如果id相同，我想获得连续2行的日期差异。
一个简单的代码是：
for(i in 2:nrow(Data))
{
if(id[i]==id[i-1]
{
dated[i]-dated[i-1]
}
}
但是这个循环需要很长时间才能执行。有没有更快的方法在超过200万行上运行这种代码？
答案 0 :(得分：2)
我们可以使用/// Server.js
var users = []
io.sockets.on('connect', (socket) => {
socket.on("joingame", (jgCallback) => {
if (users.length < 3) {
if (users.length > 0) {
users[users.length - 1].nextSocket = socket; // connect this socket to the previous one
}
users.push(socket);
if (users.length === 3) {
socket.nextSocket = users[0]; // connect this socket to the first one
}
jgCallback({ success: true })
}
else {
jgCallback({ success: false })
}
})
socket.on("sendword", (payload, swCallback) => {
if (users.length === 3 && socket.nextSocket) { // There are 3 users, and This socket has a "nextSocket" property
socket.nextSocket.emit("sendword", payload); // send to the neighbor
swCallback({success:true})
}
else {
swCallback({success:false})
}
})
});
/// Clients.js
io.emit("joingame", (jgCallbackResponse) => {
if (jgCallbackResponse.success){
// you are in the game
}
})// Join the game.
io.on("sendword", (payload) => {
// from the previous user in the chain
console.log(payload.word)
})
io.emit("sendword", { word: "my word" }, (swCallbackResponse) => {
if (swCallbackResponse.success){
// word sent
}
});
执行此操作。将'data.frame'转换为'data.table'（data.table），按'id'分组，得到'date'列的差异（setDT(df1)）
diff