将向量变成矩阵

Question

我有一个5个值的向量，包含不同的月份，如下所示：

app.get('/update', function acceptReq(req, res) {
    db.getDB().collection(collection).findOneAndUpdate({
        _id: ObjectId("5c94f444dab6e423a488d2ae")
    }, {
        $set: {
            requested: false,
            requests: {
                user: undefined,
                time: undefined
            },
            Booked: true

        }
    }, {
        returnOriginal: false
    }, (err, result) => {
        if (err)
            console.log(err);
        else {
            req.flash(
                'success_msg',
                'Request Accepted!'
            );
            res.redirect('/dashboard/requests')
        }
    });
});

如何将它们转换成如下所示的矩阵形式。

[July August September October November]

此外，我还有一个向量要附加在末尾，这样输出将如下所示。

[July August September October November]
[1,    0,       0,       0,       0    ]
[0,    1,       0,       0,       0    ]
[0,    0,       1,       0,       0    ]
[0,    0,       0,       1,       0    ]
[0,    0,       0,       0,       1    ]

Answer 1

我们可以将task = SomeTask() try: task.start() except KeyboardInterruptException: print("Keyboard interrupted") except Exception as e: print_exception(e) # To show what happened finally: task.store_results() # If Control-C is pressed here, data gets corrupted 与向量的序列以及向量本身一起使用。如果需要保持月份名称的顺序，则将其转换为table并使用factor，其中levels由vector的intersect个元素的unique和month.name指定

tbl <- table(seq_along(v1), v2, dnn = NULL)
tbl
#   July August September October November
#1    1      0         0       0        0
#2    0      1         0       0        0
#3    0      0         1       0        0
#4    0      0         0       1        0
#5    0      0         0       0        1

现在，我们可以附加第二个vector

cbind(tbl, Other)

---

或使用model.matrix

model.matrix(~ v2 - 1, data = data.frame(v2))

---

或使用mtabulate中的qdapTools

library(qdapTools)
mtabulate(v2)

数据

v1 <- c("July", "August", "September", "October", "November")
v2 <- factor(v1, levels = intersect(month.name, unique(v1)))

Answer 2

sapply(v1, function(x) as.numeric(v1 == x))
#     July August September October November
#[1,]    1      0         0       0        0
#[2,]    0      1         0       0        0
#[3,]    0      0         1       0        0
#[4,]    0      0         0       1        0
#[5,]    0      0         0       0        1

Answer 3

您也可以使用model.matrix或dummyVars

数据：

v1 <- c("July", "August", "September", "October", "November")
v1 <- as.data.frame(v1)

Model.Matrix：

v2 <- data.frame(v1 = v1)
as.data.frame(model.matrix(~.-1, v1))
  v1August v1July v1November v1October v1September
1        0      1          0         0           0
2        1      0          0         0           0
3        0      0          0         0           1
4        0      0          0         1           0
5        0      0          1         0           0

DummyVars方法：

library(caret)
m1 <- dummyVars(" ~ .", data = v1)
data.frame(predict(m1, newdata = v1))
  v1.August v1.July v1.November v1.October v1.September
1         0       1           0          0            0
2         1       0           0          0            0
3         0       0           0          0            1
4         0       0           0          1            0
5         0       0           1          0            0