主页 / user-5167711

LucaS's questions

Martin Hope
LucaS
Asked: 2025-02-06 07:43:39 +0800 CST
  • 7

我想比较以下数据的单指数和双指数拟合。单指数函数拟合得很好，但我找不到双指数函数的起始值（我也尝试过根据合理的值手动输入一些起始值）。

有人能解释为什么双指数函数不适合这些数据以及原因吗？是否有其他可能起作用的软件包，或者这是数据的限制？

谢谢

# Dummy data
dat <-  structure(list(x = c(5, 10, 30, 60, 120, 180, 240), y = c(0.000215875066736858, 
                                                          0.000213143286462547, 0.000184642176655826, 0.000154069098478106, 
                                                          0.000100308278926715, 6.56265655274e-05, 4.80349446283348e-05
)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
                                                            -7L))


# Mono-exponential model
nls_mod_mono <-  nls(y ~ statforbiology::NLS.expoDecay(x, C0, lambda1), dat)
# Predict and plot
xNew <- seq(0, 240, length.out = 100) # new grid of times
yNew <- predict(nls_mod_mono, list(x = xNew)) # predict activity
dfNew <-  data.frame(x = xNew, y = yNew)
ggplot(data = dfNew, aes(x = x, y = y*100)) +
  geom_line() +
  geom_point(data = dat, aes(x = x, y = y*100), size = 3)

# Bi-exponential model
params <-  getInitial(y ~ SSbiexp(x, intercept_initial, log_slope_initial, intercept_terminal, log_slope_terminal), data = dat)

#...failing to find initial values

nls_mod_bi <-  nls(y ~ C0*exp(-lambda1*x) + C1*exp(-lambda2*x), start = list(C0 = params[1], lambda1 = exp(params[2]), C1 = params[3], lambda2 = exp(params[4])), data = dat)

在此处输入图片描述

Martin Hope
LucaS
Asked: 2025-01-18 17:54:01 +0800 CST
  • 7

我正在尝试使用函数将指数衰减曲线拟合到一些生物数据nls()（我想多次执行此操作）。

nls()我收到了一个错误，我无法通过指定的方式解决：

Error in str2lang(x) : <text>:2:0: unexpected end of input
1: ~ 
   ^

下面是一个小的虚拟数据集和我编写的代码 - 任何帮助都将不胜感激。

dat_small <-  structure(list(actual_time = c(5, 9, 30, 59, 119, 171, 216), 
                             activity = c(7158, 7386, 5496, 3884, 1502, 819, 409)), row.names = c(NA, 
                                                                                                  -7L), class = c("tbl_df", "tbl", "data.frame"))
curve_fit <-  function(x, y, df) {
  #browser()
  # Mono-exponential model for activity using actual sampling time
  mod <-  lm(log(df[[y]]) ~ df[[x]], data = df) # get starting values from log-linear model
  C0 <-  as.numeric(exp(mod$coef[1])) # exponential of log-linear intercept
  lambda1 <-  as.numeric(abs(mod$coef[2])) # absolute value of slope of log-linear model
  nls_mod_mono <-  nls(df[[y]] ~ C0*exp(-lambda1*df[[x]]), start = c(C0 = C0, lambda1 = lambda1), data = df)
  summary(nls_mod_mono) # estimate model
  xNew <- seq(0, 240, length.out = 100) # new grid of times
  yNew <- predict(nls_mod_mono, list(x = xNew)) # predict activity
  dfNew <-  data.frame(x = xNew, y = yNew)
  p_mono <-  ggplot(dfNew, aes(x = x, y = y)) +
    geom_line() +
    geom_point(data = df, aes(x = x, y = y), size = 3) +
    xlab("Actual Sampling Time (mins)") + ylab("Activity (Bq)") +
    theme_bw(base_size = 20)
  mono_half_life <-  0.693/as.numeric(coef(nls_mod_mono)[2]) # half-life
  mono_auc <-  as.numeric(coef(nls_mod_mono)[1])/as.numeric(coef(nls_mod_mono)[2]) # AUC to infinity (can be estimated in mono case by simply dividing the intercept by lambda)
  results <-  c(p_mono, mono_half_life, mono_auc)
}
curve_fit("actual_time", "activity", dat_small)
Martin Hope
LucaS
Asked: 2024-11-10 14:11:04 +0800 CST
  • 5

我已经在这个问题上纠结了好几个小时，这有点令人沮丧。基本上，我想安排一些数据，以便 NA 基于分组结构首先出现。我可以部分实现，但我尝试的任何方法都无法让我得到想要的结果。

使用此代码，

df <-  df |> 
  group_by(AESOC, AEPT) |> 
  arrange(!is.na(AEPT), !is.na(Severity), .by_group = TRUE)

我已经能够实现图中所示的效果。

在此处输入图片描述

但是我仍想进一步安排，以便第 9-12 行出现在第 1 行之前，第 25-28 行出现在第 13 行之前（即在 AESOC 和 AEPT 确定的组的最开始处）。

这些小数据包括在这里：

df <-  structure(list(AESOC = c("Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders"
), AEPT = c("    Anaemia", "    Anaemia", "    Anaemia", "    Anaemia", 
"    Lymphopenia", "    Lymphopenia", "    Lymphopenia", "    Lymphopenia", 
NA, NA, NA, NA, "    Dizziness", "    Dizziness", "    Dizziness", 
"    Dizziness", "    Palpitations", "    Palpitations", "    Palpitations", 
"    Palpitations", "    Presyncope", "    Presyncope", "    Presyncope", 
"    Presyncope", NA, NA, NA, NA), Severity = c("        mild", 
"        moderate", "        severe", NA, "        mild", "        moderate", 
"        severe", NA, "    mild", "    moderate", "    severe", 
NA, "        mild", "        moderate", "        severe", NA, 
"        mild", "        moderate", "        severe", NA, "        moderate", 
"        mild", "        severe", NA, "    moderate", "    mild", 
"    severe", NA)), row.names = c(NA, -28L), class = c("tbl_df", 
"tbl", "data.frame"))

任何帮助都将不胜感激。

Martin Hope
LucaS
Asked: 2024-11-06 13:44:24 +0800 CST
  • 5

我正在尝试使用传递给 的函数重命名多个 df 的列purrr::map，但显然我做错了，因为我收到错误。有人能告诉我怎么做吗？

# Create new df rows
newrow1 <-  data.frame(t(c("a", rep("", 6))))
newrow2 <-  data.frame(t(c("b", rep("", 6))))
newrow3 <-  data.frame(t(c("c", rep("", 6))))
newrow4 <-  data.frame(t(c("d", rep("", 6))))
newrow5 <-  data.frame(t(c("e", rep("", 6))))
newrow6 <-  data.frame(t(c("f", rep("", 6))))
newrow7 <-  data.frame(t(c("g", rep("", 6))))
newrow8 <-  data.frame(t(c("h", rep("", 6))))
newrow9 <-  data.frame(t(c("i", rep("", 6))))
# Function to rename dfs
changenames <- function(x) {
  names(x) <- c("row","n_1","n_2","n_3","pct_1","pct_2","pct_3")
  return(x)
}
df_names <-  paste0("newrow", 1:9)
purrr::map(df_names, changenames)
Martin Hope
LucaS
Asked: 2024-09-30 07:50:13 +0800 CST
  • 5

我正在尝试学习一些基本的 purrr。以这个玩具示例为例 - 我如何将“rank”附加到使用 map_df 创建的当前变量名称中？

library(tidyverse)
# generate some random data
dat <- tibble(
  x_1 = runif(10),
  x_2 = runif(10),
  x_3 = runif(10)
)
# find the ranks
dat <-  cbind(dat, dat |> map_df(rank))

目前生产的产品如下：

> dat
          x_1        x_2       x_3 x_1 x_2 x_3
1  0.37593751 0.20597008 0.4826491   6   3   4
2  0.01805845 0.05432019 0.6732073   1   1   7
3  0.39243407 0.47535664 0.9920948   7   7  10
4  0.06361097 0.21624621 0.4309171   2   4   3
5  0.53391717 0.28778664 0.9323643   9   6   9
6  0.24612385 0.83270359 0.2719036   4  10   1
7  0.47170853 0.26261303 0.3418129   8   5   2
8  0.33821497 0.57270648 0.7606502   5   9   8
9  0.85717819 0.48097921 0.4848689  10   8   5
10 0.11503862 0.07063651 0.6391102   3   2   6
Martin Hope
LucaS
Asked: 2024-09-28 08:14:48 +0800 CST
  • 8

有人能帮我看看我哪里出错了吗？我认为在管道函数中使用花括号时，您可以调用变量而不必将其括在“”（即作为字符串）。那么我的函数哪里出错了：

library(tidyverse)
dat <-  structure(list(id = 1:6, 
                       var_1 = c("1,1,-11000", "1,1,1", "0,0,0", "1,1,0", "1,1,1", "1,1,1"), 
                       var_2 = c("0,0,-13000", "0,0,0", "-13000,-13000,-13000", "6,4,-13000", "0,0,0", "0,0,0"), 
                       var_3 = c("24,7,-13000", "0,0,0", "-13000,-13000,-13000", "0,0,-13000", "0,0,0", "0,0,0")), 
                  row.names = 1:6, class = "data.frame")

# Separate to wide and convert to long in one step
split_to_long <-  function(col){
  i <-  substr({{col}}, 5, 5)
  temp <-  dat |>
    select("id", {{col}}) |>
    separate_wider_delim({{col}}, ",", too_few = "align_start",
                         names = c(paste0({{col}},"_1"),
                                   paste0({{col}},"_2"),
                                   paste0({{col}},"_3"))) |>
    pivot_longer(2:4,
                 names_to = "visit",
                 values_to = paste0("var_", i),
                 names_prefix = paste0("var_", i, "_"))
  temp
}

# This works
split_to_long("var_1")

# But this doesn't
split_to_long(var_1)

这来自我发布的另一个问题： 如何用 purrr::map 替换这个 for 循环

Martin Hope
LucaS
Asked: 2024-08-22 08:17:03 +0800 CST
  • 5

我有一个相当简单的函数，其中包含一个 for 循环，将特定编码的缺失值转换为 NA：

make_na <- function(value, df){
  for(thisvar in names(df)){
    df[[thisvar]][df[[thisvar]] == value] <- NA
    num_NA <- length(df[[thisvar]][is.na(df[[thisvar]])])
    print(paste0(thisvar, " has ", num_NA, " NA's"))
  }  
  return(df)
}

运行该函数可以正确打印修改后的数据框，但我不知道如何将数据框放回工作区。我也尝试new_dat <- return(df)将数据框分配给一个对象，但那也不起作用。显然我遗漏了一些简单的东西。