r - 如何使用 R 中的自动过程(即 for 循环等)查找向量之间特定值序列的匹配行

Question

我有矢量 A和 B .向量长度 A是 12。向量的长度 B是 23。

A <- c(125,195,322,421,65,102,85,98,88,176,300)

B <- c(62,138,124,78,117,84,148,91,71,112,137,102,65,102,85,98,88,176,150,78,72,68,102)

我需要在自动化过程中做几件事(如果可能):

首先，我需要在 A 中找到最长的值序列满足这一点:它们是连续的，它们在 60 到 180 之间。在这个例子中，这个序列是:

A.selected <- c(65,102,85,98,88,176)

其次，我必须在 B 中找到第一行对于其中存在与 A.selected 具有最高重合度的值序列(与 A.selected 的长度相同) .我认为这样做是这样的:

# First, I create different vectors of `B` of the same length (`5` in this example) than `A.selected` starting from the first row in `B`:

B_1 <- c(B[1],B[2],B[3],B[4],B[5],B[6])
B_2 <- c(B[2],B[3],B[4],B[5],B[6],B[7])
B_3 <- c(B[3],B[4],B[5],B[6],B[7],B[8])
 .       .        .       .
 .       .        .       .
 .       .        .       .
B_13 <- c(B[13],B[14],B[15],B[16],B[17],B[18])
 .       .        .       .
 .       .        .       .

# Second, I estimate the strength of the correlation between `A.selected` and the different combinations of `B` (`B_1`,`B_2`,...,`B_13`,`B_14`, etc) using the Pearson's correlation coefficient (`R²`). I also calculated the `P-value` of this correlation to be sure is significant.

mod1 <- cor.test(A.selected,B_1)
mod2 <- cor.test(A.selected,B_2)
mod3 <- cor.test(A.selected,B_3)
 .       .        .       .
 .       .        .       .
 .       .        .       .
mod13 <- cor.test(A.selected,B_13)



coef.mod1 <- c(as.numeric(mod1[4]),as.numeric(mod1[3]))  # R² and P-value of the 1st correlation
coef.mod2 <- c(as.numeric(mod2[4]),as.numeric(mod2[3]))  # R² and P-value of the 2nd correlation
coef.mod3 <- c(as.numeric(mod3[4]),as.numeric(mod3[3]))  # R² and P-value of the first correlation
 .       .        .       .
 .       .        .       .
 .       .        .       .
coef.mod13 <- c(as.numeric(mod4[4]),as.numeric(mod4[3]))  # R² and P-value of the first correlation

 # I find the model with the highest R², but considering also that the `P-value` has to be lower than `0.05`.

Models.dataframe <- data.frame(R2 = c(coef.mod1[1],coef.mod2[1],coef.mod3[1],0.38,0.65,0.13,0.26,-0.34,0.76,0.48,0.32,0.42,coef.mod13[1]),
                               P.value = c(coef.mod1[2], coef.mod2[2], coef.mod3[2], 0.38, 0.65, 0.13, 0.26, 0.84, 0.26, 0.48, 0.32, 0.42, coef.mod13[2]))

rownames(Models.dataframe[which.max(Models.dataframe$R2) & Models.dataframe$P.value <= 0.05,])
"13" # In row 13 in `B` starts the sequence of numbers that have the highest overlap with the sequence `A.selected`

在现实世界中， A和 B有几十万的长度，所以我需要一个代码来自动完成所有这些。

有谁知道如何创建代码来自动执行此操作？

Answer 1

实现步骤 1 的方法有点麻烦:

根据OP的评论编辑:

library(tidyverse)
get_consecutive_grps <- function(x) {
  runlengths <- rle(x) %>% .$lengths
  map2(runlengths, 1:length(runlengths), ~ rep(..2, ..1)) %>% unlist()
}
tmp <- 
  enframe(A) %>% 
  mutate(
    sel = between(value, 60L, 180L),
    idx = get_consecutive_grps(sel)
  ) %>% 
  group_by(idx) %>% 
  mutate(
    idx_cnt = row_number()
  ) 
longestrun <- filter(tmp, sel) %>% pull(idx_cnt) %>% max()
longestidx <- filter(tmp, sel, idx_cnt == longestrun) %>% pull(idx)

# It's possible that there are several selected sequences of the same length;
# use the first one
A.selected <- filter(tmp, idx == longestidx[1]) %>% pull(value)

编辑:
我在第二步添加了一个同样繁琐的方法:

get_Bs <- function(start_idx, length, vec) {
  vec[start_idx:(start_idx + length - 1)]
}
offset <- 1:(length(B) - length(A.selected))
Bs <- 
  map_dfc(offset, get_Bs, length = length(A.selected), vec = B) %>% 
  setNames(str_c("Bidx_", offset)) %>% 
  mutate(relpos = row_number()) %>% 
  select(relpos, everything())

# Rearrange data and calculate correlations with `A.selected`
B_corr <- 
  Bs %>% 
  pivot_longer(
    cols = -relpos, 
    names_to = "Bidx", 
    names_prefix = "Bidx_"
  ) %>% 
  pivot_wider(
    id_cols = Bidx,
    values_from = value,
    names_from = relpos, 
    names_prefix = "relpos_"
  ) %>% 
  nest(B_snippits = starts_with("relpos")) %>% 
  mutate(
    corr = map(B_snippits, ~ cor.test(A.selected, as.numeric(..1))), 
    corr_tidy = map(corr, broom::tidy)
  ) %>% 
  unnest(corr_tidy)

# Get B-index for highest correlation
B_corr %>% 
  filter(estimate == max(B_corr$estimate), p.value <= 0.05) %>% 
  pull(Bidx)

# ==> "13" 

我相信有更直接的方法可以做到这一切......