如何用渐变色填充下方(sp)线以下的区域?
这个例子是在Inkscape中绘制的 - 但我需要垂直渐变 - 不是水平的.
从零到正的间隔==从白色到红色.
从零到负的间隔==从白色到红色.

有没有可以做到这一点的包?
我编造了一些源数据....
set.seed(1)
x<-seq(from = -10, to = 10, by = 0.25)
data <- data.frame(value = sample(x, 25, replace = TRUE), time = 1:25)
plot(data$time, data$value, type = "n")
my.spline <- smooth.spline(data$time, data$value, df = 15)
lines(my.spline$x, my.spline$y, lwd = 2.5, col = "blue")
abline(h = 0)
Run Code Online (Sandbox Code Playgroud) 我的问题很简单!如何将点xyz坐标(都属于单个平面)转换为仅xy坐标.我找不到任何R功能或R解决方案.

来源数据:
# cube with plain
library(scatterplot3d)
my.plain <- data.frame(ID = c("A","B","C","D","E","F","G","H"),
x = c(1,1,1,2,2,2,3,3),
y = c(1,1,1,2,2,2,3,3),
z = c(1,2,3,1,2,3,1,2))
scatterplot3d(my.plain$x, my.plain$y, my.plain$z,
xlim = c(0,3), ylim = c(0,3), zlim = c(0,3))
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如何获得点数据帧,其中A点为[0,0],而A和D之间的距离为sqrt(2)?
如何从Delaunay三角剖分中移除比我需要的更大的距离?
示例数据:
x<-rep(1:12, c(2,2,7,9,10,5,4,6,10,10,9,4))
y<-c(1,2,1,2,1:3,5:8,1:9,1:10,2,7:10,8:11,7:12,3:12,3:12,4:12,5,8:10)
x_plus<-seq(0.2:0.8, by=0.1)
x<-x+sample(x_plus, 78, replace=TRUE)
y<-y+sample(x_plus, 78, replace=TRUE)
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绘制地图:
plot(x,y)
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Delaunay三角测量与tri.mesh() - 包(tripack)
my.triangles<-tri.mesh(x,y)
plot(my.triangles, do.points=FALSE, lwd=0.2)
points(x,y, col = "blue", pch=20)
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我怎样才能提取更短的距离?我不需要那些大的你肯定知道我的意思.在tri.mesh()函数中是否有一些参数可以做到这一点?或者它可以在它之后完成?
距离是否存储在此对象中?
my.triangles
triangulation nodes with neigbours:
node: (x,y): neighbours
1: (1.4,1.7) [5]: 2 3 4 11 12
2: (2,3) [6]: 1 4 7 8 9 11
3: (3,1.8) [4]: 1 4 5 12
.
.
.
76: (12.4,8.8) [5]: 68 69 70 75 77
77: (12.9,9.9) [6]: 70 71 72 75 …Run Code Online (Sandbox Code Playgroud) 有没有办法在R中制作自定义点?我很熟悉pch有很多选择的论点,但是如果我需要绘制例如树形轮廓的话呢?
例如,如果我画一点作为eps.(或类似的)文件,我可以在R中使用它吗?.在复杂对象(fe树)的情况下,栅格解决方案并不好.

我需要特定格式的data.frame进行社会结构分析.如何转换包含多个事件上一起出现的个人列表的data.frame:
my.df <- data.frame(individual = c("A","B","C","B","C","D"),
time = rep(c("event_01","event_02"), each = 3))
individual time
1 A event_01
2 B event_01
3 C event_01
4 B event_02
5 C event_02
6 D event_02
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包含每对(包括[A,A]; [B,B]等对的出现的data.frame:
ind_1 ind_2 times
A A 0
A B 1
A C 1
A D 0
B A 1
B B 0
B C 2
B D 1
C A 1
C B 2
C C 0
C D 1
D A 0
D B 1
D …Run Code Online (Sandbox Code Playgroud) 我知道,愚蠢的问题,但我不是程序员,真的需要知道......
to look-for-food
if food > 0
[ set color orange + 1
set food food - 1
rt 180
stop ]
if (chemical >= 0.05) and (chemical < 2)
[ uphill-chemical ]
end
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谢谢
我在从PCA对象中提取值时遇到问题.我需要" Comp.1"中的" 方差比例"值.在此示例中,它的值为0.7056111.
我能够总结,但无法提取,因为它不存在str(pca_test).
pca_test<-data.frame(var1=rnorm(20), var2=rnorm(20))
summary(princomp(pca_test))
Importance of components:
Comp.1 Comp.2
Standard deviation 1.0200426 0.6588649
Proportion of Variance 0.7056111 0.2943889
Cumulative Proportion 0.7056111 1.0000000
# value 0.7056111 is absent here
> str(summary(princomp(pca_test)))
List of 9
$ sdev : Named num [1:2] 1.02 0.659
..- attr(*, "names")= chr [1:2] "Comp.1" "Comp.2"
$ loadings : loadings [1:2, 1:2] 0.289 0.957 -0.957 0.289
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:2] "var1" "var2"
.. …Run Code Online (Sandbox Code Playgroud) 我需要将矢量值排序为自定义间隔,然后确定哪个元素属于哪个区间.
例如,如果向量是:
x <- c(1,4,12,13,18,24)
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间隔是:
interval.vector <- c(1,7,13,19,25)
1st interval: 1 - 7
2nd interval: 7 - 13
3rd interval: 13 - 19
4th interval: 19 - 25
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...我如何结合x并interval.vector得到这个:
element: 1 4 12 13 18 24
interval: 1 1 2 2 3 4
Run Code Online (Sandbox Code Playgroud) 该调用mF@X应返回固定效应设计矩阵(对于glmm中的边际和条件R ^ 2是必需的).但它不起作用.甚至这样的矩阵列在模型结构中?
我很感激任何建议.
str(mF)
Formal class 'glmerMod' [package "lme4"] with 13 slots
..@ resp :Reference class 'glmResp' [package "lme4"] with 11 fields
.. ..$ Ptr :<externalptr>
.. ..$ mu : num [1:480] 0.168 0.356 0.168 0.356 0.284 ...
.. ..$ offset : num [1:480] 0 0 0 0 0 0 0 0 0 0 ...
.. ..$ sqrtXwt: num [1:480] 0.373 0.479 0.373 0.479 0.451 ...
.. ..$ sqrtrwt: num [1:480] 2.68 2.09 2.68 2.09 …Run Code Online (Sandbox Code Playgroud)