的函数:
def next_index(df, num):
# Returns the next index from num in the df
idx = np.searchsorted(df.index, num)
return df.index[max(0, idx)]
您可以使用 string.join()
Console.WriteLine(string.Join("<span>l</span>", words));
而不是
//No need to build resultString as well. That mean you can eliminate for loop
Console.WriteLine(resultString);
。
最后,我不得不添加两个额外的滴答:
ROM DISK = '''
* @backupPath + @backupFile + ''' WITH STANDBY = N''D:\TelcorLogDump\ROLLBACK_UNDO_Telcor.BAK
'''
order.cartItems.forEach(async (item) => {
const result = await db
.collection("items")
.updateOne(
{ _id: parseInt(item.itemId) },
{ $inc: { numInStock: -parseInt(item.quantity) } }
);
});
然后恢复numinstock
result.cartItems.forEach(async (item) => {
const result = await db
.collection("items")
.updateOne(
{ _id: parseInt(item.itemId) },
{ $inc: { numInStock: parseInt(item.quantity) } }
);
});
由于usestate是异步直接替换状态不是一个好主意。因此,最好访问先前的状态并设置新的更新值。
for ex:setState(((prev)=&gt; [... prev,... result]);
//结果是对象数组
尝试将PLIST文件中的服务名称更改为
_ABC-TXTCHAT._TCP
。 PLIST文件中的值必须符合Bonjour命名约定,而在构造McNearByserviceadvertiser
时,应继续使用abc-txtchat
。Try changing the service name in your plist file to
_abc-txtchat._tcp
. The value in the plist file has to conform to the Bonjour naming convention while you should continue usingabc-txtchat
while constructingMCNearbyServiceAdvertiser
.McNearByserviceAdvertiser服务器未发布