英文:
How to return xy-coordinates of a mathematical function into a dataframe of arbitrary length?
问题
我想获得一个包含描述两点P1(x1, y1)和P2(x2, y2)之间线性函数的n个xy坐标的数据框。
我的方法是找到两点的xy坐标，然后计算这个线性方程的斜率和截距 y = slope * x + intercept。
为了简化，让：
x1 = 0.1y1 = 0.2x2 = 1.1y2 = 0.6n = c(1:200) # 例子中我的数据的长度slope = (y2 - y1) / (x2 - x1)intercept = y1 - slope * x1
到目前为止都好，但现在我想将线性方程的各个xy坐标计算到一个名为baseline的数据框中，该数据框有length(n)行，包含x_baseline 和 y_baseline 两列。
如何解决这个问题？到目前为止，我在网上找不到令人满意的答案，非常感谢任何帮助！
为了澄清，这是我期望的输出的示意图在这里。
英文:
I want to get a dataframe containing n numbers of xy-coordinates that describe a linear function between two points P1(x1, y1) and P2(x2, y2) .
My approach would be to find the xy-coordinates of two points and calculate the slope and intercept of this linear equation y = slope * x + intercept.
For simplicity let: P
x1 = 0.1y1 = 0.2x2 = 1.1y2 = 0.6n = c(1:200) #example length of my dataslope = (y2 - y1) / (x2 - x1)intercept = y1 - slope * x1
So far so good, but now i would like to compute the individual xy-coordinates of my linear equation into a dataframe baseline with length(n) rows with the columns x_baseline and y_baseline.
How would one go about solving this problem? So far, I could not find an satisfactory answer online, any help is highly appreciated!
For clarification a drawing of my desired output here
答案1
得分: 1
如果你有x1、y1、x2、y2，你可以使用一些代数运算来计算点之间的插值。
steps <- (n-1)/(max(n)-1)dd <- data.frame(x = x1 + (x2-x1) * steps,y = y1 + (y2-y1) * steps)
然后，你可以使用以下代码来绘制它们：
plot(y~x, dd)points(c(x1,x2), c(y1, y2), col="green", pch=16)
你也可以使用approx进行插值：
dd <- data.frame(x = approx(c(0,1), c(x1, x2), seq(0, 1, length.out = length(n)))$y,y = approx(c(0,1), c(y1, y2), seq(0, 1, length.out = length(n)))$y)
或者，如果你真的想要使用斜率（slope）和截距（intercept）：
dd <- data.frame(x = seq(x1, x2, length.out = length(n))) %>%transform(y = intercept + x * slope)
英文:
If you have x1, y1, x2, y2 you can just use some some algebra to calculate the interpolation between the points
steps <- (n-1)/(max(n)-1)dd <- data.frame(x = x1 + (x2-x1) * steps,y = y1 + (y2-y1) * steps)
And we can plot them with
plot(y~x, dd)points(c(x1,x2), c(y1, y2), col="green", pch=16)
You could also do interpolation with approx
dd <- data.frame(x = approx(c(0,1), c(x1, x2), seq(0, 1, length.out = length(n)))$y,y = approx(c(0,1), c(y1, y2), seq(0, 1, length.out = length(n)))$y)
or if you really wanted to use the slope and intercept you could do
dd <- data.frame(x = seq(x1, x2, length.out = length(n))) |>transform(y = intercept + x * slope)
答案2
得分: 1
以下是翻译好的代码部分：
library(tidyverse)library(modelr)baseline <- tibble(x = c(0.1, 1.1),y = c(0.2, 0.6))mod <- lm(y ~ x, baseline)baseline |>data_grid(x = seq_range(x, 200)) |>add_predictions(mod, var = "y_baseline") |>rename(x_baseline = x)
创建于2023年3月12日，使用reprex v2.0.2
英文:
If you don't want to do the math yourself, fit a linear model and predict from it:
library(tidyverse)library(modelr)baseline <- tibble(x = c(0.1, 1.1),y = c(0.2, 0.6))mod <- lm(y ~ x, baseline)baseline |>data_grid(x = seq_range(x, 200)) |>add_predictions(mod, var = "y_baseline") |>rename(x_baseline = x)#> # A tibble: 200 × 2#> x_baseline y_baseline#> <dbl> <dbl>#> 1 0.1 0.200#> 2 0.105 0.202#> 3 0.110 0.204#> 4 0.115 0.206#> 5 0.120 0.208#> 6 0.125 0.210#> 7 0.130 0.212#> 8 0.135 0.214#> 9 0.140 0.216#> 10 0.145 0.218#> # … with 190 more rows
<sup>Created on 2023-03-12 with reprex v2.0.2</sup>
答案3
得分: 0
您的x值将由seq(x1, x2, length = 100)给出，新的y值将由这个向量乘以slope再加上intercept：
df <- data.frame(x_baseline = seq(x1, x2, length = 100),y_baseline = slope * seq(x1, x2, length = 100) + intercept)
让我们用红色绘制您的原始点，用黑色绘制数据框的点：
plot(c(x1, x2), c(y1, y2), col = "red", cex = 2)points(df$x_baseline, df$y_baseline)
2023-03-11创建，使用reprex v2.0.2。
英文:
Your x values would be given by seq(x1, x2, length = 100), and your new y values given by this vector multiplied by slope plus the intercept:
df <- data.frame(x_baseline = seq(x1, x2, length = 100),y_baseline = slope * seq(x1, x2, length = 100) + intercept)
Let's plot your original points in red, and the dataframe's points in black:
plot(c(x1, x2), c(y1, y2), col = "red", cex = 2)points(df$x_baseline, df$y_baseline)
<sup>Created on 2023-03-11 with reprex v2.0.2</sup>
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