Question

我有以下数据集。

共有三列：Pentad，A和B。

library(zoo)
library(rlang)
library(tidyverse) 

dat<-structure(list(Pentad = 50:73, A = c(152.796, 
109.678, 91.5594,115.155, 135.9, 202.441, 71.6951, 
88.3894, 261.962,135.853, 89.3425, 110.674, 100.558, 
173.507, 87.2157, 86.6425, 75.1852, 57.403, 62.5705, 
49.6846, 52.0257, 92.819, 105.419, 97.7598), 
B = c(145.402, 110.109, 83.1076, 95.3952, 148.571, 
119.178, 56.5031, 76.2635, 260.443, 109.705, 62.3749, 
100.322, 88.4134, 135.721, 63.1486, 69.7161, 62.3886, 
46.4513, 52.4546, 42.7725, 45.7643, 79.5419, 79.9434, 
87.6405)), class = "data.frame", row.names = c(NA, 
-24L))

我想在R中实现以下条件。

[1] V1 should be between 0 and 90 at the time step (excluding 0 and 90)

[2] In the succeeding FOUR time steps (including the 
timestep in [1]), V1 between 0 and 90 in AT LEAST THREE timesteps

到目前为止我所拥有的：

 test2 <- function(dat, column_name){ 
   dat %>%
   rownames_to_column() %>%
   filter((.data[[column_name]] > 0 & .data[[column_name]] < 90) & 
         rollsum(.data[[column_name]] > 0 & .data[[column_name]] < 90, 4, fill = NA, align = 
                   "left") >= 3) %>%
   slice(1) -> result
  return(result)
}

out <- colnames(dat2) %>% 
  set_names %>% 
  map_dfr(~ test2(dat2, .x), .id = 'Col_ID')

问题：

我想获取第二列（A列）和第三列（B列）都满足上述三个条件的时间步/五进制值。

即两列同时满足条件的时间步长。

预期输出为Pentad 64。

有什么想法可以在R中实现吗？

我将不胜感激。

Answer 1

这里的尝试与OP的尝试非常接近。

library(dplyr)
library(zoo)

test2 <- function(dat) {
   dat %>%
      filter_at(vars(A:B), all_vars(. > 0 & . < 90 & 
                     rollsum(. > 0 & . < 90, 4, fill = NA) >= 3)) %>%
       slice(1L)
}

test2(dat)

#  Pentad       A       B
#1     64 87.2157 63.1486

在多个条件下使用dplyr过滤器

1 个答案: