LongZhao1992
commented
1 year ago Hello, I'm not quite clear about the detail of your code. However, I wrote a simple example in R Markdown for you regarding building a loess model, predicting the value, and visualization. Hope this could be helpful with your questions.
加载R包
library(tidyverse)
library(modelr)创建伪时间
#假设5个点
atac_dist_norm <- c(0,2,5,7,10)
#准备模拟出500个点
grid <- data.frame(time = seq(0, 10, length.out = 500))定义function
atac_modle_fun <- function(value){
time = atac_dist_norm
data = tibble(value = value,time =time)
model = loess(value ~ time, data)
predict <- grid %>% add_predictions(model)
return(predict)
}创建假设的表达矩阵
atac_data <- matrix(rnorm(50,mean = 1),ncol = 5)
rownames(atac_data) <- paste0("gene",1:10)
head(atac_data)运行function
atac_modle_list <- apply(atac_data,1,atac_modle_fun)
atac_modle_df <- atac_modle_list %>% reduce(bind_rows)长列表变宽列表便于可视化
atac_modle_wide <-atac_modle_df %>%
mutate(rank = rep(1:500,10),
ID = rep(rownames(atac_data),each=500)) %>%
select(-time) %>%
pivot_wider(names_from = rank, values_from = pred)可视化
atac_modle_wide_mt <- as.matrix(atac_modle_wide[,-1])
rownames(atac_modle_wide_mt) <- atac_modle_wide$ID
pheatmap::pheatmap(atac_modle_wide_mt,cluster_cols = F,show_colnames = F)
when I used the script of fig5C(data = tibble(value = value[-c(1:2)],time =time) model = loess(value ~ time, data)), I had a error like: Error in simpleLoess(y, x, w, span, degree = degree, parametric = parametric, : invalid 'y'. And the loess funtion seems to accept only 4 variables,but my RNA-seq had more than it,what should I do? Thanks in advance for your reply.