如果需要,请仅保留唯一的值a,b
或a,b,c
在顺序:使用:
s = df.drop_duplicates(['Type','Market']).groupby('Type')['Market'].agg(tuple)
df = df[df['Type'].isin(s.index[s.isin([('A','B'),('A','B','C')])])]
print (df)
Type Market Price
0 1 A 2
1 1 B 2
2 1 B 2
7 3 A 8
8 3 B 7
9 3 B 7
10 3 C 7
另一个想法:
def f(x):
u = tuple(dict.fromkeys(x))
return (u == ('A','B')) | (u == ('A','B','C'))
df = df[df.groupby('Type').Market.transform(f)]
从问题中,我看不到带有RabbitMQ的服务。
您的服务名称是Zipkin。
尝试跑步
您还可以指定外部IP
kind: Service
apiVersion: v1
metadata:
name: zipkin
spec:
selector:
app: zipkin
ports:
- protocol: "TCP"
port: 8081
targetPort: 80
type: LoadBalancer
externalIPs:
- 192.168.64.2
,然后可以使用 http://192.168.64.2:8081/ 访问
{
"theme": "neutral"
}
是最简单的解决方法。 (在线编辑中的“ config”下。)
https://github.com/github.com/mermeraid.com/mermeraid.com/mermeraid -JS/Mermaid/essume/3121 开放了一年。
根据我对此案的信息,我可以建议您这样的灵魂。
这也许不是最漂亮的解决方案,但它很容易理解,并且可以正常工作。
public class Main {
public static void main(String[] args) {
String A = " AAA \n"
+ " A:::A \n"
+ " A:::::A \n"
+ " A:::::::A \n"
+ " A:::::::::A \n"
+ " A:::::A:::::A \n"
+ " A:::::A A:::::A \n"
+ " A:::::A A:::::A \n"
+ " A:::::A A:::::A \n"
+ " A:::::AAAAAAAAA:::::A \n"
+ " A:::::::::::::::::::::A \n"
+ " A:::::AAAAAAAAAAAAA:::::A \n"
+ " A:::::A A:::::A \n"
+ " A:::::A A:::::A \n"
+ " A:::::A A:::::A \n"
+ "AAAAAAA AAAAAAA";
String B = "\nBBBBBBBBBBBBBBBBB \n"
+ "B::::::::::::::::B \n"
+ "B::::::BBBBBB:::::B \n"
+ "BB:::::B B:::::B\n"
+ " B::::B B:::::B\n"
+ " B::::B B:::::B\n"
+ " B::::BBBBBB:::::B \n"
+ " B:::::::::::::BB \n"
+ " B::::BBBBBB:::::B \n"
+ " B::::B B:::::B\n"
+ " B::::B B:::::B\n"
+ " B::::B B:::::B\n"
+ "BB:::::BBBBBB::::::B\n"
+ "B:::::::::::::::::B \n"
+ "B::::::::::::::::B \n"
+ "BBBBBBBBBBBBBBBBB ";
List<String> arrayA = List.of(A.split("\n"));
List<String> arrayB = List.of(B.split("\n"));
for(int i = 0; i < arrayA.size(); i++){
System.out.println(arrayA.get(i)+arrayB.get(i));
}
}
运行以下代码解决了问题:
npx migrate-mongo create blacklist_the_beatles
我认为该文档只是意味着HILT需要Java 8中的语言功能,而不是将Java版本限制为1.8。我尝试将源兼容性和目标兼容性更改为11,并且效果很好。
尝试将3.6.0版添加
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"
type="text/javascript"></script>
到您的代码头中,然后再次运行灯塔报告。
导入模块 - 您不需要额外的努力就可以从模块中获取另一件事。它的缺点,例如
从少量打字中导入的冗余键入模块,可以访问模块的哪个项目。要从模块中使用新项目,您必须更新导入语句。
您尚未发布任何的代码来修改字符串 - 您只是在计数加倍的字母。
要解决您的内容,您需要在您当前处于的字符串中具有“位置”,请将该字母与下一个字母(如果下一个字母)进行比较 - 如果您需要拼接字符串,请删除字母pos+1 ,然后用字母的下一个字母替换为 pos 。
这样就会这样:
s="aabcdghhjjk"
results = []
pos = 0
# modifying s (string == immuteable) is wasteful, could use list instead
while s:
if len(s) < 2 or len(s) < pos+2:
results.append(s)
break
# only add to results if not at results's last spot
if not results or results[-1] != s:
results.append(s)
if s[pos] == s[pos+1]:
next_letter = chr(ord(s[pos])+1)
# replace stuff by list slicing, creates a new immuteable string
s = s[:pos] + next_letter + s[pos+2:]
else:
# not equal letters, advance position in string
pos += 1
print(results)
输出:
['aabcdghhjjk',
'bbcdghhjjk',
'ccdghhjjk',
'ddghhjjk',
'eghhjjk',
'egijjk',
'egikk',
'egil']
仍然有一个边缘案例,您最终可能会在单词中加倍字母如果,您以后将字母替换为与已经评估的位置相结合的东西,向上。
固定边缘情况很有趣。
保存到变量并从数组中获取数据如下
$message = $responce['value']['message'];
$phone_number = $responce['value']['metadata']['phone_number_id'];
$contact = $responce['value']['contacts'];
最简单的方法是将响应
$res = json_decode($responce);
$message = $res->value->message;
$phone_number= $res->value->metadata->phone_number_id;
$contact = $res->value->contacts;
。建议查看以下答案。
您好,我遇到了同样的问题。
我通过在不使用.withConverter()的情况下为每个类编写自己的转换器来解决它。
看起来好像:
async customConverter (snapshot) {
const { refField } = snapshot.data();
const populated = await getDoc(refField);
return {
...snapshot.data(),
refField: populated.data(),
};
}
async fetch () {
const docRef = doc(firestore, 'collection', 'id');
const snapshot = await getDoc(docRef);
const dataPopulated = await customConverter(snapshot);
return dataPopulated;
}
我知道这不是完美的,但是如果有人有更好的方式,我都会耳朵! )
您可以定义几个辅助功能以调整两个数据范围的长度和宽度:
def equalize_length(short, long):
return pd.concat(
[
short,
pd.DataFrame(
{
col: ["nan"] * (long.shape[0] - short.shape[0])
for col in short.columns
}
),
]
).reset_index(drop=True)
def equalize_width(short, long):
return pd.concat(
[
short,
pd.DataFrame({col: [] for col in long.columns if col not in short.columns}),
],
axis=1,
).reset_index(drop=True)
def equalize(df, other_df):
if df.shape[0] <= other_df.shape[0]:
df = equalize_length(df, other_df)
else:
other_df = equalize_length(other_df, df)
if df.shape[1] <= other_df.shape[1]:
df = equalize_width(df, other_df)
else:
other_df = equalize_width(other_df, df)
df = df.fillna("nan")
other_df = other_df.fillna("nan")
return df, other_df
然后,在您的代码中:
a, b = equalize(a, b)
comparevalues = a.values == b.values
rows, cols = np.where(comparevalues == False)
for item in zip(rows, cols):
a.iloc[item[0], item[1]] = " {} --> {} ".format(
a.iloc[item[0], item[1]], b.iloc[item[0], item[1]]
)
print(a) # with 'a' being shorter in lenght but longer in width than 'b'
# Output
A B C D
0 1 abcd --> dah jamba OQEWINVSKD --> nan
1 2 efgh --> fupa refresh --> dimez DKVLNQIOEVM --> nan
2 3 ijkl portobello --> pocketfresh asdlikvn --> nan
3 4 uhyee --> danju performancehigh --> reverbb asdkvnddvfvfkdd --> nan
4 5 uhuh jackalack nan
5 nan --> 6 nan --> freshhhhhhh nan --> boombackimmatouchit nan
我只是解决了我的问题。我认为您只需要在要将cookie发送回服务器时,只需添加
withcredentials:true
(使用Axios btw)。事实证明,如果您也想从服务器中获取cookie,则需要添加该属性。现在,浏览器将cookie存储在应用程序中&gt; cookie 。感谢所有帮助&lt; 3I just solved my problem. I thought you only need to add
withCredentials: true
(using axios btw) whenever you want to send the cookie back to the server. Turns out, you need to add that property if you want to get the cookie from the server as well. Now the browser is storing the cookie in Application > Cookies. Appreciate all the help <3如何使用Aspnet Core 6 Web API在浏览器中设置cookie?