我正在病例对照研究中用权重来估计生存率。因此，所有案例的权重都等于 1。在绘制估计曲线并将其与未加权估计进行比较时，我注意到案例的 KM 曲线不重叠。
这是“survival”包中的数据代码。
library(dplyr)
library(tidyverse)
library(survival)
library(broom)
library(WeightIt)
library(survey)
a <- survival::ovarian
#Calculation of weghts:
weights <- WeightIt::weightit(rx ~ age + ecog.ps + resid.ds, int = T, estimand = "ATT", data = a, method = "glm" , stabilize = F, missing = "saem")
a$weights <- weights$weights
a$ps <- weights$ps
design <- svydesign(ids = ~ 1, data = a, weights = ~weights)
KM_PFS <- survfit(Surv(futime, fustat > 0)~rx, a) # KM naive
KM_PFS_w_TT <- survfit(Surv(futime, fustat > 0)~rx, a, …我有一个数据集，显示了X国甲方和乙方的宗教信仰，以及每个国家宗教信徒的百分比。
df <- data.frame(
PartyA = c("Christian","Muslim","Muslim","Jewish","Sikh"),
PartyB = c("Jewish","Muslim","Christian","Muslim","Buddhist"),
ChristianPop = c(12,1,74,14,17),
MuslimPop = c(71,93,5,86,13),
JewishPop = c(9,2,12,0,4),
SikhPop = c(0,0,1,0,10),
BuddhistPop = c(1,0,2,0,45)
)
# PartyA PartyB ChristianPop MuslimPop JewishPop SikhPop BuddhistPop
# 1 Christian Jewish 12 71 9 0 1
# 2 Muslim Muslim 1 93 2 0 0
# 3 Muslim Christian 74 5 12 1 2
# 4 Jewish Muslim 14 86 0 0 0
# 5 Sikh Buddhist 17 13 4 10 45
借此，我想将“参与”的宗教信徒的总数加在一起。因此，第一行将得到一个等于 …
My goal is to have a generic way to traverse a std::tuple, and the following code tries to show it:
#include <iostream>
#include <tuple>
template <typename t, size_t t_idx, typename t_tuple>
concept visit_tuple_element_value = requires(t &&p_t, const t_tuple &p_tuple) {
{
p_t.template operator()<t_idx>(
std::declval<std::add_const_t<std::add_lvalue_reference_t<t_tuple>>>())
} -> std::same_as<bool>;
};
template <typename t_tuple, typename t_function, size_t t_idx = 0>
requires(visit_tuple_element_value<t_function, t_idx, t_tuple>)
void traverse_tuple_values(t_function p_function, const t_tuple &p_tuple) {
if constexpr (t_idx < std::tuple_size_v<t_tuple>) {
if (p_function.template operator()<t_idx>(p_tuple)) {
traverse_tuple_values<t_tuple, t_function, …所以，假设我们有一个这样的类：
TFieldsReadyEvent = procedure(Sender: TObject; Code, AuthorizatedID: string) of object;
TCatcherServer = class(TServer)
private
FOnFieldsReady: TFieldsReadyEvent;
FCode: string;
FAuthoID: string;
FAuthorizationType: TAuthorizationType;
function GetAuthorizationType: string;
function GetCode: string;
function GetEntityID: string;
public
property OnFieldsReady: TFieldsReadyEvent read FOnFieldsReady write FOnFieldsReady;
property Code: string read GetCode;
property EntityID: string read GetEntityID;
property AuthorizationType: string read GetAuthorizationType;
procedure Callback(Session: TIdContext; Req: TIdHTTPRequestInfo; Res: TIdHTTPResponseInfo); override;
end;
程序Callback如下所示：
procedure TCatcherServer.Callback(Session: TIdContext;
Req: TIdHTTPRequestInfo; Res: TIdHTTPResponseInfo);
begin
inherited Callback(Session, Req, Res);
if (Req.Params.Values['code'] …下面是我在 pandas 中的异常值检测代码。我正在滚动 15 个窗口，我想要的是滚动 5 个窗口，其中该窗口基于居中日期的星期几，即如果中心是星期一，则在星期一向后移动 2 个，在星期一向前移动 2 个。Rolling 对此没有任何支持。怎么做？
import pandas as pd
import numpy as np
np.random.seed(0)
dates = pd.date_range(start='2022-01-01', end='2023-12-31', freq='D')
prices1 = np.random.randint(10, 100, size=len(dates))
prices2 = np.random.randint(20, 120, size=len(dates)).astype(float)
data = {'Date': dates, 'Price1': prices1, 'Price2': prices2}
df = pd.DataFrame(data)
r = df.Price1.rolling(window=15, center=True)
price_up, price_low = r.mean() + 2 * r.std(), r.mean() - 2 * r.std()
mask_upper = df['Price1'] > price_up
mask_lower = df['Price1'] < price_low
df.loc[mask_upper, 'Price1'] = r.mean()
df.loc[mask_lower, 'Price1'] …我在 PowerBI 仪表板中有一个名为“邮政编码”的专栏，我可以使用 PowerBI 中的 python 库编写脚本并提取城市、州和国家/地区名称吗？
这是我到目前为止所尝试过的，我使用 ODBC 连接到数据库。我有一个返回 49,445 行的示例查询
SELECT
ID,
POSTAL_CODE
FROM EMPLOYEE_DATA;
这是我的 python 脚本，它是我的 PowerBI 报告的一部分
# 'dataset' holds the input data for this script
# Import the required library
from geopy.geocoders import Nominatim
from geopy.exc import GeocoderTimedOut
# Function to extract city, state, and country from postal code
def extract_location(POSTAL_CODE):
geolocator = Nominatim(user_agent="geoapi")
try:
location = geolocator.geocode(POSTAL_CODE, timeout=10) # Adjust the timeout value as needed
if location:
return location.address.split(", ")[-3:]
else:
return ["", …假设我们有一些如下所示的二元素字符串向量A, B, C, ...。我们只知道保存为 object 的元素之一second。
有没有办法使second始终成为 中的第二个元素A, B, C, ...？
A = c("time","prof") # --> must become `c("prof", "time")`
B = c("prof", "time") # --> remains Unchanged
C = c("time","ziba") # --> must become `c("ziba", "time")`
second = "time"
我有这段代码：
template <int V>
struct Constant {
constexpr operator int() const noexcept { return V; }
};
template <class T, int N>
struct Array { };
auto function(auto s) -> Array<int, s + s> {
return {};
}
auto const a = function(Constant<3>{});
让我最悲伤的是，似乎只有 Clang 接受这个代码。
哪个编译器是正确的，为什么？
Visual Visual Studio 中的 C++，“LONG”显示绿色，而“long”显示蓝色。
使用一个有什么区别吗？使用它有什么缺点吗？颜色有什么意义吗？
我只是对此感到好奇，但没有找到任何真正可以解释这一点的东西。
我正在尝试从此网站下载 .pdf 网址 PDFLINK https://www.cell.com/heliyon/pdf/S2405-8440(18)33206-7.pdf 我可以从浏览器中查看它，但使用Python 的请求不允许我下载它。
我尝试了以下操作，但无法正确下载。它会返回 403 或说 .pdf 已损坏。但我可以很好地查看链接https://www.cell.com/heliyon/pdf/S2405-8440(18)33206-7.pdf 有什么想法吗？
import requests
def download_pdf(url):
response = requests.get(url)
with open('sample.pdf', 'wb') as f:
f.write(response.content)
if __name__ == "__main__":
url = "https://www.cell.com/heliyon/pdf/S2405-8440(18)33206-7.pdf"
download_pdf(url)