查找选定列中多个点的斜率 [英] Finding the slope for multiple points in selected columns

问题描述

给出以下数据框:

structure(list(`-5` = c(0, 1, 0, 0, 9, 22), `-4` = c(1, 3, 0, 
0, 1, 17), `-3` = c(1, 3, 0, 0, 0, 12), `-2` = c(1, 3, 0, 0, 
2, 10), `-1` = c(0, 0, 0, 4, 3, 9), `0` = c(0, 1, 0, 2, 2, 21
), `1` = c(0, 1, 1, 7, 1, 21), `2` = c(1, 0, 1, 2, 1, 10), `3` = c(0, 
9, 0, 6, 1, 12), `4` = c(0, 2, 0, 5, 0, 18), `5` = c(0, 0, 0, 
3, 0, 23)), .Names = c("-5", "-4", "-3", "-2", "-1", "0", "1", 
"2", "3", "4", "5"), row.names = c(NA, 6L), class = "data.frame")

#  -5 -4 -3 -2 -1  0  1  2  3  4  5
#1  0  1  1  1  0  0  0  1  0  0  0
#2  1  3  3  3  0  1  1  0  9  2  0
#3  0  0  0  0  0  0  1  1  0  0  0
#4  0  0  0  0  4  2  7  2  6  5  3
#5  9  1  0  2  3  2  1  1  1  0  0
#6 22 17 12 10  9 21 21 10 12 18 23

我希望R给我列-5:-1的每一行中所有数据点的斜率.基本上，基于这5个数据点的线性回归趋势线的斜率.然后是列1:5的所有数据点的第二个斜率.年份0被忽略.

I would like R to give me the slope for all the data points in each row for columns -5:-1. Basically the slope for a linear regression trendline based on those 5 data points. Then a second slope for all the data points for the columns 1:5. The year 0 is ignored.

基本上就是这样(使用Excel计算的最后两列):

Basically this is what it would look like (the two last columns computed using Excel):

structure(list(`-5` = c(0, 1, 0, 0, 9, 22), `-4` = c(1, 3, 0, 
0, 1, 17), `-3` = c(1, 3, 0, 0, 0, 12), `-2` = c(1, 3, 0, 0, 
2, 10), `-1` = c(0, 0, 0, 4, 3, 9), `0` = c(0, 1, 0, 2, 2, 21
), `1` = c(0, 1, 1, 7, 1, 21), `2` = c(1, 0, 1, 2, 1, 10), `3` = c(0, 
9, 0, 6, 1, 12), `4` = c(0, 2, 0, 5, 0, 18), `5` = c(0, 0, 0, 
3, 0, 23), `Negative Years` = c(0, -2, 0, 0.8, -1.1, -3.3), `Positive Years` = c(-0.1, 
0, -0.3, -0.5, -0.3, 1.2)), .Names = c("-5", "-4", "-3", "-2", 
"-1", "0", "1", "2", "3", "4", "5", "Negative Years", "Positive Years"
), row.names = c(NA, 6L), class = "data.frame")

#  -5 -4 -3 -2 -1  0  1  2  3  4  5 Negative Years Positive Years
#1  0  1  1  1  0  0  0  1  0  0  0            0.0           -0.1
#2  1  3  3  3  0  1  1  0  9  2  0           -2.0            0.0
#3  0  0  0  0  0  0  1  1  0  0  0            0.0           -0.3
#4  0  0  0  0  4  2  7  2  6  5  3            0.8           -0.5
#5  9  1  0  2  3  2  1  1  1  0  0           -1.1           -0.3
#6 22 17 12 10  9 21 21 10 12 18 23           -3.3            1.2

推荐答案

这就是统计学家(而非数据科学家)会做的事情.

This is what a statistician (not a data scientist) would do.

让您的数据框为dat.

Y <- t(dat)  ## response matrix
t <- -5:5    ## time stamps
id <- c(rep("-", 5), NA, rep("+", 5))  ## group index (factor)
fit <- lm(Y ~ t * id)  ## mlm
m <- coef(fit)[c(2, 4), ]  ## coefficient matrix
m[2, ] <- m[2, ] + m[1, ]  ## reverse contrast
round(t(m), 2)

#     t t:id+
#1  0.0  -0.1
#2 -0.2   0.0
#3  0.0  -0.3
#4  0.8  -0.5
#5 -1.1  -0.3
#6 -3.3   1.2

将列名称更改为所需的名称.

Change column names to what you desire.

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